Vector calculus: divergence theorem! help!

**TheFirstOrder** · June 9th 2010, 06:47 AM

Using the divergence thm, evaluate the following equation

$\int\int_{s}(a^{2}x^{2}+b^{2}y^{2}+c^{2}z^{2})^{-1/2}dS$

where S is the surface of the ellipsoid $ax^{2}+by^{2}+cz^{2}=1$

I am suck on this question because I have only done them for spheres.

I am not sure of how to get the normal and then what to do from there. I know it seems like I haven't done a lot on this question, but I am struggling to see how this question would be done.

**Ackbeet** · June 9th 2010, 06:56 AM

If you're supposed to use the divergence theorem, then use the divergence theorem. You can essentially take a derivative inside the integral sign, and integrate over the volume instead.

**TheFirstOrder** · June 9th 2010, 08:11 AM

Originally Posted by Ackbeet

If you're supposed to use the divergence theorem, then use the divergence theorem. You can essentially take a derivative inside the integral sign, and integrate over the volume instead.

Okay, so I should get something like this?

$\int\int\int_{v}-1/2(a^{2}x^{2}+b^{2}y^{2}+c^{2}z^{2})^{-3/2}(2a^{2}x+2b^{2}y+2c^{2}z)dV$

**Ackbeet** · June 9th 2010, 09:20 AM

Maybe... The divergence theorem states that

$\iiint_{V}(\nabla\cdot\vec{F})\,dV=\iint_{S}\vec{F }\cdot\hat{n}\,dS$ .

You're going to have to get the surface integral you started with to look like the RHS of the divergence theorem. So I think your original question still stands: how to get the normal vector.

How do you get the normal to the surface? The gradient is your answer. According to the wiki, if you have a surface defined implicitly: $G(x,y,z)=0$ , then the normal at point $\langle t,u,v\rangle$ is given by $\nabla G(t,u,v)$ .

So, with your $G(x,y,z)=ax^{2}+by^{2}+cz^{2}-1=0$ , the gradient vector is $\langle 2ax, 2by, 2cz\rangle$ .

The formula in the divergence theorem, however, wants to see a unit normal vector. So, we take this result and divide by its length:

$\hat{n}=\frac{\langle 2ax, 2by, 2cz\rangle}{2\sqrt{a^{2}x^{2}+b^{2}y^{2}+c^{2}z^{2 }}}<br /> =\frac{\langle ax, by, cz\rangle}{\sqrt{a^{2}x^{2}+b^{2}y^{2}+c^{2}z^{2}} }$ .

Next, we want to recover $\vec{F}$ . That is, we need

$\vec{F}\cdot\hat{n}=\sqrt{a^{2}x^{2}+b^{2}y^{2}+c^ {2}z^{2}}$ (the RHS of this equation is your original integrand), or
$\vec{F}\cdot\frac{\langle ax, by, cz\rangle}{\sqrt{a^{2}x^{2}+b^{2}y^{2}+c^{2}z^{2}} }=\sqrt{a^{2}x^{2}+b^{2}y^{2}+c^{2}z^{2}}$ .

It jumps out to me that it must be that $\vec{F}=\langle ax,by,cz\rangle$ . This is then the function of which you must take the divergence, which I think you'll agree is much easier. It had to be a vector function, in any case, right?

I think you'll find the integrand of the volume integral to be very simple indeed. You might even find the problem reduced to finding the volume of an ellipsoid.

Hope this helps.

**TheFirstOrder** · June 9th 2010, 10:11 AM

Okay, that makes sense to me. I just didn't know where to start.Thanks for helping!

**Ackbeet** · June 9th 2010, 10:13 AM

You're very welcome. Have fun.

**TheFirstOrder** · June 10th 2010, 07:33 AM

Sorry, I just realised I make a typo in the original question - it is to the power of -1/2 not 1/2. But would we expect to get the same F?
EDIT: sorry, please ignore, it's 1.38 am and i'm not thinking straight. got the answer :P

**Ackbeet** · June 11th 2010, 08:57 AM

Can you please re-type the problem exactly? Thanks!

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