Find the flux integral $\int_{\delta V} F \cdot \hat n dS$ where $F = xy^2 \hat i + xe^z \hat j + z^3 \hat k$, $\delta V$ is the surface of the cylinder V bounded by $y^2 + z^2 = 1$ and the planes x = -1, x = 2 and $\hat n$ is that outwards pointing unit normal to V.
My attempt:
I have found the normal vector as:
$$\hat n = \frac{2y \hat j + 2z \hat k}{\sqrt{4y^2 + 4z^2}} = \frac{y \hat j + z \hat k}{1} $$
Now by observation of the volume i know it is a cylinder aligns parallel to the x plane. Giving a shadow on the y-z plane as $y^2 + z^2 =1$
Now I say the flux, $$\Phi = \int \int _S F \cdot \hat n dS = \int \int _S xye^z + z^4 dS $$,
But now I am unsure how to progress. I converted to polar coordinates but I am unsure what this means in terms what I am calculating.
I say,
$$x = x, y = cos(\theta), z = sin(\theta)$$
However then I need to do the Jacobian since I am changing variables. But if I calculate my Jacobian it is 0 which is problematic!
Anyone have any idea what I am doing wrong?
You have to choose the two variables to perform the surface integral. As you started posing the exercise, it seems that they have to be $x,z$ or $x,y$. But, as you saw at the integration time, maybe is better use some others. If we parametrice the surface, we have,
$\begin{cases} x=u\\ y=\cos v\\ z=\sin v \end{cases}$
Now, $d\mathbf S=\dfrac{d\mathbf r}{dv}\times\dfrac{d\mathbf r}{du}\mathbb du\mathbb dv=\begin{vmatrix}\hat i&\hat j&\hat k\\0&-\sin v&\cos v\\1&0&0\end{vmatrix}\mathbb du\mathbb dv=(\cos v\hat j+\sin v\hat k)\mathbb du\mathbb dv$
$\displaystyle\Phi = \int \int _S \mathbf F \cdot d\mathbf S = \int_{-1}^2 \int_0^{2\pi}(u\cos^2v\,\hat i + ue^{\sin v} \hat j + \sin^3v\,\hat k)·(\cos v\,\hat j+\sin v\,\hat k)\mathbb dv\mathbb du$