OBJECTIVES

When you finish this section, you should be able to:

1. Find the curl of \(\mathbf{F}\) (p. 1046)
2. Verify Stokes' Theorem (p. 1047)
3. Use Stokes' Theorem to find an integral (p. 1048)
4. Use Stokes' Theorem with conservative vector fields (p. 1049)
5. Interpret the curl of \(\mathbf{F}\) (p. 1050)

ORIGINS

Stokes' Theorem is named after the Irish mathematician George Gabriel Stokes (1819–1903). Stokes, along with Maxwell and Kelvin, contributed to the prominence of the Cambridge School of Mathematical Physics.

spanDEFINITIONspan Curl of \(\mathbf{F}\)

Let \(\mathbf{F}=\mathbf{F}(x,y,z)=P(x,y,z)\mathbf{i}+Q(x,y,z)\mathbf{j }+R(x,y,z)\mathbf{k}\) be a vector field, where the functions \(P\), \(Q\), and \(R\) have first-order partial derivatives. The \(\mathbf{curl}\, \mathbf{of}\, \mathbf{F}\), denoted by curl \(\mathbf{F}\), is defined as* \[ {\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{ \rm{curl}\, \mathbf{F}=\left( \dfrac{\partial R}{\partial y}- \dfrac{\partial Q}{\partial z}\right) \mathbf{i}-\left( \dfrac{\partial R}{ \partial x}-\dfrac{\partial P}{\partial z}\right) \,\mathbf{j}+\left( \dfrac{ \partial Q}{\partial x}-\dfrac{\partial P}{\partial y}\right) \mathbf{k}}}} \]

Finding the Curl of \(\mathbf{F}\)

Find curl \(\mathbf{F}\) if \(\mathbf{F}=x^{2}y\mathbf{i}-2xz\,\mathbf{j} +2yz\,\mathbf{k}\).

Solution  \[ \begin{eqnarray*} {\rm curl}\,\mathbf{F}& =&\left\vert \begin{array}{@{}c@{\quad}c@{\quad}c} \mathbf{i} & \mathbf{j} & \mathbf{k}\\ \dfrac{\partial }{\partial x} & \dfrac{\partial }{\partial y} & \dfrac{ \partial }{\partial z}\\ x^{2}y & -2xz & 2yz \end{array} \right\vert \\ & =&\left[ \dfrac{\partial }{\partial y}(2yz)-\dfrac{\partial }{\partial z} (-2xz)\right] \,\mathbf{i}-\left[ \dfrac{\partial }{\partial x}(2yz)-\dfrac{ \partial }{\partial z}(x^{2}y)\right] \,\mathbf{j}+\left[ \dfrac{\partial }{ \partial x}(-2xz)-\dfrac{\partial }{\partial y}(x^{2}y)\right] \mathbf{k} \\ & =&(2z+2x)\,\mathbf{i}-(0-0)\,\mathbf{j}+(-2z-x^{2})\mathbf{k}=(2z+2x)\, \mathbf{i}+(-2z-x^{2})\,\mathbf{k} \end{eqnarray*} \]

NOW WORK

Problem 11.

THEOREM Stokes' Theorem

Let \(S\) be a smooth, simply connected, orientable surface bounded by a piecewise-smooth, simple closed curve \(C\). Let \(\mathbf{F}=\mathbf{F} (x,y,z)=P(x,y,z)\mathbf{i}+Q(x,y,z)\mathbf{j}+R(x,y,z)\mathbf{k}\) be a vector field, where \(P\), \(Q\), and \(R\) have continuous first-order partial derivatives throughout a solid \(E\) containing \(S\) and \(C\). Let \(\mathbf{n}\) denote the positive unit normal to \(S\), and let \(C\) be positively oriented, as described earlier. Then \[ {\bbox[#FAF8ED,5pt]{\bbox[5px, border:1px solid black, #F9F7ED]{ \oint_{C}\mathbf{F}\,{\bf\cdot}\, d\mathbf{r}=\iint\limits_{\kern-3ptS} {\rm{curl}\, \mathbf{F}\,{\bf\cdot}\, \mathbf{n}}\,dS}}} \]

In terms of the components of each vector, Stokes' Theorem takes the form \[ {\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{ \oint\limits_{C}(P\,dx+Q\,dy+R\,dz)=\iint\limits_{\kern-3ptS}\left[ \left( \dfrac{ \partial R}{\partial y}-\dfrac{\partial Q}{\partial z}\right) dy\,dz-\left( \dfrac{\partial R}{\partial x}-\dfrac{\partial P}{\partial z} \right) \,dz\,dx+\left( \dfrac{\partial Q}{\partial x}-\dfrac{\partial P}{ \partial y}\right) dx\,dy\right]}}} \]

The proof of Stokes' Theorem is given in advanced calculus books.

Verifying Stokes' Theorem

Verify Stokes' Theorem for \(\mathbf{F}=y\mathbf{i}-x\mathbf{j}\), where the surface \(S\) is the paraboloid \(z=x^{2}+y^{2}\), with \(x^{2}+y^{2}=1\), \(z=1 \), as its boundary \(C\).

Figure 72 \(C: x^{2} + y^{2} = 1; z =1\)

Solution  Figure 72 shows the paraboloid and the circle \(C\) in the plane \(z=1\). Stokes' Theorem states the following: \[ \oint\limits_{C}\mathbf{F}\,{\bf\cdot}\, d\mathbf{r}=\iint\limits_{\kern-3ptS}{\rm{curl}\, \mathbf{F} \,{\bf\cdot}\, \mathbf{n}}\,dS \]

Parametric equations for \(C\) are \(x=\cos t\), \(y=\sin t\), \(z=1,\) \( 0\leq t\leq 2\pi\). We use the parametric equations to find the line integral \(\oint_{C}\mathbf{F}\,{\bf\cdot}\, d\mathbf{r}\) for \(\mathbf{F}=y\mathbf{i}-x \mathbf{j}\). \[ \begin{eqnarray*} \oint_{C}\mathbf{F}\,{\bf\cdot}\, d\mathbf{r}&=&\oint_{C}(y\,dx-x\,dy)=\int_{0}^{2\pi } \left[ \sin t\left( -\sin t\,dt\right) -\cos t\cos t\,dt\right]\\[4pt] &=&-\int_{0}^{2\pi }[\sin ^{2}t+\cos ^{2}t]\,dt=-\int_{0}^{2\pi }dt=-2\pi \end{eqnarray*} \]

To find the surface integral, \(\iint\limits_{\kern-3ptS}{\rm{curl}\, \mathbf{F}\,{\bf\cdot}\, \mathbf{n}}\,dS\), we find \(\rm{curl}\, \mathbf{F}\) and \(\mathbf{n}\): \[ \begin{eqnarray*} \rm{curl}\, \mathbf{F} &\mathbf{=}&\left\vert \begin{array}{@{}c@{\quad}c@{\quad}c} \mathbf{i} & \mathbf{j} & \mathbf{k} \\[3pt] \dfrac{\partial }{\partial x} & \dfrac{\partial }{\partial y} & \dfrac{\partial }{\partial z} \\[5pt] P & Q & R \end{array} \right\vert =\left\vert \begin{array}{@{}c@{\quad}c@{\quad}c} \mathbf{i} & \mathbf{j} & \mathbf{k} \\[3pt] \dfrac{\partial }{\partial x} & \dfrac{\partial }{\partial y} & \dfrac{\partial }{\partial z} \\[5pt] y & -x & 0 \end{array} \right\vert \\[4pt] &=&\left[ \dfrac{\partial }{\partial y}0-\dfrac{\partial }{\partial z}(-x) \right] \,\mathbf{i}-\left[ \dfrac{\partial }{\partial x}0-\dfrac{\partial }{ \partial z}y\right] \,\mathbf{j}+\left[ \dfrac{\partial }{\partial x}(-x)- \dfrac{\partial }{\partial y}y\right] \mathbf{k}\\[4pt] &=&0\mathbf{i}-0\mathbf{j}-2\mathbf{k}=-2\mathbf{k} \end{eqnarray*} \]

For \(z=f(x,y) =x^{2}+y^{2}\), we have \[ \begin{eqnarray*} \mathbf{n\,} &=& \tfrac{-f_{x}(x,y)\mathbf{i}-f_{y}(x,y) \mathbf{j\,}+\,\mathbf{k}}{\sqrt{[f_{x}(x,y)]^{2}+[f_{y}(x,y)]^{2}+1}} = \dfrac{-2x\mathbf{i}-2y\mathbf{j}+\mathbf{k}}{\sqrt{ 4x^{2}+4y^{2}+1}}\\[4pt] \rm{curl}\, \mathbf{F}\,{\bf\cdot}\, \mathbf{n} &=&-2\mathbf{k\,{\bf\cdot}\, }\dfrac{1}{\sqrt{ 4x^{2}+4y^{2}+1}}\left( -2x\mathbf{i}-2y\mathbf{j}+\mathbf{k}\right) =\dfrac{ -2}{\sqrt{4x^{2}+4y^{2}+1}} \end{eqnarray*} \] So,

where \(D\) is the interior of the circle \(x^{2}+y^{2}=1\).

NOW WORK

Problem 19.

Using Stokes' Theorem to Find a Surface Integral

Find the surface integral \(\iint\limits_{\kern-3ptS}{\rm{curl} \, \mathbf{F}\,{\bf\cdot}\, \mathbf{n}}\,dS\), where \(\mathbf{F}=\mathbf{F}(x,y,z)=y \mathbf{i}+x\mathbf{j}+z\mathbf{k}\) and \(S\) is the surface \( z=5-(x^{2}+y^{2}),\) \(z\geq 1\), using Stokes' Theorem.

Figure 73

Solution  Figure 73 shows the surface \(S\). The boundary curve \(C\) is the circle \(x^{2}+y^{2}=4\) that lies in the plane \(z=1\). The vector form of \(C\) is \(\mathbf{r}=\mathbf{r}(t)=2\cos t\,\mathbf{i}+2\sin t\,\mathbf{j}+ \mathbf{k}\), \(0\leq t\leq 2\pi \). Then \[ \begin{eqnarray*} d\mathbf{r}& =&(-2\,\sin t\,\mathbf{i}+2\,\cos t\,\mathbf{j})\,dt \\[4pt] \mathbf{F}& =&y\mathbf{i}+x\mathbf{j}+z\mathbf{k}=2\,\sin t\mathbf{i}+2\,\cos t \mathbf{j}+\mathbf{k} \\[4pt] \mathbf{F}\,{\bf\cdot}\, d\mathbf{r}& =&(-4\,\sin ^{2}t+4\,\cos ^{2}t)\,dt=4\,\cos (2t) \,dt \end{eqnarray*} \]

Now we use Stokes' Theorem. \[ \iint\limits_{\kern-3ptS}{\rm{curl}\, \mathbf{F}\,{\bf\cdot}\, \mathbf{n}}\,dS=\oint_{C}\mathbf{F} \,{\bf\cdot}\, d\mathbf{r}=\int_{0}^{2\pi }4\,\cos (2t) \,dt= 2 \big[ \sin(2t)\big] _{0}^{2\pi }=0 \]

NOW WORK

Problem 25.

Using Stokes' Theorem to Find a Line Integral

Find \(I=\oint_{C}[(e^{-x^{2}/2}-yz)\,dx+(e^{-y^{2}/2}+xz+2x) \,dy+(e^{-z^{2}/2}+5)\,dz]\), where \(C\) is the circle \(x=\cos t\), \(y=\sin t\), \(z=2\), \(0\leq t\leq 2\pi \).

Solution  Let \(\mathbf{F}=P\,\mathbf{i}+Q\,\mathbf{j}+R\,\mathbf{k} \), with \(P=e^{-x^{2}/2}-yz\), \(\ Q=e^{-y^{2}/2}+xz+2x\), and \( R=e^{-z^{2}/2}+5\). It can be verified that \[ {\rm{curl }}\,{\bf{F}} = - x{\bf{i}} - y{\bf{j}}{\rm{ + (2 + 2}}z){\bf{k}} \]

To use Stokes' Theorem, take \(S\) to be a plane region enclosed by the circle \(C\) in the plane \(z\rm{=2}\) so that \(\mathbf{n}=\mathbf{k}\). Then \(\rm{curl}\, \mathbf{F}\,{\bf\cdot}\, \mathbf{n}=6\) on \(S\), and \[ I=\iint\limits_{\kern-3pt S}{\rm{curl}\, \mathbf{F}\,{\bf\cdot}\, \mathbf{n}}\,dS=6\iint \limits_{\kern-3pt S}\,dx\,dy=6\,\pi \]

NOW WORK

Problem 27.

THEOREM Conservative Vector Fields in Space

Let \(\mathbf{F}=P(x,y,z)\,\mathbf{i}+Q(x,y,z)\,\mathbf{j}+R(x,y,z)\, \mathbf{k}\) be a vector field whose components \(P\), \(Q\), and \(R\) have continuous first-order partial derivatives throughout a solid \(E\) whose boundary is a surface \(S\) satisfying the conditions of Stokes' Theorem. Then \(\mathbf{F}\) is a conservative vector field if and only if \(\rm{curl}\, \mathbf{F}=\mathbf{0}\) throughout \(S\).

Partial Proof

We provide only an outline of the proof. Suppose \(\rm{curl}\, \mathbf{F}= \mathbf{0}\). Then, by Stokes' Theorem, \[ \begin{equation*} \oint_{C}\mathbf{F}\,{\bf\cdot}\, d\mathbf{r}=\iint\limits_{\kern-3ptS}{\rm{curl}\, \mathbf{F} \,{\bf\cdot}\, \mathbf{n}}\,dS=0 \end{equation*} \]

for any closed curve \(C\). That is, \(\mathbf{F}\) is a conservative vector field.

Conversely, suppose \(\mathbf{F}\) is conservative. Then \({\mathbf{F} = \bf\nabla} f\) for some function \(f\). Since \(\mathbf{F}= \mathbf{F}(x,y,z)= P(x,y,z)\mathbf{i} + Q(x,y,z)\,\mathbf{j}+R(x,y,z)\,\mathbf{k},\) and \({\bf\nabla}f(x,y,z) = \dfrac{\partial }{\partial x} f(x,y,z)\mathbf{i} + \dfrac{\partial }{\partial y} f(x,y,z)\mathbf{j} + \dfrac{\partial }{\partial z}(x,y,z)\mathbf{k}\), then \[ \begin{eqnarray*} \rm{curl}\, \mathbf{F} &\mathbf{=}&\left\vert \begin{array}{@{}c@{\quad}c@{\quad}c} \mathbf{i} & \mathbf{j} & \mathbf{k} \\[3pt] \dfrac{\partial }{\partial x} & \dfrac{\partial }{\partial y} & \dfrac{\partial }{\partial z} \\[7pt] P & Q & R \end{array} \right\vert =\left\vert \begin{array}{@{}c@{\quad}c@{\quad}c} \mathbf{i} & \mathbf{j} & \mathbf{k} \\[3pt] \dfrac{\partial }{\partial x} & \dfrac{\partial }{\partial y} & \dfrac{\partial }{\partial z} \\[5pt] \dfrac{\partial }{\partial x}f(x,y,z) & \dfrac{\partial }{\partial y}f(x,y,z) & \dfrac{\partial }{\partial z}f(x,y,z) \end{array} \right\vert \\[6pt] &=& \left(\dfrac{\partial^{2}f }{\partial{y} \partial{z}} - \dfrac{\partial^{2}f }{\partial{z} \partial{y}}\right)\,\mathbf{i} - \left(\dfrac{\partial^{2}f }{\partial{x} \partial{z}} - \dfrac{\partial^{2}f }{\partial{z} \partial{x}}\right)\,\mathbf{j} + \left(\dfrac{\partial^{2}f }{\partial{x} \partial{y}} - \dfrac{\partial^{2}f }{\partial{y} \partial{x}}\right)\mathbf{k}\\[6pt] &=&0\mathbf{i}+0\mathbf{j}+0\mathbf{k}= \mathbf{0} \end{eqnarray*} \]

NEED TO REVIEW?

The Equality of Mixed Partials Theorem is discussed in Section 12.3, p. 835.

Showing That \(\mathbf{F}\) Is a Conservative Vector Field

Show that \(\mathbf{F}=\left( \dfrac{3}{5}y^{5}+2z^{2}\right) \mathbf{i} +3xy^{4}\,\mathbf{j}+4xz\,\mathbf{k}\) is a conservative vector field in space.

Solution  We find \(\rm{curl}\, \mathbf{F}\). \[ \begin{equation*} \rm{curl}\, \mathbf{F}=\left\vert \begin{array}{@{}c@{\quad}c@{\quad}c} \mathbf{i} & \mathbf{j} & \mathbf{k} \\[5pt] \dfrac{\partial }{\partial x} & \dfrac{\partial }{\partial y} & \dfrac{ \partial }{\partial z} \\[5pt] \dfrac{3}{5}y^{5}+2z^{2} & 3xy^{4} & 4xz \end{array} \right\vert =(0-0)\,\mathbf{i}-(4z-4z)\,\mathbf{j}+(3y^{4}-3y^{4})\mathbf{k}= \mathbf{0} \end{equation*} \]

Since \(\rm{curl}\,\mathbf{F=0}\), \(\mathbf{F}\) is conservative.

NOW WORK

Problem 31.

THEOREM Curl and Angular Velocity

Let \(\mathbf{F}\) be the velocity field of a fluid rotating about a fixed axis and let \({\bf\omega }\) be the constant angular velocity. Then \[ {\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{ \rm{curl}\, \mathbf{F}=2 \bf\omega }}} \]

NEED TO REVIEW?

Angular velocity is discussed in Section 10.5, pp. 729-730.

Proof

Since the motion of a fluid is simply a rotation about a given fixed axis in space, the angular velocity \({\bf\omega }\) can be represented by a constant vector: \[ {\bf\omega }=\omega _{1}\mathbf{i}+\omega _{2}\mathbf{j}+\omega _{3}\mathbf{k} \]

where the magnitude of \({\bf\omega }\) is the angular speed and the direction of \({\bf\omega }\) is along the direction of the axis of rotation, in accordance with the right-hand rule. If the origin of a rectangular coordinate system is on the fixed axis and \(\mathbf{r}=x\mathbf{i }+y\mathbf{j}+z\mathbf{k}\), then the velocity vector \(\mathbf{F}\) is \[ \mathbf{F}={\bf\omega} \times \mathbf{r} \] and \[ \rm{curl}\, \mathbf{F}=\left\vert \begin{array}{@{}c@{\quad}c@{\quad}c} \mathbf{i} & \mathbf{j} & \mathbf{k} \\[3pt] \dfrac{\partial }{\partial x} & \dfrac{\partial }{\partial y} & \dfrac{\partial }{\partial z} \\[3pt] \omega _{2}z-\omega _{3}y & \omega _{3}x-\omega _{1}z & \omega _{1}y-\omega _{2}x \end{array} \right\vert \]

Since \({\bf\omega }\) is constant, then \[ \rm{curl}\, \mathbf{F}=2\omega _{1}\mathbf{i}+2\omega _{2}\mathbf{j}+2\omega _{3}\mathbf{k}=2 {\bf\omega } \]

IN WORDS

If a fluid is rotating, the curl of the velocity vector is a constant vector that equals twice the angular velocity vector.