SCalcET$7$

$16\mathrm{.}8.$AE$.002.$

EXAMPLE $2$ Use Stokes' Theorem to compute the integral $$ curl $FdS,$ where $F(x,y,z)=zyzjxyk$ and $S$ is the sphere $x^2{y}^2{z}^2=49$ that lies inside the cylinder $x^2{y}^2=36$ and above the $xy-$plane. $($See the figure.$)$

SOLUTION To find the boundary curve $C$ we solve the equations $x^2{y}^2{z}^2=49$ and $x^2{y}^2=36.$ Subtracting, we get $z^2=13$ and so $z=\sqrt[\phantom{2}]{13}.$ A vector equation of $C$ is

$r(t)=6cos(t)i6sin(t)j,k$;$0t2$

$r^{\text{'}}(t)=-6sin(t)i6cos(t)j,$

Also, we have

$F(r(t))=$

$$

$i6\sqrt[\phantom{2}]{13}sin(t)j36cos(t)sin(t)k$

Therefore, by Stokes' Theorem,

$curlF*dS={\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}F*dr={\displaystyle \underset{0}{\overset{2}{}}}F(r(t))*r^{\text{'}}(t)dt$

$={\displaystyle \underset{0}{\overset{2}{}}}(\frac{?}{26\sqrt[\phantom{2}]{13}sin(t)cos(t)}dt)$

$=36\sqrt[\phantom{2}]{13}{\displaystyle \underset{0}{\overset{2}{}}},dt=0$