Question

Use Stoke's Theorem to evaluate ∫CF⋅dr where F(x,y,z)=xi+yj+2(x2+y2)k and C is the boundary of the part of the paraboloid where z=81−x2−y2 which lies above the xy-plane and C is oriented counterclockwise when viewed from above.

Answer 1

Parameterize the surface (call it $S$) that has $C$ as its boundary by

$\vec s(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec\jmath+(81-u^2)\,\vec k$

with $0\le u\le9$ and $0\le v\le2\pi$.

Take the normal vector to $S$ to be

$\vec n=\dfrac{\partial\vec s}{\partial u}\times\dfrac{\partial\vec s}{\partial v}=2u^2\cos v\,\vec\imath+2u^2\sin v\,\vec\jmath+u\,\vec k$

Compute the curl of $\vec F(x(u,v),y(u,v),z(u,v))=\vec F(u,v)$. We have

$\nabla\times\vec F(x,y,z)=4y\,\vec\imath-4x\,\vec\jmath$

$\implies\nabla\times\vec F(u,v)=4u\sin v\,\vec\imath-4u\cos v\,\vec\jmath$

Then by Stoke's theorem, the line integral is

$\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\iint_S(\nabla\times\vec F(x,y,z))\cdot\mathrm d\vec S$

$=\displaystyle\iint_S(\nabla\times\vec F(u,v))\cdot\vec n\,\mathrm du\,\mathrm dv$

$=\displaystyle\int_0^{2\pi}\int_0^90\,\mathrm du\,\mathrm dv=\boxed{0}$

We can verify this result by computing the line integral directly. Parameterize $C$ by

$\vec r(t)=9\cos t\,\vec\imath+9\sin t\,\vec\jmath$

with $0\le t\le2\pi$. Then

$\displaystyle\int_C\vec F(x,y,z)\cdot\mathrm d\vec r=\int_0^{2\pi}\vec F(t)\cdot\frac{\mathrm d\vec r}{\mathrm dt}\,\mathrm dt$

$=\displaystyle\int_0^{2\pi}(9\cos t\,\vec\imath+9\sin t\,\vec\jmath+162\,\vec k)\cdot(-9\sin t\,\vec\imath+9\cos t\,\vec\jmath)\,\mathrm dt$

$=\displaystyle\int_0^{2\pi}0\,\mathrm dt=\boxed{0}$