# Thread: A few flux questions

1. ## A few flux questions

1. Let S be the surface $\displaystyle z=x, x^2+y^2<=1$ Find

$\displaystyle \iint (x^2 +y^2)dS$

I know the answer is supposed to be $\displaystyle \sqrt{2}*\pi/ 2$ but can't seem to get there. I've tried evaluating it with the divergence theorem, but nothing I do seems to work correctly. I know that $\displaystyle x^2+y^2dS$ is equal to $\displaystyle F \cdot n$ and I have $\displaystyle n = <-1/\sqrt{2},0,1/\sqrt{2}>$, which would make $\displaystyle F=<-\sqrt{2} x^2,0,\sqrt{2} y^2>$. Then the integral becomes $\displaystyle \sqrt{2} \int_0^{2\pi}} \int_0^{1} \int_0^{rcos\theta} } -r^2cos(\theta)^2+r^2sin(\theta)^2 rdzdrd\theta$.

Am I doing this right? I can't figure out where I'm messing up.

These next two I have the same issues with:

2. Let S be the surface $\displaystyle z = x + y, 0\leqslant x \leqslant 1, 0\leqslant x \leqslant 1$. Find the upward flux of the vector field $\displaystyle F = <z,x,y>$ across S.

I think that the divergence is 0, and wouldn't that make the whole integral 0? But I'm told the correct answer is -1. Why is this?

3. Let S be the portion of the cylinder given by $\displaystyle 0\leqslant z \leqslant 3, r=1, 0\leqslant \theta \leqslant \pi/2$.

Orient S by normal vectors pointing away from the z-axis and compute the flux of $\displaystyle F = <2x,y,-3z>$ across S.

In this case, wouldn't the divergence also be 0, since $\displaystyle \nabla \cdot F = 2 + 1 -3 = 0$?

But I know the answer to be $\displaystyle 9\pi/4$.

I'm obviously missing the same sort of thing in all these problems so any help would be appreciated very much!

2. For the first one the divergence theorem does not apply it is a scalar surface integral.

$\displaystyle dS=\sqrt{ \left(\frac{\partial z}{\partial x} \right)^2+ \left(\frac{\partial z}{\partial y} \right)^2+1}=\sqrt{2}$

This gives the integral in cylindrical coordinates

$\displaystyle \sqrt{2}\int_{0}^{2\pi}\int_{0}^{1}r^2 rdrd\theta$

for 2 the divergence theorem does not apply because the surface is not closed.
use the definition of a surface integral

for 3 again the surface is not closed so the divergence theorem does not apply.
This can be parametrized by

$\displaystyle \mathbf{r}(\theta,z)=\cos(\theta)\mathbf{i}+\sin( \theta )\mathbf{j}+z\mathbf{k}$

3. Okay, I sort of understand this, but I still have some questions.

For the second problem, since $\displaystyle F = zi + xj + yk$ and $\displaystyle z = x+ y$

Then $\displaystyle \frac{\partial z}{\partial x} =1$ and $\displaystyle \frac{\partial z}{\partial y} = 1$, so for the definition of a surface integral I have

$\displaystyle \iint F\cdot dS = \iint (-P \frac{\partial z}{\partial x} - Q \frac{\partial z}{\partial y} + R) dA$ for $\displaystyle F = P i + Q j + Rk.$

So my integral is $\displaystyle \iint -2x dxdy$

But how do I set up the limits?

And for the 3rd one, if $\displaystyle \mathbf{r}(\theta,z)=\cos(\theta)\mathbf{i}+\sin( \theta )\mathbf{j}+z\mathbf{k}$, then I'm getting {$\displaystyle r}_{\theta } = <-sin(\theta),cos(\theta),0>$ and $\displaystyle {r}_{z} = <0,0,1>,$ and their cross product is $\displaystyle <cos(\theta),sin(\theta),0>.$ How do I proceed from there?

4. For the 2nd problem the limits are given to you! You typed them in the problem statement.

To the third problem dot the vector you obtianed by the cross product with the vector field and use the limits you gave in the problem statement.

5. Thank you very much!