Divergence

**arbolis** · July 18th 2009, 11:42 AM

Verify the divergence theorem for $F(x,y,z)=(xz,yz,3z^2)$ in the solid $E$ limited by the paraboloid $z=x^2+y^2$ and the plane $z=1$ .
My attempt : They ask me the check that $\int_{\partial E} F \bold n dE= \iiint _E div F dE$ holds.
I've already calculated the right hand side, which gave me $\frac{8\pi}{3}$ .
I'm stuck on the left side. I know that $\int_{\partial E} F \bold n dE$ is a double integral. I think that I must parametrize the surface $E$ and calculate $\bold n$ . How do I find $\bold n$ ? By calculating the gradient of $F$ ? I'm not sure at all about this approach though. Can you confirm it?

**HallsofIvy** · July 18th 2009, 12:20 PM

Originally Posted by arbolis

Verify the divergence theorem for $F(x,y,z)=(xz,yz,3z^2)$ in the solid $E$ limited by the paraboloid $z=x^2+y^2$ and the plane $z=1$ .
My attempt : They ask me the check that $\int_{\partial E} F \bold n dE= \iiint _E div F dE$ holds.
I've already calculated the right hand side, which gave me $\frac{8\pi}{3}$ .
I'm stuck on the left side. I know that $\int_{\partial E} F \bold n dE$ is a double integral. I think that I must parametrize the surface $E$ and calculate $\bold n$ . How do I find $\bold n$ ? By calculating the gradient of $F$ ? I'm not sure at all about this approach though. Can you confirm it?

Yes, you can think of x and y as parameters but I think it is better, because of the symmetry, to go to polar coordinates immediately and write the surface of the paraboid as $x= rcos(\theta)$ , $y= rsin(\theta)$ , z= $x^2+ y^2= r^2$ . The "position vector" for a point on that surface is $\vec{r}(r,\theta)= rcos(\theta)\vec{i}+ rsin(\theta)\vec{j}+ r^2\vec{k}$ . Then, differentiating with respect to r and $\theta$ , in turn, $\vec{r}_r= cos(\theta)\vec{i}+ sin(\theta)\vec{j}+ 2r\vec{k}$ and $\vec{r}_\theta= -rsin(\theta)\vec{j}+ rcos(\theta)\vec{j}+ 2r\vec{k}$ are two vectors in the tangent plane.

Now, here is the important point: the "fundamental vector product", the cross product of those two tangent vectors, gives the surface differential.
The cross product is:
$\left|\begin{array}{ccc}\vec{i} & \vec{j} & \vec{k} \\ cos(\theta) & sin(\theta) & 2r \\ -rsin(\theta) & rcos(\theta) & 0\end{array}\right|$ $= -2r^2cos(\theta)\vec{i}- 2r^2sin(\theta)\vec{j}+ r\vec{k}$ which is orthogonal to the surface. The sign, of course, depends on the order in which we write those to vectors in the cross product and really depends whether we want the "outward" normal or the "inward" normal
The divergence theorem requires the use of the "outward" normal. $d\vec{S}= \vec{n}dS= (2r^2cos(\theta)\vec{i}+ 2r^2sin(\theta)\vec{j}- r\vec{k})drd\theta$ . Also, the paraboloid by itself is not a "closed" surface. The circular surface is given by "z= 1" which has "position vector" $rcos(\theta)\vec{i}+ rsin(\theta)\vec{j}+ \vec{k}$ so its "fundamental vector product is
$\left|\begin{array}{ccc}\vec{i} & \vec{j} & \vec{k} \\ cos(\theta) & sin(\theta) & 0\\ -rsin(\theta) & rcos(\theta) & 0\end{array}\right|= r\vec{k}$ , the outward being $r\vec{k}$ upward, of course, and the differential is $d\vec{S}= \vec{n}dS= r\vec{k}drd\theta$ .

With $\vec{F}= r^2cos(\theta)\vec{i}+ r^2sin(\theta)\vec{j}+ 3r^4\vec{k}$ , your surface integral will be the sum of
$\int_{r= 0}^{1}\int_{y= 0}^{2\pi}(r^2cos(\theta)\vec{i}+ r^2sin(\theta)\vec{j}+ 3r^4\vec{k})\cdot(2r^2cos(\theta)\vec{i}+ 2r^2sin(\theta)\vec{j}- r^4\vec{k})drd\theta$ $= \int_{r=0}^1\int_{\theta= 0}^{2\pi}2r^3- r^8 d\theta dr$ and $\int_{r= 0}^{1}\int_{\theta= 0}^{2\pi}(r^2cos(\theta)\vec{i}+ r^2sin(\theta)\vec{j}+ 3r^4\vec{k}) \cdot r\vec{k}drd\theta$ .

**arbolis** · July 19th 2009, 08:01 PM

Thank you very much HallofIvy for the detailed answer.
I think you made a sign error in the first determinant, but this is not really important.
However I'm confused about when you wrote

With

, your surface integral will be the sum of

, it seems to me that you considered $z=r$ and then $z=r^2$ . I don't know which to chose!

After I complete the problem I'll read it like 10 times, drawing the solids, etc. I need to grasp all this!

**HallsofIvy** · July 20th 2009, 02:36 AM

Where did I use z= r? I clearly said $z= x^2+ y^2= r^2$ on the parabola.

I did say that the "fundamental vector product" is $<br /> = -2r^2cos(\theta)\vec{i}- 2r^2sin(\theta)\vec{j}+ r\vec{k}<br />$ which has "r" times $\vec{k}$ but that has nothing to do with the value of z.

**arbolis** · July 20th 2009, 06:46 AM

Here : $\vec{F}= r^2cos(\theta)\vec{i}+ r^2sin(\theta)\vec{j}+ 3r^4\vec{k}$

$F(x,y,z)=(xz,yz,3z^2)$ . $x=r\cos (\theta) \vec i$ , $z=r^2$ so the first component of $F$ is $r^3 \cos (\theta) \vec i$ . (You wrote $r^2 \cos (\theta) \vec i$ .) And the " $k$ " component is indeed what you wrote, $3(r^2)^2=3r^4$ .
Thanks for your help.

Thread: Divergence

LinkBack

Thread Tools

Display

Divergence

Similar Math Help Forum Discussions

divergence

divergence of {an} - help please

Divergence ~ ln(ln(ln x))

Divergence

divergence

Search Tags