surface integral with normal unit vector

Question

Let $\vec f(\vec r) = (y,-x, zxy)$ and let $S$ be the surface $x^2+y^2+3z^2=1$ , $z≤0$, with unit normal vector $\vec n$ pointing in the positive $z$-direction. The value of the surface integral $$\iint_S(\nabla \times \vec f) \cdot \vec n\,\mathrm dS$$ is $n\pi$ where $n$ is an integer. What is the value of $n$?

what i know:

I know you need to use stokes theorem when calculating this. I have calculated the curl of the vector field as $(xz, -yz, -2)$. I dont know where to go from this, any help would be appreciated.

Answer 1

If you want to use Stokes' theorem, then there's no need to actually compute the curl of $\vec f$. The theorem lets you exchange integrating the curl of $\vec f$ over $S$ with the line integral of $\vec f$ over the boundary of $S$, which is the ellipse/circle $x^2+y^2=1$ in the plane $z=0$.

So you have

$$\begin{align} \iint_S(\nabla\times\vec f)\cdot\vec n\,\mathrm dS&=\oint_{\partial S}\vec f\cdot\mathrm d\vec r\\[1ex] &=\oint_{x^2+y^2\le1}(y,-x,xyz)\cdot\mathrm d\vec r\\[1ex] &=\int_0^{2\pi}(\sin t,-\cos t,0)\cdot(-\sin t,\cos t,0)\,\mathrm dt\\[1ex] &=-\int_0^{2\pi}\mathrm dt \end{align}$$

where we convert to polar coordinates $(x,y,z)=(\cos t,\sin t,0)$ in the third line. Determining the value of $n$ from this point is trivial.

Answer 2

You mention you need to use Stokes Theorem but that is not correct. You can calculate it without that too but usually it is simpler to find surface integral use Stokes Theorem.

You already found curl of $\vec{F}$ as $(xz, -yz, -2)$.

Parametrizing our surface $z = -\frac{1}{\sqrt3} \sqrt {1 - x^2 -y^2}$.

$\frac{\partial z}{\partial x} = -\frac{x}{3z}, \frac{\partial z}{\partial y} = -\frac{x}{3z}$

So $d\vec{S} = (-\frac{\partial z}{\partial x}, -\frac{\partial z}{\partial y}, 1) = (\frac{x}{3z},\frac{y}{3z},1)$

So integral becomes $I = \displaystyle \frac{1}{3}\int_{-1}^{1} \int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}} (xz, -yz, -2).(\frac{x}{z}, \frac{y}{z}, 3) dx \, dy$

$ = \displaystyle \frac{1}{3}\int_{-1}^{1} \int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}} (x^2 - y^2 - 6) \, dx \, dy = -2 \pi$