# Problem on Stokes' Theorem

I'm really struggling to understand Stokes' Theorem. I tried this exercise:

Let D be the portion of $$z=1-x^2-y^2$$ above the xy-plane, oriented up, and let $$\vec{F}=\langle xy^2,-x^2y,xyz\rangle$$. Compute $$\iint_{D}^{}(\nabla\times \vec{F})\cdot \hat{n}dS$$ Here is my work: $$\nabla\times \vec{F}=\langle xz,-yz,-4xy\rangle$$

$$\vec{f}(r,\theta) = \bigl\langle r\cos\theta ,r\sin\theta ,1-r^2 \bigr\rangle$$

$$\frac{\partial\vec{f} }{\partial r}= \langle\cos\theta,\sin\theta,-2r\rangle$$

$$\frac{\partial \vec{f}}{\partial\theta }= \langle -r\sin\theta ,r\cos\theta ,0 \rangle$$

$$\frac{\partial\vec{f} }{\partial r}\times \frac{\partial \vec{f}}{\partial \theta}=\left \langle 2r^2\cos\theta ,2r^2\sin\theta ,r\right \rangle$$

$$\left \|\frac{\partial\vec{f} }{\partial r}\times \frac{\partial \vec{f}}{\partial \theta } \right \|=r\sqrt{3}$$ $$\widehat{n}= \biggl\langle \frac{2r\cos\theta}{\sqrt{3}} ,\frac{2r\sin\theta }{\sqrt{3}},\frac{1}{\sqrt{3}}\biggr\rangle$$

Integrating, I have $$\frac{1}{\sqrt{3}}\int_{0}^{2\pi }\int_{0}^{1}2r^2\cos^2\theta (1-r^2)-2r^2\sin^2\theta (1-r\cos\theta )-4r\sin\theta\cos\theta\,dr\,d\theta$$

$$=\frac{1}{\sqrt{3}}\int_{0}^{2\pi }\int_{0}^{1}-2r^2\sin^2\theta (1-r\cos\theta )+2r^2\cos\theta\, \theta (1-r^2)-2r\sin(2\theta )\,dr\,d\theta$$ After splitting the integral into three integrals, I have $$\frac{1}{\sqrt{3}}\int_{0}^{2\pi }\int_{0}^{1}-2r^2\sin^2\theta (1-r\cos\theta )\,dr\,d\theta=-\frac{2\pi }{3\sqrt{3}}$$

$$\frac{1}{\sqrt{3}}\int_{0}^{2\pi }\int_{0}^{1}2r^2\cos\theta (1-r^2)\,dr\,d\theta =0$$

$$\frac{1}{\sqrt{3}}\int_{0}^{2\pi }\int_{0}^{1}-2r\sin(2\theta )\,dr\,d\theta =0$$

$$=-\frac{2\pi }{3\sqrt{3}}+0+0$$

But the answer is zero. What am I doing wrong?

• You want to calculate using double integral instead of using the line integral around its boundary? Also you seem to have some mistakes in your calculations. Commented Dec 16, 2020 at 20:08
• You didn't use Stokes' Theorem!
– user801306
Commented Dec 16, 2020 at 20:09
• You need to identify the boundary of your surface (call this $C$) and then evaluate $\int_CF\cdot dr$ while making sure the normal vector given in the original problem induces the orientation you prescribe to $C$
– user801306
Commented Dec 16, 2020 at 20:19
• @CalebWilliamsUIC it's a great book! you must get it :) Commented Dec 16, 2020 at 20:40

If you don't apply Stokes' theorem (that is, you insist on computing the integral of the curl over the surface $$D$$), then you would have

\begin{align} \iint_D(\nabla\times\vec F)\cdot\mathrm d\vec S&=\iint_D\langle xz,-yz,-4xy\rangle\cdot\vec n\,\mathrm dS\\[1ex] &=\iint_D \langle r(1-r^2)\cos\theta,-r(1-r^2)\sin\theta,-4r^2\cos\theta\sin\theta\rangle\cdot\vec n\,\mathrm dS \end{align}

where all I've done here is compute the curl of $$\vec F$$ and composed it with $$\vec f$$ to replace $$x\to r\cos\theta$$, $$y\to r\sin\theta$$, and $$z\to1-r^2$$. The normal vector is

$$\vec n=\frac{\partial\vec f}{\partial r}\times\frac{\partial\vec f}{\partial\theta}=\langle2r^2\cos\theta,2r^2\sin\theta,r\rangle$$

So the surface integral reduces to

$$2\int_0^{2\pi}\int_0^1 \left(r^3(\cos(2\theta)-\sin(2\theta))-r^5\cos(2\theta)\right)\,\mathrm dr\,\mathrm d\theta$$

Note that you do more work than necessary, since

$$\vec n\,\mathrm dS=\left\|\frac{\partial\vec f}{\partial r}\times\frac{\partial\vec f}{\partial\theta}\right\|\frac{\left(\frac{\partial\vec f}{\partial r}\times\frac{\partial\vec f}{\partial\theta}\right)}{\left\|\frac{\partial\vec f}{\partial r}\times\frac{\partial\vec f}{\partial\theta}\right\|}\,\mathrm dr\,\mathrm d\theta=\left(\frac{\partial\vec f}{\partial r}\times\frac{\partial\vec f}{\partial\theta}\right)\,\mathrm dr\,\mathrm d\theta$$

so you don't strictly need to normalize the normal vector.

If you do wish to apply Stokes' theorem, the integral is trivial by comparison (I've omitted the details):

$$\int_C \vec F\cdot\mathrm d\vec r=-\int_0^{2\pi}\left(\cos^3\theta\sin\theta-\cos\theta\sin^3\theta\right)\,\mathrm d\theta$$

and both do indeed have the same value.

• I see now. I forgot that $dS$ is literally defined by the magnitude of the cross product of the vector partials times the area element. Commented Dec 16, 2020 at 21:05