# 6.7 Stokes’ theorem

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Calculate the curl of electric field E if the corresponding magnetic field is $\text{B}\left(t\right)=⟨tx,ty,-2tz⟩,0\le t<\infty .$

$\text{curl}\phantom{\rule{0.2em}{0ex}}\text{E}=⟨x,y,-2z⟩$

Notice that the curl of the electric field does not change over time, although the magnetic field does change over time.

## Key concepts

• Stokes’ theorem relates a flux integral over a surface to a line integral around the boundary of the surface. Stokes’ theorem is a higher dimensional version of Green’s theorem, and therefore is another version of the Fundamental Theorem of Calculus in higher dimensions.
• Stokes’ theorem can be used to transform a difficult surface integral into an easier line integral, or a difficult line integral into an easier surface integral.
• Through Stokes’ theorem, line integrals can be evaluated using the simplest surface with boundary C .
• Faraday’s law relates the curl of an electric field to the rate of change of the corresponding magnetic field. Stokes’ theorem can be used to derive Faraday’s law.

## Key equations

• Stokes’ theorem
${\int }_{C}\text{F}·d\text{r}={\iint }_{S}\text{curl}\phantom{\rule{0.2em}{0ex}}\text{F}·d\text{S}$

For the following exercises, without using Stokes’ theorem, calculate directly both the flux of $\text{curl}\phantom{\rule{0.2em}{0ex}}\text{F}·\text{N}$ over the given surface and the circulation integral around its boundary, assuming all are oriented clockwise.

$\text{F}\left(x,y,z\right)={y}^{2}\text{i}+{z}^{2}\text{j}+{x}^{2}\text{k}\text{;}$ S is the first-octant portion of plane $x+y+z=1.$

$\text{F}\left(x,y,z\right)=z\text{i}+x\text{j}+y\text{k}\text{;}$ S is hemisphere $z={\left({a}^{2}-{x}^{2}-{y}^{2}\right)}^{1\text{/}2}.$

${\iint }_{S}\left(\text{curl}\phantom{\rule{0.2em}{0ex}}\text{F}·\text{N}\right)dS=\pi {a}^{2}$

$\text{F}\left(x,y,z\right)={y}^{2}\text{i}+2x\text{j}+5\text{k}\text{;}$ S is hemisphere $z={\left(4-{x}^{2}-{y}^{2}\right)}^{1\text{/}2}.$

$\text{F}\left(x,y,z\right)=z\text{i}+2x\text{j}+3y\text{k}\text{;}$ S is upper hemisphere $z=\sqrt{9-{x}^{2}-{y}^{2}}.$

${\iint }_{S}\left(\text{curl}\phantom{\rule{0.2em}{0ex}}\text{F}·\text{N}\right)dS=18\pi$

$\text{F}\left(x,y,z\right)=\left(x+2z\right)\text{i}+\left(y-x\right)\text{j}+\left(z-y\right)\text{k}\text{;}$ S is a triangular region with vertices (3, 0, 0), (0, 3/2, 0), and (0, 0, 3).

$\text{F}\left(x,y,z\right)=2y\text{i}-6z\text{j}+3x\text{k}\text{;}$ S is a portion of paraboloid $z=4-{x}^{2}-{y}^{2}$ and is above the xy -plane.

${\iint }_{S}\left(\text{curl}\phantom{\rule{0.2em}{0ex}}\text{F}·\text{N}\right)dS=-8\pi$

For the following exercises, use Stokes’ theorem to evaluate ${\iint }_{S}\left(\text{curl}\phantom{\rule{0.2em}{0ex}}\text{F}·\text{N}\right)dS$ for the vector fields and surface.

$\text{F}\left(x,y,z\right)=xy\text{i}-z\text{j}$ and S is the surface of the cube $0\le x\le 1,0\le y\le 1,0\le z\le 1,$ except for the face where $z=0,$ and using the outward unit normal vector.

$\text{F}\left(x,y,z\right)=xy\text{i}+{x}^{2}\text{j}+{z}^{2}\text{k}\text{;}$ and C is the intersection of paraboloid $z={x}^{2}+{y}^{2}$ and plane $z=y,$ and using the outward normal vector.

${\iint }_{S}\left(\text{curl}\phantom{\rule{0.2em}{0ex}}\text{F}·\text{N}\right)dS=0$

$\text{F}\left(x,y,z\right)=4y\text{i}+z\text{j}+2y\text{k}$ and C is the intersection of sphere ${x}^{2}+{y}^{2}+{z}^{2}=4$ with plane $z=0,$ and using the outward normal vector

Use Stokes’ theorem to evaluate $\underset{C}{\int }\left[2x{y}^{2}zdx+2{x}^{2}yzdy+\left({x}^{2}{y}^{2}-2z\right)dz\right],$ where C is the curve given by $x=\text{cos}\phantom{\rule{0.2em}{0ex}}t,y=\text{sin}\phantom{\rule{0.2em}{0ex}}t,z=\text{sin}\phantom{\rule{0.2em}{0ex}}t,0\le t\le 2\pi ,$ traversed in the direction of increasing t .

${\int }_{C}\text{F}·d\text{S}=0$

[T] Use a computer algebraic system (CAS) and Stokes’ theorem to approximate line integral $\underset{C}{\int }\left(ydx+zdy+xdz\right),$ where C is the intersection of plane $x+y=2$ and surface ${x}^{2}+{y}^{2}+{z}^{2}=2\left(x+y\right),$ traversed counterclockwise viewed from the origin.

[T] Use a CAS and Stokes’ theorem to approximate line integral $\underset{C}{\int }\left(3ydx+2zdy-5xdz\right),$ where C is the intersection of the xy -plane and hemisphere $z=\sqrt{1-{x}^{2}-{y}^{2}},$ traversed counterclockwise viewed from the top—that is, from the positive z -axis toward the xy -plane.

${\int }_{C}\text{F}·d\text{S}=-9.4248$

[T] Use a CAS and Stokes’ theorem to approximate line integral $\underset{C}{\int }\left[\left(1+y\right)zdx+\left(1+z\right)xdy+\left(1+x\right)ydz\right],$ where C is a triangle with vertices $\left(1,0,0\right),$ $\left(0,1,0\right),$ and $\left(0,0,1\right)$ oriented counterclockwise.

Use Stokes’ theorem to evaluate ${\iint }_{S}\text{curl}\phantom{\rule{0.2em}{0ex}}\text{F}·d\text{S},$ where $\text{F}\left(x,y,z\right)={e}^{xy}\text{cos}\phantom{\rule{0.2em}{0ex}}z\text{i}+{x}^{2}z\text{j}+xy\text{k},$ and S is half of sphere $x=\sqrt{1-{y}^{2}-{z}^{2}},$ oriented out toward the positive x -axis.

$\underset{S}{\iint }\text{curl}\phantom{\rule{0.2em}{0ex}}\text{F}·d\text{S}=0$

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