Stokes Theorem How To Find Normal Vector

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Our last variant of the fundamental theorem of calculus is Stokes'¹ theorem, which is like Light-green's theorem, but in three dimensions. It relates an integral over a finite surface in \(\bbbr^3\) with an integral over the curve bounding the surface.

Sir George Gabriel Stokes (1819–1903) was an Irish physicist and mathematician. In improver to Stokes' theorem, he is known for the Navier-Stokes equations of fluid dynamics and for his work on the wave theory of light. He gave show to the Regal Commission on the Employ of Iron in Railway Structures afterward the Dee span disaster of 1847.

Theorem 4.iv.1 . Stokes' Theorem.

Permit

• \(S\) exist a piecewise smooth oriented surface (i.e. a unit normal \(\hn\) has been chosen at each indicate of \(S\) and this choice depends continuously on the bespeak)

• the purlieus, \(\partial Southward\text{,}\) of the surface \(Due south\) consist of a finite number of piecewise shine, simple curves that are oriented consistently with \(\hn\) in the sense that

  • if y'all walk along \(\fractional S\) in the management of the arrow on \(\partial S\text{,}\)
  • with the vector from your feet to your caput having management \(\hn\)
  • and so \(S\) is on your left hand side.

  

• \(\vF\) be a vector field that has continuous first partial derivatives at every signal of \(Due south\text{.}\)

Then

\begin{equation*} \oint_{\partial Southward}\vF\cdot \dee{\vr} =\dblInt_{S}\vnabla\times\vF\cdot\hn\ \dee{Due south} \end{equation*}

Notation that

• in Stokes' theorem, \(Due south\) must exist an oriented surface. In particular, \(Due south\) may not be a Möbius strip. (See Case three.5.three.)
• If \(S\) is part of the \(xy\)-aeroplane, then Stokes' theorem reduces to Green's theorem. Our proof of Stokes' theorem volition consist of rewriting the integrals and then every bit to allow an application of Green'south theorem.

• If \(\partial S\) is a unproblematic airtight bend and

  • when you look at \(\partial South\) from high on the \(z\)-centrality, it is oriented counterclockwise (expect at the figure in Theorem iv.four.1), then
  • \(\hn\) is upward pointing, i.e. has positive \(z\)-component, at least near \(\partial S\text{.}\)

Proof.

Write \(\vF=F_1\,\hi+F_2\,\hj+F_3\,\hk\text{.}\) Both integrals involve \(F_1\) terms and \(F_2\) terms and \(F_3\) terms. We shall show that the \(F_1\) terms in the 2 integrals agree. In other words, we shall assume that \(\vF=F_1\howdy\text{.}\) The proofs that the \(F_2\) and \(F_3\) terms also hold are similar. For simplicity, we'll presume² that the boundary of \(S\) consists of merely a single bend, and that we tin can

• pick a parametrization of \(S\) with

  \begin{equation*} S=\Prepare{\vr(u,v)=\large(x(u,v),y(u,v),z(u,v)\big)}{ (u,v)\text{ in } R\subset\bbbr^2} \terminate{equation*}

  and with \(\vr(u,five)\) orientation preserving in the sense that \(\hn\,\dee{Due south} = +\frac{\partial \vr}{\partial u} \times \frac{\partial \vr}{\partial v}\,\dee{u}\,\dee{v}\text{.}\) Too
• pick a parametrization of the curve, \(\fractional R\text{,}\) bounding \(R\) equally \(\big( u(t), 5(t)\big)\text{,}\) \(a\le t\le b\text{,}\) in such a way that when you walk along \(\partial R\) in the management of increasing \(t\text{,}\) then \(R\) is on your left.

So the curve \(\partial South\) bounding \(S\) tin can exist parametrized as \(\vR(t)=\vr\big(u(t),v(t)\big)\text{,}\) \(a\le t\le b\text{.}\)

Otherwise, decompose \(S\) into simpler pieces, analogously to what we did in the proof of the divergence theorem.

The orientation of \(\vR(t)\text{:}\)

We'll now verify that the direction of increasing \(t\) for the parametrization \(\vR(t)\) of \(\fractional S\) is the direction of the arrow on \(\partial S\) in the figure on the left above.

By continuity, it suffices to bank check the orientation at a single bespeak.

Find a betoken \((u_0,v_0)\) on \(\fractional R\) where the frontwards pointing tangent vector is a positive multiple of \(\,\how-do-you-do\text{.}\) The horizontal arrow on \(\partial R\) in the figure on the left below is at such a indicate. Suppose that \(t=t_0\) at this signal — in other words, suppose that \((u_0,v_0)=\large(u(t_0),v(t_0)\large)\text{.}\) Considering the forward pointing tangent vector to \(\partial R\) at \((u_0,v_0)\text{,}\) namely \(\large(u'(t_0),v'(t_0)\big)\text{,}\) is a positive multiple of \(\,\hullo\text{,}\) we have \(u'(t_0) \gt 0\) and \(v'(t_0)=0\text{.}\) The tangent vector to \(\fractional S\) at \(\vR(t_0)=\vr\big(u_0,v_0\big)\text{,}\) pointing in the management of increasing \(t\text{,}\) is

\begin{align*} \vR'(t_0)&=\diff{ }{t}\vr\big(u(t),five(t)\large)\big|_{t=t_0} =u'(t_0)\frac{\partial \vr}{\partial u}(u_0,v_0) +v'(t_0) \frac{\partial \vr}{\partial v}(u_0,v_0)\\ &=u'(t_0)\frac{\fractional \vr}{\partial u}(u_0,v_0) \end{align*}

and then is a positive multiple of \(\frac{\partial \vr}{\partial u}(u_0,v_0)\text{.}\) Run into the figure on the right below.

If we now walk along a path in the \(uv\)-airplane which starts at \((u_0,v_0)\text{,}\) holds \(u\) fixed at \(u_0\) and increases \(v\text{,}\) nosotros move into the interior of \(R\) starting at \((u_0,v_0)\text{.}\) Correspondingly, if we walk along the path, \(\vr(u_0,v)\text{,}\) in \(\bbbr^iii\) with \(v\) starting at \(v_0\) and increasing, we move into the interior of \(S\text{.}\) The forrard tangent to this new path, \(\frac{\fractional \vr}{\partial v}(u_0,v_0)\text{,}\) points from \(\vr(u_0,v_0)\) into the interior of \(Southward\text{.}\) It's the bluish arrow in the figure on the correct below.

Now imagine that you are walking forth \(\partial Due south\) in the management of increasing \(t\text{.}\) At time \(t_0\) you are at \(\vR(t_0)\text{.}\) Y'all betoken your right arm straight alee of you. So information technology is pointing in the direction \(\frac{\fractional \vr}{\partial u}(u_0,v_0)\text{.}\) You point your left arm out sideways into the interior of \(S\text{.}\) It is pointing in the direction \(\frac{\partial \vr}{\partial v}(u_0,v_0)\text{.}\) If the direction of increasing \(t\) is the aforementioned as the forward management of the orientation of \(\partial Southward\text{,}\) then the vector from our feet to our head, which is \(\frac{\partial \vr}{\partial u}(u_0,v_0) \times \frac{\fractional \vr}{\partial v}(u_0,v_0)\text{,}\) should exist pointing in the aforementioned direction as \(\hn\text{.}\) And since \(\hn\,\dee{S} = +\frac{\partial \vr}{\partial u} \times \frac{\partial \vr}{\partial v}\,\dee{u}\,\dee{v}\text{,}\) it is.

Now, with our parametrization and orientation sorted out, we tin can examine the integrals.

The surface integral:

Since \(F=F_1\,\hi\text{,}\) so that

\brainstorm{align*} \vnabla\times\vF &=\det\left[\begin{matrix} \hi & \hj & \hk \\ \frac{\partial }{\partial x} & \frac{\partial }{\fractional y} & \frac{\partial }{\fractional z} \\ F_1 & 0 & 0 \finish{matrix}\right] = \left(0,\frac{\partial F_1}{\partial z}, -\frac{\partial F_1}{\partial y}\right) \end{marshal*}

and

\begin{align*} \hn\,\dee{Due south} &=\frac{\fractional \vr}{\partial u} \times \frac{\partial \vr}{\fractional v} \ \dee{u}\,\dee{five} =\det\left[\begin{matrix} \hi & \hj & \hk \\ \frac{\partial x}{\fractional u} & \frac{\partial y}{\fractional u} & \frac{\partial z}{\partial u} \\ \frac{\partial ten}{\partial 5} & \frac{\partial y}{\fractional v} & \frac{\partial z}{\partial v} \stop{matrix}\right]\\ &=\left(\frac{\partial y}{\fractional u} \frac{\fractional z}{\partial v} -\frac{\partial z}{\partial u} \frac{\partial y}{\fractional 5}\correct)\howdy + \left(\frac{\fractional z}{\partial u} \frac{\fractional x}{\partial v} -\frac{\partial x}{\partial u} \frac{\fractional z}{\partial v}\correct)\hj\\ &\hskip1in +\left(\frac{\partial x}{\partial u} \frac{\partial y}{\partial v} -\frac{\partial y}{\fractional u} \frac{\partial 10}{\fractional five}\correct) \hk \terminate{align*}

and

\begin{align*} &\dblInt_{Due south}\vnabla\times\vF\cdot\hn\,\dee{S} =\dblInt_{R}\left(0,\frac{\partial F_1}{\partial z}, -\frac{\partial F_1}{\partial y}\right) \cdot\frac{\partial \vr}{\partial u} \times \frac{\partial \vr}{\partial v}\ \dee{u}\,\dee{v}\cr &\hskip.15in=\dblInt_{R}\left\{\frac{\partial F_1}{\fractional z} \left(\frac{\partial z}{\partial u} \frac{\partial 10}{\partial five} -\frac{\partial x}{\partial u} \frac{\partial z}{\partial 5}\correct) -\frac{\partial F_1}{\fractional y} \left(\frac{\partial x}{\fractional u} \frac{\partial y}{\fractional v} -\frac{\partial y}{\partial u} \frac{\partial 10}{\fractional v}\right) \correct\}\,\dee{u}\,\dee{v} \cease{align*}

Now we examine the line integral and evidence that information technology equals this ane.

The line integral:

\brainstorm{align*} &\oint_{\partial S}\vF\cdot \dee{\vr} =\int_a^b\vF\Big(\vr\big(u(t),v(t)\big)\Big)\cdot\unequal{ }{t} \vr\big(u(t),five(t)\big)\ \dee{t}\\ &\hskip0.1in=\int_a^b\vF\Big(\vr\big(u(t),five(t)\large)\Big)\cdot\Big[ \frac{\partial\vr}{\partial u}\big(u(t),v(t)\big)\diff{u}{t}(t) +\frac{\fractional\vr}{\partial v}\large(u(t),v(t)\big)\diff{5}{t}(t)\Large]\ \dee{t} \end{align*}

Nosotros can write this every bit the line integral

\begin{marshal*} &\oint_{\partial R}M(u,v)\ \dee{u}+Due north(u,5)\ \dee{v}\\ &\hskip1in=\int_a^b \Big[M\big(u(t),v(t)\big)\ \diff{u}{t}(t) +N\large(u(t),v(t))\ \unequal{v}{t}(t)\Big]\ \dee{t} \end{marshal*}

around \(\partial R\text{,}\) if we choose

\begin{alignat*}{two} M(u,five)&=\vF\big(\vr(u,v)\large)\cdot\frac{\partial\vr}{\partial u}(u,v) &&=F_1\large({x(u,v),y(u,five),z(u,v)}\big)\frac{\partial x}{\partial u}(u,five)\\ N(u,five)&=\vF\large(\vr(u,5)\big)\cdot\frac{\partial\vr}{\partial v}(u,five) &&=F_1\big({x(u,v),y(u,v),z(u,v)}\big)\frac{\partial x}{\partial v}(u,v) \end{alignat*}

Finally, we evidence that the surface integral equals the line integral:

Past Dark-green's Theorem, we have

\brainstorm{align*} \oint_{\fractional S}\vF\cdot \dee{\vr} &=\oint_{\fractional R}Thousand(u,v)\ \dee{u}+North(u,v)\ \dee{v}\cr =\dblInt_{R}\bigg\{&\frac{\fractional North}{\partial u}-\frac{\partial M}{\partial v} \bigg\}\ \dee{u} \dee{v}\cr =\dblInt_{R}\bigg\{& \frac{\fractional }{\partial u}\big[ F_1\big({x(u,v),y(u,v),z(u,5)}\big) \big]\frac{\partial ten}{\partial v} +F_1\frac{\partial^2 x}{\partial u\partial v}\\ -& \frac{\partial }{\partial v}\big[ F_1\large({10(u,v),y(u,v),z(u,v)}\big) \big]\frac{\partial x}{\partial u} -F_1\frac{\partial^two x}{\partial v\partial u} \bigg\}\ \dee{u} \dee{v}\\ =\dblInt_{R}\bigg\{& {\color{blue}{\Big(\frac{\partial F_1}{\partial x} \frac{\partial x}{\partial u}}} +\frac{\partial F_1}{\partial y} \frac{\partial y}{\partial u} +\frac{\partial F_1}{\partial z} \frac{\partial z}{\partial u} \Large)\frac{\partial x}{\fractional 5} {\color{red}{ +F_1\frac{\partial^two x}{\partial u\partial v}}}\cr -&\Big( {\color{blueish}{\frac{\fractional F_1}{\partial x} \frac{\partial ten}{\partial v}}} +\frac{\partial F_1}{\fractional y} \frac{\partial y}{\fractional 5} +\frac{\partial F_1}{\partial z} \frac{\partial z}{\partial v} \Big)\frac{\partial ten}{\fractional u} {\color{red}{ -F_1\frac{\partial^2 x}{\fractional v\fractional u}}} \bigg\}\ \dee{u} \dee{v}\cr =\dblInt_{R}\bigg\{&\Big(\frac{\partial F_1}{\partial y} \frac{\partial y}{\fractional u} +\frac{\partial F_1}{\partial z} \frac{\partial z}{\fractional u} \Big)\frac{\fractional x}{\partial v} -\Big(\frac{\fractional F_1}{\partial y} \frac{\partial y}{\partial 5} +\frac{\partial F_1}{\partial z} \frac{\partial z}{\partial v} \bigg)\frac{\fractional ten}{\partial u} \Big\}\,\dee{u}\, \dee{v}\\ &\hskip-33pt=\dblInt_{Due south}\vnabla\times\vF\cdot\hn\,\dee{S} \stop{align*}

which is the conclusion that we wanted.

Before we motion on to some examples, here are a couple of remarks.

• Stokes' theorem says that \(\oint_{C}\vF\cdot \dee{\vr} =\dblInt_{S}\vnabla\times\vF\cdot\hn\ \dee{Southward}\) for whatsoever (suitably oriented) surface whose boundary is \(C\text{.}\) So if \(S_1\) and \(S_2\) are two unlike (suitably oriented) surfaces having the same boundary curve \(C\text{,}\) and so

  \begin{equation*} \dblInt_{S_1}\vnabla\times\vF\cdot\hn\ \dee{South} =\dblInt_{S_2}\vnabla\times\vF\cdot\hn\ \dee{S} \end{equation*}

  For example, if \(C\) is the unit circle

  \brainstorm{equation*} C=\Ready{(x,y,z)}{x^2+y^ii=i,\ z=0} \end{equation*}

  oriented counterclockwise when viewed from in a higher place, so both

  \brainstorm{marshal*} S_1&= \Prepare{(x,y,z)}{x^2+y^ii\le one,\ z=0}\\ S_2&= \Set{(x,y,z)}{z\ge0,\ x^2+y^ii+z^ii=ane} \cease{align*}

  with up pointing unit of measurement normal vectors, have boundary \(C\text{.}\) And then Stokes' tells us that \(\dblInt_{S_1}\vnabla\times\vF\cdot\hn\ \dee{S} =\dblInt_{S_2}\vnabla\times\vF\cdot\hn\ \dee{S}\text{.}\)

  

  It should not be a surprise that \(\dblInt_{S_1}\vnabla\times\vF\cdot\hn\ \dee{S} =\dblInt_{S_2}\vnabla\times\vF\cdot\hn\ \dee{S}\text{,}\) for the following reason. Let

  \begin{equation*} V= \Set{(10,y,z)}{x^two+y^2+z^2\le 1,\ z\ge 0} \\ \end{equation*}

  be the solid between \(S_1\) and \(S_2\text{.}\) The purlieus \(\partial V\) of \(V\) is the wedlock of \(S_1\) and \(S_2\text{.}\)

  

  Just beware that the outward pointing normal to \(\partial V\) (call it \(\hN\)) is \(+\hn\) on \(S_2\) and \(-\hn\) on \(S_1\text{.}\) So the divergence theorem gives

  \begin{align*} &\dblInt_{S_2}\vnabla\times\vF\cdot\hn\ \dee{S} -\dblInt_{S_1}\vnabla\times\vF\cdot\hn\ \dee{Due south}\\ &\hskip0.5in= \dblInt_{S_2}\vnabla\times\vF\cdot\hN\ \dee{Southward} +\dblInt_{S_1}\vnabla\times\vF\cdot\hN\ \dee{S}\\ &\hskip0.5in= \dblInt_{\partial 5}\vnabla\times\vF\cdot\hN\ \dee{Due south}\\ &\hskip0.5in= \tripInt_{Five}\vnabla\cdot\big(\vnabla\times\vF\big)\ \dee{Five}\\ &\hskip2in\qquad\text{by the divergence theorem}\\ &\hskip0.5in=0 \finish{align*}

  by the vector identity Theorem four.1.vii.a.

• As a 2nd remark, suppose that the vector field \(\vF\) obeys \(\vnabla\times\vF=\vZero\) everywhere. Then Stokes' theorem forces \(\oint_{C}\vF\cdot \dee{\vr}=0\) are around all closed curves \(C\text{,}\) which implies that \(\vF\) is conservative, by Theorem 2.4.7. So Stokes' theorem provides another proof of Theorem 2.4.8.

Here is an easy instance which shows that Stokes' can be very useful when \(\vnabla\times\vF\) simplifies.

Case 4.4.2 .

Evaluate \(\oint_C\vF\cdot\dee{\vr}\) where \(\vF= \big[2z+\sin\big(x^{146}\big)\big]\,\hi-5z\,\hj -5y\,\hk\) and the curve \(C\) is the circle \(x^ii+y^2=4\text{,}\) \(z=1\text{,}\) oriented counterclockwise when viewed from higher up.

Solution

The \(x^{146}\) in \(\vF\) will probably make a direct evaluation of the integral difficult. So nosotros'll use Stokes' theorem. To practise then we need a surface \(S\) with \(\fractional S=C\text{.}\) The simplest is just the flat deejay

\begin{equation*} S = \Set up{(x,y,z)}{ x^2+y^2\le 4,\ \ z=1} \end{equation*}

Since

\begin{align*} \vnabla\times\vF & = \det\left[\begin{matrix} \howdy & \hj & \hk \\ \frac{\partial }{\partial x} & \frac{\partial }{\partial y} & \frac{\fractional }{\fractional z} \\ 2z+\sin\large(x^{146}\large) & -5z & -5y \cease{matrix}\right]\\ & =\hello \det\left[\begin{matrix} \frac{\partial }{\partial y} & \frac{\partial }{\partial z} \\ -5z & -5y \stop{matrix}\right] -\hj\det\left[\begin{matrix} \frac{\partial }{\partial x} & \frac{\partial }{\partial z} \\ 2z+\sin\big(x^{146}\large) & -5y \end{matrix}\right]\\ &\hskip1in +\hk\det\left[\begin{matrix} \frac{\fractional }{\partial x} & \frac{\fractional }{\partial y} \\ 2z+\sin\big(ten^{146}\big) & -5z \end{matrix}\correct]\\ &=2\hj \end{align*}

and the normal to \(S\) is \(\hk\text{,}\) Stokes' theorem gives

\brainstorm{marshal*} \oint_C\vF\cdot\dee{\vr} & = \dblInt_S \vnabla\times\vF\cdot\hn\,\dee{S} = \dblInt_S (ii\hj)\cdot\hk\,\dee{S} =0 \cease{align*}

At present we'll repeat the last example with a harder bend.

Example 4.4.iii .

Evaluate \(\oint_C\vF\cdot\dee{\vr}\) where \(\vF= \big[2z+\sin\big(ten^{146}\big)\large]\,\hi-5z\,\hj -5y\,\hk\) and the curve \(C\) is the intersection of \(ten^two+y^2+z^2=iv\) and \(z=y\text{,}\) oriented counterclockwise when viewed from above.

Solution

The surface \(x^two+y^2+z^2=four\) is the sphere of radius \(2\) centred on the origin and \(z=y\) is a plane which contains the origin. So \(C\text{,}\) being the intersection of a sphere with a plane through the eye of the sphere, is a circle, with centre \((0,0,0)\) and radius \(ii\text{.}\) The part of the circle in the starting time octant is sketched on the left below.

The \(ten^{146}\) in \(\vF\) volition probably make a straight evaluation of the integral hard. So nosotros'll use Stokes' theorem. To exercise then we need a surface \(Southward\) with \(\fractional S=C\text{.}\) The simplest is the apartment disk

\brainstorm{equation*} South = \Set{(x,y,z)}{ x^ii+y^ii+z^2\le 4,\ \ z=y} \end{equation*}

The first octant of \(Southward\) is shown in the effigy on the right above. Nosotros saw in the concluding Example 4.4.2 that

\begin{align*} \vnabla\times\vF &=2\hj \end{align*}

And so Stokes' theorem gives

\begin{marshal*} \oint_C\vF\cdot\dee{\vr} & = \dblInt_S \vnabla\times\vF\cdot\hn\,\dee{S} = two\dblInt_S \hj\cdot\hn\,\dee{S} \end{marshal*}

We'll evaluate the integral \(2\dblInt_S \hj\cdot\hn\,\dee{S}\) in two means. The starting time way is more efficient, but besides requires more insight. Since \(\vnabla(z-y)=\hk-\hj\text{,}\) the upward unit normal to the plane \(z-y=0\text{,}\) and hence to \(S\text{,}\) is \(\hn = \frac{ane}{\sqrt{ii}}(\hk-\hj)\text{.}\) Consequently the integrand

\begin{assemble*} \hj\cdot\hn=\hj\cdot\Large(\frac{-\hj+\hk}{\sqrt{2}}\Big)=-\frac{1}{\sqrt{2}} \terminate{gather*}

is a constant and we do not need a formula for \(\hn\,dS\text{:}\)

\begin{align*} \oint_C\vF\cdot\dee{\vr} & = two\dblInt_S \hj\cdot\hn\,\dee{South} =-\sqrt{2} \dblInt_S \dee{S} = -\sqrt{two}\text{Surface area}(S) =-\sqrt{2}\pi\, 2^ii\\ &=-4\sqrt{2}\pi \cease{marshal*}

Alternatively, we tin evaluate the integral \(\dblInt_S \hj\cdot\hn\,\dee{S}\) using our normal protocol. Equally \(Southward\) is part of the aeroplane \(z=f(x,y)=y\text{,}\)

\begin{gather*} \hn\,dS = \pm\big(-f_x\,\howdy-f_y\,\hj+\hk\large)\,\dee{x}\dee{y} = \pm (-\hj+\hk)\,\dee{x}\dee{y} \end{get together*}

To get the upwards pointing normal pointing normal, nosotros accept the \(+\) sign so that \(\hn\,dS= (-\hj+\hk)\,\dee{10}\dee{y}\text{.}\) Every bit \((ten,y,z)\) runs over

\begin{marshal*} S &= \Fix{(10,y,z)}{ x^2+y^2+z^2\le 4,\ \ z=y}\\ &= \Set up{(x,y,z)}{ x^2+2y^2\le 4,\ \ z=y}\\ &= \Set{(ten,y,z)}{ \tfrac{x^ii}{four} +\tfrac{y^2}{2}\le 1,\ \ z=y} \end{align*}

\((x,y)\) runs over the elliptical deejay \(R=\Set{(x,y)}{\frac{10^ii}{4} +\frac{y^2}{2}\le 1}\text{.}\) The part of this ellipse in the first octant is the shaded region in the figure below.

This ellipse has semiaxes \(a=2\) and \(b =\sqrt{two}\) and hence area \(\pi a b = 2\sqrt{2} \pi\text{.}\) And so

\begin{marshal*} \oint_C\vF\cdot\dee{\vr} & = 2\dblInt_S \hj\cdot\hn\,\dee{Due south} =two\dblInt_R \hj\cdot(-\hj+\hk)\,\dee{10}\dee{y} =-2 \dblInt_R \dee{10}\dee{y}\\ &= -2\text{Area}(R)\\ &=-4\sqrt{two}\pi \end{align*}

Instance 4.4.4 .

Evaluate \(\oint_C\vF\cdot\dee{\vr}\) where \(\vF= (x+y)\,\hi+2(x-z)\,\hj +(y^2+z)\,\hk\) and \(C\) is the oriented curve obtained by going from \((2,0,0)\) to \((0, 3, 0)\) to \((0, 0, 6)\) and back to \((2, 0, 0)\) along straight line segments.

Solution one

In this first solution, nosotros'll evaluate the integral directly. The first line segment (\(C_1\) in the figure higher up) may exist parametrized as

\brainstorm{equation*} \vr(t)=(ii,0,0)+t\big\{(0, 3, 0)-(2,0,0)\big\} =\big(ii-2t\,,\,3t\,,\,0\large)\qquad 0\le t\le ane \stop{equation*}

And so the integral along this segment is

\begin{align*} \int_0^1 \vF(\vr(t))\cdot\diff{\vr}{t}\ \dee{t} &=\int_0^1 (2+t\,,\, ii(2-2t)\,,\, (3t)^2)\cdot(-two\,,\, iii\,,\, 0)\ \dee{t}\\ &=\int_0^ane (8-14t)\ \dee{t}\\ &=\Big[8t-7t^2\Big]_0^one=1 \end{align*}

The second line segment (\(C_2\) in the figure above) may be parametrized every bit

\begin{equation*} \vr(t)=(0,3,0)+t\big\{(0, 0, half dozen)-(0,iii,0)\big\} =\big(0\,,\,three-3t\,,\,6t\big)\qquad 0\le t\le 1\text{.} \end{equation*}

And then the integral along this segment is

\begin{align*} \int_0^1 \vF(\vr(t))\cdot\unequal{\vr}{t}\ \dee{t} &=\int_0^1 \large(3(one-t)\,,\, - 12t\,,\, 9(1-t)^2+ 6t\large)\cdot(0, -three, 6)\ \dee{t}\\ &=\int_0^1 [36t+54(i-t)^two+36t]\ \dee{t}\\ &=\Big[18t^ii-18(1-t)^3+18t^2\Big]_0^1\\ &=54 \end{align*}

The last line segment (\(C_3\) in the figure above) may be parametrized as

\begin{equation*} \vr(t)=(0,0,6)+t\large\{(ii,0,0) -(0,0,6)\big\} = (2t\,,\,0\,,\,half dozen-6t) \qquad 0\le t\le 1 \end{equation*}

So the line integral along this segment is

\begin{marshal*} \int_0^1 \vF(\vr(t))\cdot\diff{\vr}{t}\ \dee{t} &=\int_0^1 \big(2t\,,\, 4t - 12(i-t)\,,\, 6(1-t)\large)\cdot(ii, 0, -6)\ \dee{t}\\ &=\int_0^1 [4t-36(1-t)]\ \dee{t} =\Large[2t^2+18(1-t)^2\Big]_0^1 =-16 \end{align*}

The full line integral is

\begin{equation*} \oint_C\vF\cdot \dee{\vr}=1+54-sixteen=39 \end{equation*}

Solution ii

This time we shall apply Stokes' Theorem. The coil of \(\vF\) is

\brainstorm{align*} \vnabla\times\vF &=\det\left[\begin{matrix} \hello & \hj &\hk \\ \frac{\partial }{\fractional x} & \frac{\fractional }{\partial y} & \frac{\partial }{\partial z} \\ x+y & 2(ten-z) & y^2+z \terminate{matrix} \right]\\ & =(2y+two)\hullo-(0-0)\hj+(2-i)\hk\\ &=2(y+one)\hi+\hk \end{align*}

The curve \(C\) is a triangle and so is contained in a plane. Any aeroplane has an equation of the grade \(Ax+By+Cz=D\text{.}\) Our plane does not laissez passer through the origin (await at the figure above) so the \(D\) must be nonzero. Consequently we may divide \(Ax+By+Cz=D\) through by \(D\) giving an equation of the form \(ax+by+cz=1\text{.}\)

• Because \((2,0,0)\) lies on the plane, \(a=\frac{i}{2}\text{.}\)
• Because \((0,3,0)\) lies on the plane, \(b=\frac{ane}{3}\text{.}\)
• Because \((0,0,6)\) lies on the plane, \(c=\frac{1}{6}\text{.}\)

So the triangle is independent in the plane \(\frac{x}{2}+\frac{y}{three}+\frac{z}{vi}=1\text{.}\) Information technology is the boundary of the surface \(South\) that consists of the portion of the aeroplane \(\frac{ten}{2}+\frac{y}{3}+\frac{z}{half dozen}=ane\) that obeys \(ten\ge 0\text{,}\) \(y\ge 0\) and \(z\ge 0\text{.}\) Rewrite the equation of the plane equally \(z=6-3x-2y\text{.}\) For this surface

\begin{equation*} \hn\ \dee{Due south}=(iii\hi+2\hj+\hk)\,\dee{x}\,\dee{y} \end{equation*}

past 3.3.2, and we can write

\begin{align*} S&=\Ready{(10,y,z)}{x\ge0,\ y\ge0,\ z\ge0,\ z=6-3x-2y}\\ &=\Set{(x,y,z)}{x\ge0, y\ge0,\ 6-3x-2y\ge0,\ z=6-3x-2y} \end{align*}

Equally \((x,y,z)\) runs over \(S\text{,}\) \((x,y)\) runs over the triangle

\begin{align*} R&=\Set{(x,y,z)}{10\ge0,\ y\ge0,\ 3x+2y\le vi}\\ &=\Prepare{(10,y,z)}{x\ge0,\ 0\le y \le \tfrac{3}{ii}(2-x)} \end{marshal*}

Using horizontal strips every bit in the effigy on the left below,

\begin{align*} \oint_C\vF\cdot \dee{\vr}&=\dblInt_S\vnabla\times\vF\cdot\hn\,\dee{S}\\ &=\dblInt_R [2(y+1)\hullo+\hk]\cdot[3\hullo+2\hj+\hk]\ \dee{x}\,\dee{y}\cr &=\dblInt_R [6y+7]\ \dee{x}\,\dee{y}\\ & =\int_0^iii \dee{y}\int_0^{{ane\over 3}(6-2y)} \dee{x}\ [6y+7]\\ &=\int_0^3 \dee{y}\ \frac{1}{3}[6y+7][half-dozen-2y]\\ & =\frac{i}{3}\int_0^iii \dee{y}\ [-12y^2+22y+42]\\ &=\frac{1}{3}\Big[-4y^iii+11y^ii+42y\Big]_0^3\\ & =\large[-iv\times 9 +11\times 3 +42\big] =39 \finish{align*}

Alternatively, using vertical strips as in the figure on the right above,

\begin{marshal*} \oint_C\vF\cdot \dee{\vr} &=\dblInt_R [6y+seven]\ \dee{x}\,\dee{y}\\ &=\int_0^2 \dee{x}\int_0^{{3\over ii}(2-ten)} \dee{y}\ [6y+7]\\ &=\int_0^2 \dee{x}\ \Big[3\frac{three^ii}{2^2}(two-ten)^ii+7\frac{three}{2}(ii-x)\Big]\\ & =\Big[-\frac{27}{iv}\,\frac{1}{3}(2-ten)^three -\frac{21}{2}\,\frac{one}{2}(two-ten)^two\Big]_0^2\\ &=\frac{9}{4}\,8+\frac{21}{4}\,four =39 \end{align*}

Example 4.four.5 .

Evaluate \(\oint_C\vF\cdot\dee{\vr}\) where \(\vF= (\cos x +y+z)\,\hi+(x+z)\,\hj +(x+y)\,\hk\) and \(C\) is the intersection of the surfaces

\begin{equation*} 10^ii+\frac{y^2}{2}+\frac{z^two}{3}=1\qquad\text{and}\qquad z=ten^ii+2y^2 \finish{equation*}

oriented counterclockwise when viewed from above.

Solution

Outset, let's sketch the curve. \(x^2+\frac{y^ii}{ii}+\frac{z^2}{3}=1\) is an ellipsoid centred on the origin and \(z=x^2+2y^two\) is an upward opening paraboloid that passes through the origin. They are sketched in the effigy below. The paraboloid is red.

Their intersection, the curve \(C\text{,}\) is the blue curve in the effigy. It looks like a plain-featuredⁱⁱⁱ circumvolve.

Ane could imagine parametrizing \(C\text{.}\) For example, substituting \(x^2 = z - 2y^2\) into the equation of the ellipsoid gives \(-\frac{3}{2}y^ii + \frac{i}{three}(z+\frac{3}{2}\big)^2 = \frac{7}{iv}\text{.}\) This can exist solved to give \(y\) as a function of \(z\) and so \(10^2=z-2y^2\) also gives \(10\) as a function of \(z\text{.}\) Even so this would clearly yield, at best, a really messy integral. So let'south try Stokes' theorem.

In fact, since

\begin{align*} \vnabla\times\vF &=\det\left[\begin{matrix}\hi&\hj&\hk \\ \frac{\partial }{\partial x} &\frac{\partial }{\partial y} &\frac{\fractional }{\partial z} \\ \cos x + y+z&x+z&ten+y\end{matrix}\right]\\ &=\hi\big(1-1\big)-\hj\big(1-ane\big)+\hk\large(ane-1\big)\\ &=\vZero \terminate{marshal*}

This \(\vF\) is bourgeois! (In fact \(\vF=\vnabla\big(\sin ten + xy + xz +yz\big)\text{.}\)) As \(C\) is a airtight curve, \(\oint_C\vF\cdot\dee{\vr}=0 \text{.}\)

Example four.4.6 .

Evaluate \(\dblInt_S\vG\cdot\hn\,\dee{S}\) where \(\vG= (2x)\,\how-do-you-do+(2z-2x)\,\hj +(2x-2z)\,\hk\) and

\begin{equation*} S=\Set{(x,y,z)}{z=\big(ane-ten^2-y^ii\big)(1-y^3)\cos x\ e^{y},\ x^2+y^2\le 1} \end{equation*}

with upwardly pointing normal

Solution ane

The surface \(South\) is sketched below. It is a pretty weird surface. Most the

only simple matter most information technology is that its boundary, \(\partial Due south\text{,}\) is the circle \(x^2+y^2=1\text{,}\) \(z=0\text{.}\) It is clear that nosotros should not try to evaluate the integral directly⁴. In this solution we will combine the difference theorem with the ascertainment that

\begin{gather*} \vnabla\cdot\vG = \frac{\partial }{\partial ten}(2x)+ \frac{\partial }{\partial y}(2z-2x)+ \frac{\partial }{\partial z}(2x-2z) =0 \end{get together*}

to avoid ever having work with the surface \(S\text{.}\) Here is an outline of what we will do.

• We offset select a elementary surface \(Due south'\) whose boundary \(\partial S'\) is also the circumvolve \(x^2+y^2=1\text{,}\) \(z=0\text{.}\) A prissy unproblematic choice of \(S'\text{,}\) and the surface that we will use, is the disk

  \begin{equation*} S' =\Set{(x,y,z)}{x^2+y^ii=1,\ z=0} \finish{equation*}

• Then we define \(V\) to be the solid whose elevation surface is \(Due south\) and whose bottom surface is \(Due south'\text{.}\) So the boundary of \(V\) is the union of \(Southward\) and \(Due south'\text{.}\)

  

• For \(S'\text{,}\) we will employ the upward pointing normal \(\hn=\hk\text{,}\) which is minus the outward pointing normal to \(\fractional V\) on \(S'\text{.}\) So the difference theorem says that

  \begin{equation*} \tripInt_V \vnabla\cdot\vG\,\dee{V} =\dblInt_S\vG\cdot\hn\,\dee{S}-\dblInt_{S'}\vG\cdot\hn\,\dee{Southward} \end{equation*}

  The left hand side is zero because, as we accept already seen, \(\vnabla\cdot\vG=0\text{.}\) So

  \brainstorm{equation*} \dblInt_S\vG\cdot\hn\,\dee{South} =\dblInt_{S'}\vG\cdot\hn\,\dee{Due south} \end{equation*}

• Finally, we compute \(\dblInt_{S'}\vG\cdot\hn\,\dee{S}\text{.}\)

We saw an argument like this (with \(\vG=\vnabla\times\vF\)) in the starting time remark following the proof of Theorem four.4.1.

And so all that nosotros accept to do now is compute

\begin{align*} \dblInt_S\vG\cdot\hn\,\dee{S} &=\dblInt_{Due south'}\vG\cdot\hn\,\dee{S} =\dblInt_{S'}\vG\cdot\hk\,\dee{Southward} =\dblInt_{\Atop{x^2+y^2\le 1}{z=0}}(2x-2z)\,\dee{ten}\dee{y}\\ &=\dblInt_{\Atop{10^2+y^2\le one}{z=0}}(2x)\,\dee{x}\dee{y}\\ &=0 \terminate{align*}

but because the integrand is odd nether \(10\rightarrow-x\text{.}\)

Solution 2

In this second solution we'll employ Stokes' theorem instead of the divergence theorem. To do so, we have to express \(\vG\) in the course \(\vnabla\times\vF\text{.}\) So the first thing to do is to cheque if \(\vG\) passes the screening examination, Theorem 4.1.12, for the existence of vector potentials. That is, to bank check if \(\vnabla\cdot\vG=0\text{.}\) It is. We saw this in Solution one above.

Adjacent, we take to find a vector potential. In fact, we have already constitute, in Example 4.1.15, that

\begin{equation*} \vF = (z^2-2xz) \hullo +(10^2-2xz)\hj \end{equation*}

is a vector potential for \(\vG\text{,}\) which nosotros can quickly check.

Parametrizing \(C\) past \(\vr(t) = \cos t\,\how-do-you-do+\sin t\,\hj\text{,}\) \(0\le t\le ii\pi\text{,}\) Stokes' theorem gives (recalling that \(z=0\) on \(C\) so that \(\vF\big(\vr(t)\large)=x^ii\,\hj\Large|_{x=\cos t}=\cos^2t\))

\brainstorm{align*} \dblInt_S\vG\cdot\hn\,\dee{Due south} &= \dblInt_S\vnabla\times\vF\cdot\hn\,\dee{South} = \oint_C \vF\cdot\dee{\vr} =\int_0^{ii\pi} \vF\big(\vr(t)\big)\cdot\unequal{\vr}{t}\ \dee{t}\\ &= \int_0^{2\pi} \large(\cos^2 t\big)(\cos t)\ \dee{t} \terminate{align*}

Of course this integral can be evaluated by using that one antiderivative of the integrand \(\cos^3 t =\big(one-\sin^2t\large)\cos t\) is \(\sin t-\frac{1}{3}\sin^3 t\) and that this antiderivative is zero at \(t=0\) and at \(t=2\pi\text{.}\) Just information technology is easier to detect that the integral of whatever odd ability of \(\sin t\) or \(\cos t\) over whatsoever total period is nada. Look, for example, at the graphs of \(\sin^3x\) and \(\cos^3x\text{,}\) below.

Either way

\begin{assemble*} \dblInt_S\vG\cdot\hn\,\dee{Due south} =0 \stop{gather*}

Example 4.four.7 .

In this case we compute, in three different ways, \(\oint_C\vF\cdot \dee{\vr}\) where

\begin{equation*} \vF=(z-y)\,\hullo-(10+z)\,\hj-(x+y)\,\hk \end{equation*}

and \(C\) is the curve \(10^2+y^2+z^2=4\text{,}\) \(z=y\) oriented counterclockwise when viewed from to a higher place.

Solution ane

Direct Computation:

In this first computation, we parametrize the bend \(C\) and compute \(\oint_C\vF\cdot \dee{\vr}\) directly. The airplane \(z=y\) passes through the origin, which is the heart of the sphere \(x^two+y^two+z^2=four\text{.}\) So \(C\) is a circumvolve which, like the sphere, has radius \(two\) and heart \((0,0,0)\text{.}\) We utilise a parametrization of the course

\begin{equation*} \vr(t)=\vc+\rho\cos t\,\hello'+ \rho\sin t\,\hj' \qquad 0\le t\le two\pi \end{equation*}

where

• \(\vc=(0,0,0)\) is the middle of \(C\text{,}\)
• \(\rho=2\) is the radius of \(C\) and

• \(\howdy'\) and \(\hj'\) are two vectors that

  1. are unit vectors,
  2. are parallel to the plane \(z=y\) and
  3. are mutually perpendicular.

The trickiest function is finding suitable vectors \(\hi'\) and \(\hj'\text{:}\)

• The point \((2,0,0)\) satisfies both \(x^2+y^two+z^2=4\) and \(z=y\) and so is on \(C\text{.}\) We may choose \(\hi'\) to be the unit vector in the direction from the eye \((0,0,0)\) of the circle towards \((2,0,0)\text{.}\) Namely \(\howdy'=(1,0,0)\text{.}\)
• Since the plane of the circle is \(z-y=0\text{,}\) the vector \(\vnabla(z-y)=(0,-1,1)\) is perpendicular to the plane of \(C\text{.}\) So \(\hk'=\frac{ane}{\sqrt{2}}(0,-1,1)\) is a unit vector normal to \(z=y\text{.}\) Then \(\hj'=\hk'\times\hi'=\frac{1}{\sqrt{two}}(0,-1,1)\times(ane,0,0) =\frac{one}{\sqrt{two}}(0,1,1)\) is a unit vector that is perpendicular to \(\hi'\) and \(\hk'\text{.}\) Since \(\hj'\) is perpendicular to \(\hk'\text{,}\) it is parallel to \(z=y\text{.}\)

Substituting in \(\vc=(0,0,0)\text{,}\) \(\rho=2\text{,}\) \(\hello'=(1,0,0)\) and \(\hj'=\frac{1}{\sqrt{2}}(0,1,1)\) gives

\begin{align*} &\vr(t)=2\cos t\,(1,0,0)+ ii\sin t\,\frac{1}{\sqrt{2}}(0,1,i) =2\Big(\cos t, \frac{\sin t}{\sqrt{2}},\frac{\sin t}{\sqrt{ii}}\Big)\\ &0\le t\le 2\pi \stop{align*}

To bank check that this parametrization is correct, notation that \(x=ii\cos t\text{,}\) \(y=\sqrt{2}\sin t\text{,}\) \(z=\sqrt{2}\sin t\) satisfies both \(x^2+y^2+z^2=4\) and \(z=y\text{.}\)

At \(t=0\text{,}\) \(\vr(0)=(2,0,0)\text{.}\) As \(t\) increases, \(z(t)=\sqrt{ii}\sin t\) increases and \(\vr(t)\) moves upwards towards \(\vr\big(\frac{\pi}{two}\large)=(0,\sqrt{ii},\sqrt{two})\text{.}\) This is the desired counterclockwise direction (when viewed from in a higher place). Now that we have a parametrization, we tin can prepare up the integral.

\begin{align*} \vr(t)&=\big(2\cos t, \sqrt{2}\sin t,\sqrt{two}\sin t\large)\cr \vr\,'(t)&=\big(-2\sin t,\sqrt{2}\cos t,\sqrt{ii}\cos t\big)\cr \vF\big(\vr(t)\big)&=\large(z(t)-y(t),-10(t)-z(t),-x(t)-y(t)\big)\cr &=\large(\sqrt{2}\sin t-\sqrt{two}\sin t, -ii\cos t-\sqrt{2}\sin t,-2\cos t-\sqrt{2}\sin t\big)\cr &=-\big(0, 2\cos t+\sqrt{2}\sin t,2\cos t+\sqrt{two}\sin t\big)\cr \vF\big(\vr(t)\large)\cdot \vr\,'(t) &=-\large[4\sqrt{two}\cos^2 t+four\cos t\sin t\large]\\ &=-\big[two\sqrt{2}\cos(2t)+2\sqrt{2}+2\sin(2t)\large] \end{align*}

by the double angle formulae \(\sin(2t)=two\sin t\,\cos t\) and \(\cos(2t) = 2\cos^2t-i\text{.}\) So

\begin{align*} \oint_C\vF\cdot \dee{\vr} &=\int_0^{2\pi}\vF\big(\vr(t)\big)\cdot \vr\,'(t)\ \dee{t}\\ &=\int_0^{2\pi}-\big[2\sqrt{2}\cos(2t)+2\sqrt{2}+2\sin(2t)\big]\ \dee{t}\\ &=-\Big[\sqrt{2}\sin(2t)+two\sqrt{2}t-\cos(2t)\Big]_0^{2\pi}\\ &=-4\sqrt{2}\pi \finish{align*}

Oof! Let's practice it an easier manner.

Solution ii

Stokes' Theorem:

To apply Stokes' theorem we demand to limited \(C\) as the boundary \(\fractional S\) of a surface \(S\text{.}\) As

\brainstorm{equation*} C=\Set up{(x,y,z)}{x^2+y^ii+z^two=4,\ z=y} \stop{equation*}

is a airtight bend, this is possible. In fact there are many possible choices of \(S\) with \(\partial S=C\text{.}\) Three possible \(South\)'s (sketched beneath) are

\brainstorm{align*} S&=\Set{(x,y,z)}{x^2+y^2+z^2\le 4,\ z=y}\cr S'&=\Set{(x,y,z)}{10^2+y^two+z^two= 4,\ z\ge y}\cr S''&=\Set{(10,y,z)}{x^ii+y^2+z^two= four,\ z\le y} \finish{marshal*}

The beginning of these, which is part of a plane, is likely to lead to simpler computations than the last two, which are parts of a sphere. So we choose what looks like the simpler style.

In preparation for application of Stokes' theorem, we compute \(\vnabla\times\vF\) and \(\hn\,\dee{S}\text{.}\) For the latter, we apply the formula \(\hn\,\dee{S}=\pm(-f_x,-f_y,1)\,\dee{x}dy\) (of Equation 3.3.ii) to the surface \(z=f(x,y)=y\text{.}\) Nosotros utilise the \(+\) sign to requite the normal a positive \(\hk\) component.

\begin{align*} \vnabla\times\vF &=\det\left[\begin{matrix}\how-do-you-do&\hj&\hk \\ \frac{\fractional }{\partial x} &\frac{\partial }{\partial y} &\frac{\fractional }{\partial z} \\ z-y&-x-z&-x-y\terminate{matrix}\right]\\ &=\hi\big(-1-(-ane)\big)-\hj\big(-1-one\big)+\hk\large(-1-(-ane)\large)\\ &=ii\,\hj\\ \hn\,\dee{South}&=(0,-ane,one)\,\dee{x}\dee{y}\\ \vnabla\times\vF\cdot\hn\,\dee{S}&=(0,2,0)\cdot(0,-ane,1)\,\dee{10}\dee{y} =-2\,\dee{x}\dee{y} \finish{align*}

The integration variables are \(10\) and \(y\) and, by definition, the domain of integration is

\begin{equation*} R=\Set{(10,y)}{(ten,y,z)\text{ is in }S\text{ for some }z} \stop{equation*}

To determine precisely what this domain of integration is, we observe that since \(z=y\) on \(S\text{,}\) \(x^2+y^2+z^2\le 4\) is the same as \(x^two+2y^2\le iv\) on \(Southward\text{,}\)

\begin{equation*} Southward=\Fix{(ten,y,z)}{x^2+2y^2\le 4,\ z=y} \implies R=\Set{(x,y)}{x^2+2y^ii\le four} \end{equation*}

And then the domain of integration is an ellipse with semimajor axis \(a=2\text{,}\) semiminor axis \(b=\sqrt{two}\) and surface area \(\pi a b=2\sqrt{two}\pi\text{.}\) The integral is then

\begin{equation*} \oint_C\vF\cdot \dee{\vr} =\dblInt_S \vnabla\times\vF\cdot\hn\,\dee{S} =\dblInt_R (-two)\,\dee{x}\dee{y} =-2\ \text{Area}\,(R) =-4\sqrt{2}\pi \terminate{equation*}

Remark (Limits of integration):

If the integrand were more complicated, we would accept to evaluate the integral over \(R\) past expressing it equally an iterated integrals with the correct limits of integration. Start suppose that nosotros slice upwardly \(R\) using thin vertical slices. On each such slice, \(ten\) is essentially abiding and \(y\) runs from \(-\sqrt{(4-x^ii)/2}\) to \(\sqrt{(4-x^2)/2}\text{.}\) The leftmost such slice would take \(x=-2\) and the rightmost such piece would have \(x=2\text{.}\) Then the right limits with this slicing are

\begin{equation*} \dblInt_R f(x,y)\,\dee{x}\dee{y} =\int_{-2}^2\dee{x}\int_{-\sqrt{(four-10^2)/2}}^{\sqrt{(4-x^2)/ii}} \dee{y}\ f(x,y) \cease{equation*}

If, instead, nosotros piece up \(R\) using thin horizontal slices, so, on each such slice, \(y\) is substantially constant and \(ten\) runs from \(-\sqrt{iv-2y^two}\) to \(\sqrt{4-2y^2}\text{.}\) The bottom such slice would have \(y=-\sqrt{2}\) and the elevation such slice would have \(y=\sqrt{2}\text{.}\) And so the right limits with this slicing are

\begin{equation*} \dblInt_R f(x,y)\,\dee{x}\dee{y} =\int_{-\sqrt{ii}}^{\sqrt{ii}}\dee{y}\int_{-\sqrt{4-2y^2}}^{\sqrt{4-2y^2}} \dee{x}\ f(x,y) \end{equation*}

Note that the integral with limits

\brainstorm{equation*} \int_{-\sqrt{2}}^{\sqrt{2}}\dee{y}\int_{-2}^{2} \dee{x}\ f(x,y) \terminate{equation*}

corresponds to a slicing with \(x\) running from \(-2\) to \(2\) on {\bf every} slice. This corresponds to a rectangular domain of integration, not what we have hither.

Stokes' Theorem, Once again:

Since the integrand is just a constant (after Stoking — not the original integrand) and \(S\) is so elementary (considering we chose it wisely), we tin evaluate the integral \(\dblInt_S \vnabla\times\vF\cdot\hn\,\dee{South}\) without ever determining \(\dee{S}\) explicitly and without ever setting upwardly any limits of integration. Nosotros already know that \(\vnabla\times\vF=ii\,\hj\text{.}\) Since \(Due south\) is the level surface \(z-y=0\text{,}\) the slope \(\vnabla(z-y)=-\hj+\hk\) is normal to \(Due south\text{.}\) So \(\hn = \frac{one}{\sqrt{2}}(-\hj+\hk)\) and

\brainstorm{align*} \oint_C\vF\cdot \dee{\vr} &=\dblInt_S \vnabla\times\vF\cdot\hn\,\dee{Southward} =\dblInt_S (2\hj)\cdot \frac{1}{\sqrt{ii}}(-\hj+\hk)\,\dee{S}\\ &=\dblInt_S -\sqrt{2}\,\dee{S} =-\sqrt{two}\ {\rm Area}\,(S) \end{align*}

As \(Southward\) is a circle of radius \(ii\text{,}\) \(\oint_C\vF\cdot \dee{\vr}=-4\sqrt{2}\pi\text{,}\) nonetheless again.

Case iv.4.viii .

In Warning iv.1.17, nosotros stated that if a vector field fails to pass the screening examination \(\vnabla\cdot\vB=0\) at even a single signal, for instance because the vector field is non defined at that point, so \(\vB\) tin fail to accept a vector potential. An instance is the point source

\brainstorm{equation*} \vB(ten,y,z) = \frac{\hat\vr(x,y,z)}{r(x,y,z)^2} \end{equation*}

of Example 3.4.2. Hither, every bit usual,

\begin{equation*} r(x,y,z) = \sqrt{x^2+y^2+z^2}\qquad \hat\vr(x,y,z) = \frac{x\hullo + y\hj + z\hk}{\sqrt{x^two+y^ii+z^2}} \cease{equation*}

This vector field is divers on all of \(\bbbr^3\text{,}\) except for the origin, and its divergence

\begin{align*} \vnabla\cdot\vB &=\frac{\partial }{\fractional x} \left(\frac{x}{(x^ii+y^ii+z^ii)^{3/ii}}\right) +\frac{\partial }{\partial y} \left(\frac{y}{(x^2+y^2+z^2)^{3/2}}\right)\\ &\hskip1in +\frac{\fractional }{\partial z} \left(\frac{z}{(ten^2+y^2+z^2)^{3/2}}\correct)\\ &=\left(\frac{1}{(x^2+y^2+z^two)^{iii/2}}-\frac{3x^ii}{(x^ii+y^2+z^two)^{5/2}}\correct)\\ &\hskip1in +\left(\frac{1}{(ten^2+y^two+z^2)^{3/2}}-\frac{3y^ii}{(ten^ii+y^two+z^ii)^{v/2}}\right)\\ &\hskip1in +\left(\frac{1}{(x^2+y^2+z^ii)^{3/2}}-\frac{3z^2}{(x^ii+y^ii+z^2)^{5/two}}\right)\\ &=\frac{3}{(x^2+y^2+z^two)^{three/2}}-\frac{three(x^2+y^2+z^ii)}{(x^two+y^2+z^ii)^{v/two}} \end{align*}

is zero everywhere except at the origin, where it is not defined.

This vector field cannot take a vector potential on its domain of definition, i.e. on \(\bbbr^3\setminus\{(0,0,0)\} =\Gear up{(ten,y,z)}{(x,y,z)\ne(0,0,0)} \text{.}\) To see this, suppose to the contrary that information technology did take a vector potential \(\vA\text{.}\) So its flux through whatsoever airtight surface^v (i.e. surface without a boundary) \(S\) would be

\brainstorm{gather*} \dblInt_S\vB\cdot\hn\,\dee{S} = \dblInt_S\vnabla\times\vA\cdot\hn\,\dee{South} =\oint_{\partial S}\vA\cdot\dee{\vr} =0 \terminate{assemble*}

by Stokes' theorem, since \(\partial S\) is empty. Just we found in Example three.iv.2, with \(grand=i\text{,}\) that the flux of \(\vB\) through any sphere centred on the origin is \(4\pi\text{.}\)

If you lot are uncomfortable with the surface not having a boundary, poke a very small hole in the surface, giving it a very small purlieus. Then take the limit every bit the pigsty tends to zero.

Subsection 4.4.1 The Estimation of Div and Curl Revisited

In sections four.1.4 and 4.1.5 nosotros derived interpretations of the divergence and of the gyre. Now that we have the divergence theorem and Stokes' theorem, we can simplify those derivations a lot.

Subsubsection 4.4.1.1 Divergence

Let \(\veps \gt 0\) be a tiny positive number, and then permit

\begin{equation*} B_\veps(x_0,y_0,z_0) =\Set{(ten,y,z)}{(x-x_0)^2+(y-y_0)^ii+(z-z_0)^2 \lt \veps^2} \end{equation*}

be a tiny brawl of radius \(\veps\) centred on the betoken \((x_0,y_0,z_0)\text{.}\) Announce by

\begin{equation*} S_\veps(x_0,y_0,z_0) =\Prepare{(x,y,z)}{(x-x_0)^2+(y-y_0)^2+(z-z_0)^2=\veps^2} \terminate{equation*}

its surface. Because \(B_\veps(x_0,y_0,z_0)\) is actually small, \(\vnabla\cdot \vv\) is substantially abiding in \(B_\veps(x_0,y_0,z_0)\) and we essentially have

\begin{equation*} \tripInt_{B_\veps(x_0,y_0,z_0)} \vnabla\cdot \vv\ \dee{V} =\vnabla\cdot \vv(x_0,y_0,z_0)\ \text{Vol}\big(B_\veps(x_0,y_0,z_0)\big) \terminate{equation*}

Of class we are really making an approximation here, based on the assumption that \(\vv(x,y,z)\) is continuous and so takes values very close to \(\vv(x_0,y_0,z_0)\) everywhere on the domain of integration. The approximation gets amend and better as \(\veps\rightarrow 0\) and a more precise statement is

\begin{equation*} \vnabla\cdot\vv(x_0,y_0,z_0)=\lim_{\veps\rightarrow 0} \frac{\tripInt_{B_\veps(x_0,y_0,z_0)} \vnabla\cdot \vv\ \dee{V}} {\text{Vol}\large(B_\veps(x_0,y_0,z_0)\big)} \stop{equation*}

By the difference theorem, we also have

\begin{equation*} \tripInt_{B_\veps(x_0,y_0,z_0)} \vnabla\cdot \vv\ \dee{V} =\dblInt_{S_\veps(x_0,y_0,z_0)} \vv\cdot\hn \ \dee{South} \stop{equation*}

Call back of the vector field \(\vv\) as the velocity of a moving fluid which has density one. We take already seen, in §three.four, that the flux integral for a velocity field has the interpretation

\begin{align*} \dblInt_{S_\veps(x_0,y_0,z_0)}\hskip-10 pt \vv\cdot\hn \ \dee{S} &=\left\{ \begin{array}{l} \text{the volume of fluid leaving $B_\veps(x_0,y_0,z_0)$}\\ \text{through $S_\veps(x_0,y_0,z_0)$ per unit time} \end{array} \right. \end{align*}

We conclude that, as we said in four.1.xix,

\begin{marshal*} \vnabla\cdot\vv(x_0,y_0,z_0) &=\lim_{\veps\rightarrow 0} \frac{\text{the rate at which fluid is exiting $B_\veps(x_0,y_0,z_0)$}} {\text{Vol}\big(B_\veps(x_0,y_0,z_0)\big)}\\ &=\left\{ \begin{assortment}{50} \text{charge per unit at which fluid is exiting an }\\ \text{infinitesimal sphere centred at $(x_0,y_0,z_0)$, }\\ \text{per unit time, per unit volume} \end{array} \correct.\\ &=\text{forcefulness of the source at $(x_0,y_0,z_0)$} \end{align*}

If our earth is filled with an incompressible fluid, a fluid whose density is constant and so never expands or compresses, nosotros volition have \(\vnabla\cdot\vv=0\text{.}\)

Subsubsection four.4.1.2 Whorl

Again allow \(\veps \gt 0\) be a tiny positive number and let \(D_\veps(x_0,y_0,z_0)\) be a tiny flat circular disk of radius \(\veps\) centred on the betoken \((x_0,y_0,z_0)\) and denote by \(C_\veps(x_0,y_0,z_0)\) its boundary circle. Let \(\hn\) be a unit normal vector to \(D_\veps\text{.}\) It tells us the orientation of \(D_\veps\text{.}\) Requite the circumvolve \(C_\veps\) the corresponding orientation using the right hand rule. That is, if the fingers of your right hand are pointing in the corresponding direction of motion along \(C_\veps\) and your palm is facing \(D_\veps\text{,}\) then your thumb is pointing in the direction \(\hn\text{.}\)

Because \(D_\veps(x_0,y_0,z_0)\) is really small, \(\vnabla\!\times\vv\) is essentially constant on \(D_\veps(x_0,y_0,z_0)\) and we essentially accept

\begin{align*} \dblInt_{D_\veps(x_0,y_0,z_0)} \vnabla\!\times \vv\cdot\hn\ \dee{S} &=\vnabla\!\times \vv(x_0,y_0,z_0)\cdot\hn\ \text{Surface area}\large(D_\veps(x_0,y_0,z_0)\big)\\ &=\pi\veps^ii\ \vnabla\times\! \vv(x_0,y_0,z_0)\cdot\hn \finish{marshal*}

Again, this is really an guess argument which gets better and better as \(\veps\rightarrow 0\text{.}\) A more precise statement is

\begin{assemble*} \vnabla\times\! \vv(x_0,y_0,z_0)\cdot\hn =\lim_{\veps\rightarrow 0} \frac{\dblInt_{D_\veps(x_0,y_0,z_0)} \vnabla\!\times \vv\cdot\hn\ \dee{S}} {\pi\veps^2} \end{gather*}

By Stokes' theorem, we also take

\begin{gather*} \dblInt_{D_\veps(x_0,y_0,z_0)} \vnabla\!\times \vv\cdot\hn\ \dee{S} =\oint_{C_\veps(x_0,y_0,z_0)} \vv\cdot d\vr \end{assemble*}

Again, think of the vector field \(\vv\) as the velocity of a moving fluid. Then \(\oint_{C_\veps} \vv\cdot d\vr\) is chosen the apportionment of \(\vv\) around \(C_\veps\text{.}\)

To measure the apportionment experimentally, place a small paddle wheel in the fluid, with the axle of the paddle wheel pointing along \(\hn\) and each of the paddles perpendicular to \(C_\veps\) and centred on \(C_\veps\text{.}\)

Each paddle moves tangentially to \(C_\veps\text{.}\) It would similar to move with the same speed as the tangential speed \(\vv\cdot\hvt\) (where \(\hvt\) is the forrad pointing unit tangent vector to \(C_\veps\) at the location of the paddle) of the fluid at its location. Just all paddles are rigidly fixed to the beam of the paddle wheel and so must all move with the same speed. That common speed will be the boilerplate value of \(\vv\cdot\hvt\) effectually \(C_\veps\text{.}\) If \(\dee{s}\) represents an element of arc length of \(C_\veps\text{,}\) the boilerplate value of \(\vv\cdot\hvt\) around \(C_\veps\) is

\begin{equation*} \overline{v_T}=\frac{1}{2\pi\veps}\oint_{C_\veps} \vv\cdot \hvt\ \dee{due south} =\frac{1}{2\pi\veps}\oint_{C_\veps} \vv\cdot \dee{\vr} \end{equation*}

since \(\dee{\vr}\) has direction \(\hvt\) and length \(\dee{s}\) so that \(\dee{\vr} =\hvt \dee{s}\text{,}\) and since \(2\pi\veps\) is the circumference of \(C_\veps\text{.}\) If the paddle wheel rotates at \(\Om\) radians per unit time, each paddle travels a distance \(\Om\veps\) per unit time (recall that \(\veps\) is the radius of \(C_\veps\)). That is, \(\overline{v_T}=\Om\veps\text{.}\) Combining all this information,

\begin{align*} \vnabla\times\! \vv(x_0,y_0,z_0)\cdot\hn &=\lim_{\veps\rightarrow 0} \frac{\dblInt_{D_\veps(x_0,y_0,z_0)} \vnabla\!\times \vv\cdot\hn\ \dee{Southward}} {\pi\veps^2}\\ &=\lim_{\veps\rightarrow 0} \frac{\oint_{C_\veps} \vv\cdot \dee{\vr}} {\pi\veps^2}\\ &=\lim_{\veps\rightarrow 0} \frac{two\pi\veps\ \overline{v_T}} {\pi\veps^two}\\ &=\lim_{\veps\rightarrow 0} \frac{2\pi\veps\ (\Omega \veps)} {\pi\veps^2}\\ &=ii\Om \finish{marshal*}

and so that

\begin{equation*} \Om = \half \vnabla\times\! \vv(x_0,y_0,z_0)\cdot\hn \terminate{equation*}

The component of \(\vnabla\times\!\vv(x_0,y_0,z_0)\) in any management \(\hn\) is twice the rate at which the paddle wheel turns when it is put into the fluid at \((x_0,y_0,z_0)\) with its axle pointing in the direction \(\hn\text{.}\) The management of \(\vnabla\times\!\vv(x_0,y_0,z_0)\) is the axle direction which gives maximum rate of rotation and the magnitude of \(\vnabla\times\!\vv(x_0,y_0,z_0)\) is twice that maximum rate of rotation. For this reason, \(\vnabla\times \vv\) is called the "vorticity".

Subsection 4.4.2 Optional — An Awarding of Stokes' Theorem — Faraday'south Law

Magnetic induction refers to a physical process whereby an electrical voltage is created ("induced") past a fourth dimension varying magnetic field. This process is exploited in many applications, including electric generators, induction motors, consecration cooking, induction welding and anterior charging. Michael Faraday⁶ is generally credited with the discovery of magnetic consecration. Faraday's law is the post-obit. Let \(S\) be an oriented surface with purlieus \(C\text{.}\) Permit \(\vE\) and \(\vB\) exist the (time dependent) electrical and magnetic fields and define

Michael Faraday (1791–1867) was an English physicist and chemist. He ended upwardly beingness an extremely influential scientist despite having only the nearly basic of formal educations.

\begin{align*} \oint_C\vE\cdot \dee{\vr}&=\text{voltage around }C\\ \dblInt_S\vB\cdot \hn \,\dee{S}&=\text{magnetic flux through }S \finish{align*}

Then the voltage around \(C\) is the negative of the rate of modify of the magnetic flux through \(S\text{.}\) Every bit an equation, Faraday's Law is

\begin{equation*} \oint_C\vE\cdot \dee{\vr}=-\frac{\fractional }{\partial t}\dblInt_S\vB\cdot\hn\,\dee{S} \end{equation*}

We can reformulate this as a fractional differential equation. By Stokes' Theorem,

\begin{equation*} \oint_C\vE\cdot \dee{\vr}=\dblInt_S(\vnabla\times\vE)\cdot \hn\,\dee{Due south} \finish{equation*}

so Faraday'due south law becomes

\begin{equation*} \dblInt_S\Big(\vnabla\times\vE+\frac{\fractional\vB}{\fractional t}\Big)\cdot\hn\,\dee{S}=0 \stop{equation*}

This is true for all surfaces \(S\text{.}\) And so the integrand, assuming that it is continuous, must be zero.

To see this, let \(\vG=\Big(\vnabla\times\vE+\frac{\partial\vB}{\partial t}\Big)\text{.}\) Suppose that \(\vG(\vx_0)\ne 0\text{.}\) Pick a unit vector \(\hn\) in the direction of \(\vG(\vx_0)\text{.}\) Allow \(S\) be a very small flat disk centered on \(\vx_0\) with normal \(\hn\) (the vector we picked). Then \(\vG(\vx_0)\cdot\hn \gt 0\) and, by continuity, \(\vG(\vx)\cdot\hn \gt 0\) for all \(\vx\) on \(Southward\text{,}\) if we accept picked \(S\) small enough. Then \(\dblInt_S\Big(\vnabla\times\vE+\frac{\fractional\vB}{\partial t}\Big)\cdot \hn\,\dee{S} \gt 0\text{,}\) which is a contradiction. So \(\vG=\vZero\) everywhere and we conclude that

\begin{equation*} \vnabla\times\vE+\frac{\fractional\vB}{\fractional t}=0 \end{equation*}

This is one of Maxwell'due south electromagnetic field equations^vii.

For the others, see Example 4.1.ii

Exercises four.4.3 Exercises

Exercises — Stage 1

1.

Each of the figures beneath contains a sketch of a surface \(Due south\) and its boundary \(\partial Southward\text{.}\) Stokes' theorem says that \(\oint_{\partial South}\vF\cdot \dee{\vr} =\dblInt_S\vnabla\times\vF\cdot\hn\,\dee{South}\) if \(\hn\) is a correctly oriented unit of measurement normal vector to \(S\text{.}\) Add to each sketch a typical such normal vector.

2.

Let

• \(R\) be a finite region in the \(xy\)-airplane,

• the purlieus, \(C\text{,}\) of \(R\) consist of a single piecewise shine, uncomplicated airtight bend

  • that is oriented (i.due east. an arrow is put on \(C\)) consistently with \(R\) in the sense that if yous walk along \(C\) in the direction of the pointer, then \(R\) is on your left

  

• \(F_1(x,y)\) and \(F_2(10,y)\) have continuous beginning partial derivatives at every bespeak of \(R\text{.}\)

Apply Stokes' theorem to show that

\begin{equation*} \oint_{C} \big[F_1(10,y)\,\dee{10} +F_2(x,y)\,\dee{y}\big] =\dblInt_{R}\Big(\frac{\partial F_2}{\fractional 10} - \frac{\partial F_1}{\partial y}\Big)\ \dee{x}\dee{y} \cease{equation*}

i.e. to show Green'south theorem.

3.

Verify the identity \(\ \oint_C\phi\vnabla\psi\cdot \dee{\vr}=-\oint_C\psi\vnabla\phi\cdot \dee{\vr}\ \) for any continuously differentiable scalar fields \(\phi\) and \(\psi\) and curve \(C\) that is the boundary of a piecewise polish surface.

Exercises — Phase 2

4.

Permit \(C\) be the curve of intersection of the cylinder \(x^two+y^two=one\) and the surface \(z=y^2\) oriented in the counterclockwise direction equally seen from \((0,0,100)\text{.}\) Let \(\vF=(x^2-y\,,\,y^2+x\,,\,one)\text{.}\) Calculate \(\oint_C\vF\cdot \dee{\vr}\)

1. by directly evaluation
2. by using Stokes' Theorem.

5.

Evaluate \(\oint_C \vF\cdot \dee{\vr}\) where \(\vF=ye^x\,\howdy+(10+east^x)\,\hj+z^2\,\hk\) and \(C\) is the curve

\begin{equation*} \vr(t)=(1+\cos t)\,\how-do-you-do+(one+\sin t)\,\hj+(1-\sin t-\cos t)\,\hk\qquad 0\le t\le 2\pi \stop{equation*}

6. ✳.

Notice the value of \(\dblInt_S\vnabla\times\vF\cdot\hn\,\dee{Due south}\) where \(\vF = \large(z - y\,,\, x\,,\, -x\big)\) and \(S\) is the hemisphere

\begin{equation*} \Fix{(x, y, z) \in\bbbr^3 }{ 10^2 + y^2 + z^2 = four,\ z \ge 0} \end{equation*}

oriented and then the surface normals signal away from the centre of the hemisphere.

7. ✳.

Permit \(\cS\) be the part of the surface \(z=xvi-{(10^2+y^2)}^two\) which lies above the \(xy\)-plane. Let \(\vF\) be the vector field

\begin{equation*} \vF=x\ln(1+z)\,\hi+10(iii+y)\,\hj+y\cos z\,\hk \end{equation*}

Calculate

\begin{equation*} \dblInt_{\cS}\vnabla\times\vF\cdot\hn\,\dee{S} \stop{equation*}

where \(\hn\) is the upward normal on \(\cS\text{.}\)

8. ✳.

Let \(\cC\) exist the intersection of the paraboloid \(z=iv-x^2-y^2\) with the cylinder \(x^2+(y-1)^2=1\text{,}\) oriented counterclockwise when viewed from high on the \(z\)-axis. Let \(\vF=xz\,\hullo+x\,\hj+yz\,\hk\text{.}\) Find \(\oint_{\cC}\vF\cdot \dee{\vr}\text{.}\)

9.

Permit \(\vF = - ye^z\,\how-do-you-do + x^three\cos z\,\hj + z\sin(xy)\,\hk\text{,}\) and let \(S\) be the role of the surface \(z = (1-10^two)(1-y^2)\) that lies above the square \(-i\le x\le one\text{,}\) \(-ane\le y\le i\) in the \(xy\)-plane. Find the flux of \(\vnabla\times \vF\) upward through \(Due south\text{.}\)

10.

Evaluate the integral \(\oint_C \vF\cdot \dee{\vr}\text{,}\) in which \(\vF = (east^{ten^ii} - yz\,,\,\sin y - yz \,,\,xz + 2y)\) and \(C\) is the triangular path from \((1, 0, 0)\) to \((0, one, 0)\) to \((0, 0, 1)\) to \((ane, 0, 0)\text{.}\)

xi. ✳.

Let \(\vF(x,y,z)=-z\,\how-do-you-do+x\,\hj+y\,\hk\) exist a vector field. Use Stokes' theorem to evaluate the line integral \(\oint_C\vF\cdot \dee{\vr}\) where \(C\) is the intersection of the plane \(z=y\) and the ellipsoid \(\frac{x^2}{4}+\frac{y^2}{ii}+\frac{z^two}{2}=1\text{,}\) oriented counter-clockwise when viewed from high on the \(z\)-axis.

12. ✳.

Consider the vector field \(\vF(x,y, z) = z^2 \,\hi + ten^two \,\hj + y^2\,\hk\) in \(\bbbr^three\text{.}\)

1. Compute the line integral \(I_1 = \int_{C_1}\vF\cdot\dee{\vr}\) where \(C_1\) is the curve consisting of three line segments, \(L_1\) from \((ii, 0, 0)\) to \((0, 2, 0)\text{,}\) then \(L_2\) from \((0, two, 0)\) to \((0, 0, 2)\text{,}\) finally \(L_3\) from \((0, 0, 2)\) to \((2, 0, 0)\text{.}\)
2. A unproblematic closed curve \(C_2\) lies on the plane \(E\,:\, 10 + y + z = 2\text{,}\) enclosing a region \(R\) on the airplane of area \(3\text{,}\) and oriented in a counterclockwise direction every bit observed from the positive \(ten\)-axis. Compute the line integral \(I_2 = \int_{C_2}\vF\cdot\dee{\vr}\text{.}\)

13. ✳.

Let \(C = C_1 + C_2 + C_3\) be the curve given past the wedlock of the three parameterized curves

\begin{alignat*}{2} \vr_1(t) &= \big(2\cos t, 2\sin t, 0\big), &\qquad &0 \le t \le \pi/ii\\ \vr_2(t) &= \large(0, 2\cos t, 2\sin t\big), & &0 \le t \le \pi/2\\ \vr_3(t) &= \large(2\sin t, 0, two\cos t\large), & &0 \le t \le \pi/ii \stop{alignat*}

1. Draw a motion picture of \(C\text{.}\) Clearly marker each of the curves \(C_1\text{,}\) \(C_2\text{,}\) and \(C_3\) and bespeak the orientations given by the parameterizations.
2. Find and parameterize an oriented surface \(S\) whose purlieus is \(C\) (with the given orientations).
3. Compute the line integral \(\int_C \vF\cdot\dee{\vr}\) where

   \begin{equation*} \vF = \Big(y + \sin(x^ii)\,,\, z - 3x + \ln(ane + y^ii)\,,\, y + eastward^{z^2}\Big) \end{equation*}

fourteen. ✳.

We consider the cone with equation \(z = \sqrt{10^2 + y^2}\text{.}\) Annotation that its tip, or vertex, is located at the origin \((0, 0, 0)\text{.}\) The cone is oriented in such a way that the normal vectors signal downwards (and away from the \(z\) centrality). In the parts below, both \(S_1\) and \(S_2\) are oriented this mode.

Allow \(\vF = \big(-zy, zx, xy \cos(yz)\big)\text{.}\)

1. Let \(S_1\) be the role of the cone that lies between the planes \(z = 0\) and \(z = 4\text{.}\) Note that \(S_1\) does not include any part of the plane \(z = 4\text{.}\) Employ Stokes' theorem to determine the value of

   \begin{equation*} \dblInt_{S_1} \vnabla\times\vF \cdot \hn\,\dee{South} \stop{equation*}

   Make a sketch indicating the orientations of \(S_1\) and of the contour(s) of integration.
2. Let \(S_2\) be the role of the cone that lies beneath the plane \(z = 4\) and above \(z = one\text{.}\) Notation that \(S_2\) does not include whatever part of the planes \(z = i\) and \(z = 4\text{.}\) Decide the flux of \(\vnabla\times\vF\) across \(S_2\text{.}\) Justify your answer, including a sketch indicating the orientations of \(S_2\) and of the contour(southward) of integration.

fifteen. ✳.

Consider the curve \(C\) that is the intersection of the plane \(z = x + iv\) and the cylinder \(x^2 + y^2 = iv\text{,}\) and suppose \(C\) is oriented so that it is traversed clockwise every bit seen from above.

Let \(\vF(ten, y, z) = \big(x^3 + 2y\,,\, \sin(y) + z\,,\, x + \sin(z^2)\big)\text{.}\)

Use Stokes' Theorem to evaluate the line integral \(\oint_C\vF\cdot\dee{\vr}\text{.}\)

xvi. ✳.

1. Consider the vector field \(\vF(x, y, z) = (z^ii , x^ii , y^ii)\) in \(\bbbr^iii\text{.}\) Compute the line integral \(\oint_C \vF\cdot\dee{\vr}\text{,}\) where \(C\) is the bend consisting of the iii line segments, \(L_1\) from \((2, 0, 0)\) to \((0, two, 0)\text{,}\) and then \(L_2\) from \((0, 2, 0)\) to \((0, 0, 2)\text{,}\) and finally \(L_3\) from \((0, 0, 2)\) to \((2, 0, 0)\text{.}\)
2. A elementary airtight curve \(C\) lies in the airplane \(ten + y + z = 2\text{.}\) The surface this bend \(C\) surrounds within the plane \(ten + y + z = 2\) has area \(3\text{.}\) The curve \(C\) is oriented in a counterclockwise management every bit observed from the positive x-axis. Compute the line integral \(\oint_C\vF\cdot \dee{\vr}\) , where \(\vF\) is as in (a).

17. ✳.

Evaluate the line integral

\begin{equation*} \int_C\left(z+\frac{1}{1+z}\right)\dee{x} +xz\,\dee{y} +\left(3xy-\frac{x}{(z+1)^2}\right)\dee{z} \terminate{equation*}

where \(C\) is the curve parameterized by

\begin{equation*} \vr(t) = \big(\cos t\,,\, \sin t\,,\, ane - \cos^2 t \sin t\big)\qquad 0 \le t \le 2\pi \stop{equation*}

18. ✳.

A simple closed bend \(C\) lies in the plane \(x + y + z = 1\text{.}\) The surface this curve \(C\) surrounds inside the plane \(ten + y + z = 1\) has area \(v\text{.}\) The bend \(C\) is oriented in a clockwise direction as observed from the positive \(z\)-axis looking down at the plane \(x + y + z = 1\text{.}\)

Compute the line integral of \(\vF (x, y, z) = (z^ii , x^ii , y^2)\) around \(C\text{.}\)

19. ✳.

Let \(C\) be the oriented curve consisting of the 5 line segments which class the paths from \((0, 0, 0)\) to \((0, one, 1)\text{,}\) from \((0, ane, 1)\) to \((0, 1, two)\text{,}\) from \((0, 1, two)\) to \((0, 2, 0)\text{,}\) from \((0, ii, 0)\) to \((ii, 2, 0)\text{,}\) and from \((2, ii, 0)\) to \((0, 0, 0)\text{.}\) Allow

\begin{equation*} \vF = (-y+due east^x\sin ten)\,\howdy +y^4\,\hj +\sqrt{z}\tan z\,\hk \end{equation*}

Evaluate the integral \(\int_C\vF\cdot\dee{\vr}\text{.}\)

xx. ✳.

Suppose the bend \(C\) is the intersection of the cylinder \(x^two + y^ii = 1\) with the surface \(z = xy^2\text{,}\) traversed clockwise if viewed from the positive z-axis, i.east. viewed "from higher up". Evaluate the line integral

\begin{equation*} \int_C (z + \sin z) \,\dee{x} + (x^3 - x^two y) \,\dee{y} + (x \cos z - y) \,\dee{z} \cease{equation*}

21. ✳.

Evaluate \(\dblInt_S \vnabla\times\vF\cdot\hn\,\dee{S}\) where \(S\) is that part of the sphere \(x^2+y^2+z^2=2\) above the aeroplane \(z=one\text{,}\) \(\hn\) is the up unit normal, and

\begin{gather*} \vF(x,y,z) = -y^2\,\hi +x^three\,\hj + \large(east^ten + e^y +z\large)\,\hk \end{gather*}

22. ✳.

Let

\brainstorm{equation*} \vF = x \sin y\,\hullo - y \sin x\,\hj + (10 - y)z^ii\,\hk \end{equation*}

Use Stokes' theorem to evaluate

\begin{equation*} \int_C\vF\cdot \dee{\vr} \end{equation*}

along the path consisting of the straight line segments successively joining the points \(P_0 = (0, 0, 0)\) to \(P_1 = (\pi/2, 0, 0)\) to \(P_2 = (\pi/two, 0, one)\) to \(P_3 = (0, 0, 1)\) to \(P_4 = (0, \pi/two, ane)\) to \(P_5 = (0, \pi/2, 0)\text{,}\) and dorsum to \((0, 0, 0)\text{.}\)

23. ✳.

Let

\begin{equation*} \vF=\left(\frac{2z}{1+y}+\sin(10^2)\,,\, \frac{3z}{1+x}+\sin(y^2)\,,\, v(x+one)(y+two)\right) \end{equation*}

Permit \(C\) be the oriented curve consisting of four line segments from \((0,0,0)\) to \((2,0,0)\text{,}\) from \((2,0,0)\) to \((0,0,2)\text{,}\) from \((0,0,ii)\) to \((0,3,0)\text{,}\) and from \((0,iii,0)\) to \((0,0,0)\text{.}\)

1. Describe a moving picture of \(C\text{.}\) Clearly signal the orientation on each line segment.
2. Compute the work integral \(\int_C\vF\cdot\dee{\vr}\text{.}\)

24. ✳.

Evaluate \(\displaystyle\dblInt_S\vnabla\times\vF\cdot\hn\,\dee{Due south}\) where \(\vF=y\,\hi+2z\,\hj+3x\,\hk\) and \(S\) is the surface \(z=\sqrt{1-x^2-y^2}\text{,}\) \(z\ge 0\) and \(\hn\) is a unit normal to \(S\) obeying \(\hn\cdot\hk\ge 0\text{.}\)

25. ✳.

Let \(\cS\) be the curved surface below, oriented by the outward normal:

\begin{equation*} x^ii + y^ii + 2(z-ane)^2 = six,\qquad z\ge 0. \end{equation*}

(E.grand., at the high betoken of the surface, the unit normal is \(\hk\text{.}\))

Define

\begin{equation*} \vG = \nabla\times\vF, \quad\text{where}\quad \vF = (xz - y^3\cos z)\,\hi + 10^3 e^z\,\hj + xyze^{10^2+y^ii+z^2}\,\hk. \cease{equation*}

Find \(\dblInt_\cS \vG\cdot \hn\dee{S}\text{.}\)

26. ✳.

Allow \(C\) be a circle of radius \(R\) lying in the plane \(10+y+z=3\text{.}\) Use Stokes' Theorem to summate the value of

\begin{equation*} \oint_C \vF\cdot \dee{\vr} \end{equation*}

where \(\vF=z^2\how-do-you-do+ten^ii\hj+y^2\hk\text{.}\) (You may utilise either orientation of the circle.)

27.

Let \(S\) exist the oriented surface consisting of the top and 4 sides of the cube whose vertices are \((\pm 1,\pm1,\pm1)\text{,}\) oriented outward. If \(\vF(ten,y,z)=(xyz,xy^two,x^2yz)\text{,}\) discover the flux of \(\vnabla\times\vF\) through \(S\text{.}\)

28.

Permit \(S\) denote the function of the spiral ramp (that is helicoidal surface) parametrized past

\brainstorm{equation*} 10=u\cos 5,\ y=u\sin v,\ z=five\qquad 0\le u\le ane,\ 0\le v\le 2\pi \end{equation*}

Let \(C\) announce the boundary of \(South\) with orientation specified by the upward pointing normal on \(Southward\text{.}\) Detect

\brainstorm{equation*} \int_C y\,\dee{x}-x\,\dee{y}+ xy\,\dee{z} \cease{equation*}

Exercises — Stage 3

29.

Let \(C\) be the intersection of \(ten+2y-z=7\) and \(x^2-2x+4y^2=15\text{.}\) The curve \(C\) is oriented counterclockwise when viewed from high on the \(z\)-axis. Permit

\begin{equation*} \vF=\big(due east^{10^2}+yz\big)\,\hi +\big(\cos(y^2)-x^2\large)\,\hj +\big(\sin(z^2)+xy\big)\,\hk \end{equation*}

Evaluate \(\oint_C\vF\cdot \dee{\vr}\text{.}\)

30. ✳.

1. Notice the curl of the vector field \(\vF = \big(2 + ten^2 + z\,,\, 0\,,\, 3 + x^2 z\large)\text{.}\)
2. Let \(C\) exist the curve in \(\bbbr^3\) from the signal \((0, 0, 0)\) to the point \((2, 0, 0)\text{,}\) consisting of 3 sequent line segments connecting the points \((0, 0, 0)\) to \((0, 0, iii)\text{,}\) \((0, 0, three)\) to \((0, 1, 0)\text{,}\) and \((0, 1, 0)\) to \((ii, 0, 0)\text{.}\) Evaluate the line integral

   \begin{equation*} \int_C \vF\cdot\dee{\vr} \cease{equation*}

   where \(\vF\) is the vector field from (a).

31. ✳.

1. Let \(Southward\) be the bucket shaped surface consisting of the cylindrical surface \(y^2 + z^2 = 9\) between \(x = 0\) and \(x = v\text{,}\) and the disc inside the \(yz\)-airplane of radius \(3\) centered at the origin. (The bucket \(S\) has a lesser, but no lid.) Orient \(S\) in such a fashion that the unit normal points outward. Compute the flux of the vector field \(\vnabla\times\vG\) through \(S\text{,}\) where \(\vG = (x, -z, y)\text{.}\)
2. Compute the flux of the vector field \(\vF = (2 + z, xz^2 , x \cos y)\) through \(South\text{,}\) where \(S\) is as in (a).

32. ✳.

Let

\begin{equation*} \vF(x, y, z) = \Big(\frac{y}{ten} +x^{one+ten^2}\,,\, x^2-y^{ane+y^2}\,,\, \cos^5(\ln z)\Big) \end{equation*}

1. Write downwards the domain \(D\) of \(\vF\text{.}\)

2. Circle the correct argument(s):

   1. D is continued.
   2. D is but connected.
   3. D is disconnected.
3. Compute \(\vnabla\times\vF\text{.}\)
4. Let \(C\) be the square with corners \((3 \pm ane, 3 \pm 1)\) in the aeroplane \(z = 2\text{,}\) oriented clockwise (viewed from in a higher place, i.east. down \(z\)-axis). Compute

   \begin{equation*} \int_C \vF \cdot \dee{\vr} \end{equation*}

5. Is \(\vF\) conservative?

33. ✳.

A physicist studies a vector field \(\vF(x, y, z)\text{.}\) From experiments, it is known that \(\vF\) is of the grade

\begin{equation*} \vF(10, y, z) = xz\,\hi + (axe^y z + byz)\,\hj + (y^two - xe^y z^2 )\,\hk \end{equation*}

for some real numbers \(a\) and \(b\text{.}\) It is further known that \(\vF = \vnabla\times\vG\) for some differentiable vector field \(\vG\text{.}\)

1. Determine \(a\) and \(b\text{.}\)
2. Evaluate the surface integral

   \begin{equation*} \dblInt_S\vF\cdot\hn\,\dee{S} \end{equation*}

   where \(S\) is the function of the ellipsoid \(x^2 + y^two + \frac{1}{four} z^2 = one\) for which \(z \ge 0\text{,}\) oriented and then that its normal vector has a positive \(z\)-component.

34. ✳.

Let \(C\) exist the bend in the \(xy\)-plane from the point \((0, 0)\) to the betoken \((5, 5)\) consisting of the ten line segments consecutively connecting the points \((0,0)\text{,}\) \((0,1)\text{,}\) \((1,1)\text{,}\) \((ane,ii)\text{,}\) \((2,two)\text{,}\) \((2,3)\text{,}\) \((3,3)\text{,}\) \((3,iv)\text{,}\) \((iv,4)\text{,}\) \((4,5)\text{,}\) \((5,5)\text{.}\) Evaluate the line integral

\brainstorm{assemble*} \int_C \vF \cdot\dee{\vr} \terminate{get together*}

where

\begin{gather*} \vF = y\,\hi + (2x - 10)\,\hj \end{gather*}

35. ✳.

Let \(\vF = \large(\sin x^2\,,\,xz\,,\,z^ii\big)\text{.}\) Evaluate \(\oint_C\vF\cdot\dee{\vr}\) around the curve \(C\) of intersection of the cylinder \(x^2+y^2=4\) with the surface \(z=10^two\text{,}\) traversed counter clockwise as viewed from loftier on the \(z\)-axis.

36. ✳.

Explain how ane deduces the differential course

\brainstorm{equation*} \vnabla\times\vE = -\frac{1}{c}\frac{\partial\vH}{\fractional t} \finish{equation*}

of Faraday's police from its integral form

\begin{equation*} \oint_C\vE\cdot\dee{\vr} = -\frac{i}{c}\ \unequal{ }{t}\dblInt_S\vH\cdot\hn\,\dee{S} \end{equation*}

37. ✳.

Let \(C\) exist the curve given by the parametric equations:

\brainstorm{equation*} x=\cos t,\ y=\sqrt{two}\sin t,\ z=\cos t,\ 0\le t\le 2\pi \terminate{equation*}

and let

\brainstorm{equation*} \vF=z\,\hullo+ten\,\hj+y^3z^three\,\hk \end{equation*}

Use Stokes' theorem to evaluate

\brainstorm{equation*} \oint_C \vF\cdot \dee{\vr} \cease{equation*}

38. ✳.

Use Stokes' theorem to evaluate

\begin{equation*} \oint_C z\,\dee{x}+x\,\dee{y}-y\,\dee{z} \end{equation*}

where \(C\) is the closed curve which is the intersection of the plane \(x+y+z=1\) with the sphere \(x^2+y^two+z^2=i\text{.}\) Assume that \(C\) is oriented clockwise equally viewed from the origin.

39. ✳.

Let \(Due south\) be the role of the half cone

\begin{equation*} z=\sqrt{x^2+y^2},\quad y\ge 0, \terminate{equation*}

that lies below the plane \(z=ane\text{.}\)

1. Find a parametrization for \(Due south\text{.}\)
2. Summate the flux of the velocity field

   \begin{equation*} \vv=x\,\howdy + y\, \hj -2 z\,\hk \end{equation*}

   downward through \(S\text{.}\)
3. A vector field \(\vF\) has roll \(\vnabla\times \vF= x\, \howdy + y\, \hj -2 z\,\hk\text{.}\) On the \(xz\)-airplane, the vector field \(\vF\) is constant with \(\vF(x,0,z)=\hj\text{.}\) Given this data, calculate

   \begin{equation*} \int_\cC \vF\cdot \dee{\vr}, \stop{equation*}

   where \(\cC\) is the half circle

   \brainstorm{equation*} x^2+y^2=1,\ z=1,\ y\ge 0 \end{equation*}

   oriented from \((-one,0,i)\) to \((1,0,1)\text{.}\)

forty.

Consider \(\dblInt_S(\vnabla\times\vF)\cdot\hn\,dS\) where \(S\) is the portion of the sphere \(ten^two+y^2+z^ii=1\) that obeys \(x+y+z\ge 1\text{,}\) \(\hn\) is the upwards pointing normal to the sphere and \(\vF=(y-z)\hi+(z-x)\hj+(x-y)\hk\text{.}\) Find another surface \(S'\) with the property that \(\dblInt_S(\vnabla\times\vF)\cdot\hn\,\dee{Southward} =\dblInt_{South'}(\vnabla\times\vF)\cdot\hn\,\dee{S}\) and evaluate \(\dblInt_{Southward'}(\vnabla\times\vF)\cdot\hn\,\dee{South}\text{.}\)

Source: https://personal.math.ubc.ca/~CLP/CLP4/clp_4_vc/sec_StokesThm.html

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