Stokes' theorem on unit disc

Question

I'm stuck with this question:

"Consider $\iint _S(\vec \nabla \times \vec F)\cdot d\vec S$, where $\vec F=xyz\vec i+y^2z\vec j+xy^2\vec k$ and $S$ is the surface of the unit disc $0\le x^2+z^2\le 1$ on the plane $y=1$, with the normal to the plane taken in the negative $y$ direction. Calculate the integral. You may use one of the integral theorems if you wish."

So I decided to use Stokes' theorem and this is my working so far:

$\frac {\partial}{\partial x}=yz$

$\frac {\partial}{\partial y}=2yz$

$\frac {\partial}{\partial z}=0$

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$\vec \nabla =yz\vec i+2yz\vec j$

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$\vec \nabla \times \vec F= \left| \begin{array}\ {\vec i}&{\vec j}&{\vec k}\\ yz&2yz&0\\ xyz&y^2z&xy^2 \end{array} \right| $

$\vec \nabla \times \vec F=2xy^3z\vec i -xy^3z\vec j +(y^3z^2-2xy^2z^2)\vec k$

$(\vec \nabla \times \vec F)\cdot \vec n = xy^3z$

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So I've set up my integral as $\iint _Sxy^3z\,dS$ but I'm confused about what to set the integral limits to. The example that I was looking at on page 2 of this pdf sort just doesn't set any limits http://www.math.uiuc.edu/~franklan/Math241_168_StokesThm.pdf

Would it be $\int_0^1\int_0^1xy^3z\,dx\,dz$?

Answer 1

The Curl of $\vec F$ (see that embedded NOTE) is

$$\nabla \times \vec F=\hat x (2yx-y^2)+\hat y(xy-y^2)+\hat z(-xz)$$

Since the surface normal point in the direction of $-\hat y$, and $y=1$ on the surface, we have

$$\left.\left(\nabla \times \vec F\cdot (-\hat y)\right)\right|_{y=1}=1-x$$

Therefore, we have

$$\begin{align} \int_S \nabla \times \vec F \cdot \hat n\,dS&=\int_{-1}^{1}\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} (1-x)\,dx\,dz\\\\ &=\pi \end{align}$$

where we exploited the facts that $\int_{-1}^{1}\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}$ gives the area of a circle of unit radius (i.e., $\pi$) and

$$\int_{-1}^{1}\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}x\,dx\,dz=0$$

since in the inner integral we are integrating an odd function, namely $x$, over symmetric limits.

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NOTE:

$\vec \nabla \times \vec F= \left| \begin{array}\ {\vec i}&{\vec j}&{\vec k}\\ \frac{\partial}{\partial x}&\frac{\partial}{\partial y}&\frac{\partial}{\partial z}\\ xyz&y^2z&xy^2 \end{array} \right| $

Therefore,

$$\nabla \times \vec F=\hat x (2yx-y^2)+\hat y(xy-y^2)+\hat z(-xz)$$