Flux of $v=(y_1e^{y_2}-e^{y_1},y_2e^{y_3}-e^{y_2},y_3e^{y_1}-e^{y_3})$ through a sphere

Question

$v=(y_1e^{y_2}-e^{y_1},y_2e^{y_3}-e^{y_2},y_3e^{y_1}-e^{y_3})$

$Y = (y_1^2+y_2^2+y_3^2=r^2, y_2>=0, y_3>=0)$

Calculate flux of v through Y.

My first intuition was to use spherical coordinates so I set $\varphi(x)=(r\sin x_2\cos x_1, r\sin x_2 \sin x_1, r\cos x_2)$, with $x_1$ between $0$ and $\pi$ and $x_2$ between $0$ and $\pi/2$.

Then I wanted to use the Stokes-Ampère theorem, but I don't really know how to use it. I saw that div v = $0$ and found $\omega_v^2$. Then I found $\omega_u^1$ with $u=(0,y_3e^{y_1}-y_1e^{y_3},y_1y_2e^{y_3}-y_1e^{y_2})$.

But what now? How do I set the integral? In class we saw an example with polar coordinates and a simple u and it seemed easy with just one variable. But here I have spherical coordinates and the vector u is very confusing.

Could someone please explain how to use this theorem correctly and how do I integrate from here?

Answer 1

I just got the correction back and this is how we were supposed to do the exercise. Note that we only saw the divergence theorem this week, so we were supposed to solve this without using it.

So, as I wrote before, the divergence of $v$ is $0$.

Since the divergence of $v$ is $0$, there exists a $u$ such that rot$u=v$. And, by Stokes-Ampère, $\Phi=\int_{\partial Y}\omega_u^1$. We're also going to be using Green-Riemann: $\int_{\varphi (K)}d\omega = \int_{\varphi (\partial K)}\omega$

So we have:

$\omega_v^2 = (y_3e^{y_1}-e^{y_3})dy_1\wedge dy_2 - (y_2e^{y_3}-e^{y_2})dy_1\wedge dy_3+(y_1e^{y_2}-e^{y_1})dy_2\wedge dy_3$ = $d(y_3e^{y_1}dy_2)+d(y_1e^{y_2}dy_3)+d(y_2e^{y_3}dy_1)$

And so $\omega_u^1=y_2e^{y_3}dy_1+y_3e^{y_1}dy_2+y_1e^{y_2}dy_3$

We need to calculate $\Phi=\int_{\partial Y}\omega_u^1$.

How do we parametrise $\partial Y$?

$\gamma_1 = r(-\cos t, \sin t,0)$, $\gamma_2 = r(\cos t, 0, \sin t)$, $t=(0,\pi)$

And so:

$\int_{\partial Y}\omega_u^1$ = $\int_{\gamma_1}\omega_u^1 + \int_{\gamma_2}\omega_u^1$=$r^2\int_0^{\pi}((\sin t) e^0 (\sin t) +0 +0)dt+r^2\int_0^{\pi}(0+0+(\cos t) e^0 (\cos t) dt$ = $r^2\int_0^{\pi}(\cos^2 t +\sin^2 t)dt = \pi r^2$

Answer 2

$\vec F =(xe^y - e^x, ye^z - e^y, z e^x - e^z)$

Given surface is $S: x^2+y^2+z^2 = r^2; \ y, z \geq 0$

We can apply divergence theorem by closing the surface with semi-disks in plane $z = 0$ and $y = 0$.

$\nabla \cdot \vec F = e^y - e^x + e^z - e^y + e^x - e^z = 0$

As the divergence of the vector field is zero, its surface integral over closed surface is zero. Now we find the flux through semi-disks at $y = 0$ and at $z=0$ and subtract from the total flux which is zero. That would give us the flux through surface $S$.

For disk at $y = 0$, outward normal vector is $(0, -1, 0)$ and we can parametrize the surface as $(\rho \cos \theta, 0, \rho\sin \theta), 0 \leq \rho \leq r, 0 \leq \theta \leq \pi$.

For disk at $z = 0$, outward normal vector is $(0, 0, -1)$ and we can parametrize the surface as $(\rho \cos \theta, \rho\sin \theta, 0), 0 \leq \rho \leq r, 0 \leq \theta \leq \pi$.

It is easy to see that the total outward flux through both semi-disks is equal to the area of a circle of radius $r$.

Can you take it from here, fill in all the details and complete the working?

Answer 3

Since the divergence is zero, we can close the surface and say

$$\iint\limits_S\cdots + \iint\limits_{y_2=0}\cdots +\iint\limits_{y_3=0}\cdots = 0 \implies \iint\limits_S \cdots= -\iint\limits_{y_2=0}\cdots-\iint\limits_{y_3=0}\cdots$$

Those two integrals are easy to compute, they are constants being integrated over half circles

$$-\iint\limits_{y_1^2+y_3^2=r^2\cap y_3\geq 0}(\cdots, 0 - e^0,\cdots)\cdot (0,-1,0)dA - \iint\limits_{y_1^2+y_2^2=r^2\cap y_2\geq 0}(\cdots, \cdots, 0 - e^0)\cdot (0,0,-1)dA $$

$$=\iint\limits_{y_1^2+y_3^2=r^2\cap y_3\geq 0}1\:dA + \iint\limits_{y_1^2+y_2^2=r^2\cap y_2\geq 0}1\:dA = \frac{\pi r^2}{2}+\frac{\pi r^2}{2} = \pi r^2$$