# *screams in anger* ok, divergence theorem problem

1. Oct 14, 2005

### schattenjaeger

question says answer in whichever would be easier, the surface integral or the triple integral, then gives me(I'm in a mad hurry, excuse the lack of formatting...stuff)

the triple integral of del F over the region x^2+y^2+z^2>=25
F=((x^2+y^2+z^2)(xi+yj+zk)), so del F would be 3(x^2+y^2+z^2)?

In this case, the region is a sphere, it'd be easier to do it with the triple integral over the volume, right? Well, regardless I tried it that way, converting to spherical coordinates(my book mixes up the traditional phi and theta placement, but whatever)

triple integral of 3r^4*sin(theta)drd(theta)d(phi), and the limits of integration, going from the right integral to the left, 0-5, 0-pi, 0-2pi?

And somewhere before there is where I messed up 'cuz I can do the integral I have there easily enough and I get like some huge square of 5 times pi, and the answer is 100pi

-_-
:'(

2. Oct 14, 2005

### rachmaninoff

is that actually a $<=$ ? (you say the region is a sphere...)
I don't understand your notation. If you what you meant was $$\left( x^3 \hat{i} + y^3 \hat{j} + z^3 \hat{k} \right)$$, then your $$\nabla \mathbf{\vec{F}}$$ is correct. But what you wrote is $$\left( x^2 + y^2 + z^2 \right)\left( x \hat{i} + y \hat{j} + z \hat{k} \right)$$
$$=\left( x \left( x^2 + y^2 + z^2 \right) \hat{i} + y \left( x^2 + y^2 + z^2 \right) \hat{j} + z \left( x^2 + y^2 + z^2 \right)\hat{k} \right)$$, which has a different divergence.

Last edited by a moderator: Oct 14, 2005
3. Oct 14, 2005

### schattenjaeger

Oh I love you, you're fast
yes I mean <=
but yah, the latter is what I wrote and meant, but I'm confused, isn't the (x^2+y^2+z^2) like...if you just had a constant A, times (xi+yj+zk)(or A<x,y,z>)it's be like <Ax,Ay,Az>, so can you not do that with the (x^2+y^2+z^2) out front?
am I doing something stupid in the midst of night?

4. Oct 14, 2005

### schattenjaeger

F=((x^2+y^2+z^2)(xi+yj+zk)), so del F would be 3(x^2+y^2+z^2)?

I'll throw in the intermediate step I did too,

so F=<x^3+y^2+z^2,x^2+y^3+z^3,similarforz>

and the divergence would be the same in that case as <x^3,y^3,z^3>

5. Oct 14, 2005

### rachmaninoff

Not quite:

$$\left(x^2+y^2+z^2 \right)(x \mathbf{\hat{i}} \right))=\left(x^3+\mathbf{x}y^2+\mathbf{x}z^2 \right) \mathbf{\hat{i}}$$
and similarly...

6. Oct 14, 2005

### schattenjaeger

Well that was a dumb mistake. Remember kids, sleep is gooood

7. Oct 14, 2005

### schattenjaeger

So with that in mind

how the devil is the answer what it is?

8. Oct 14, 2005

### schattenjaeger

because using the correct divergence and going through it I'm getting like 12500pi

9. Dec 19, 2005

### neutrino

schattenjaeger,
I hope you're awake after a long :zzz: . I was having trouble with the exact problem,from Boas' book I presume, and a search for "divergence theorem" brought me here. Anyway, I arrived at the answer given at the back of the book,viz. $$4\pi5^5$$, by solving the surface integral. You also get it when
$$\nabla.\vec{F} = 3(x^2+y^2+z^2)$$,in the case of a volume integral, which is obviously wrong. So did you eventually solve it using the triple integral?

Anyone's help will be appreciated.

Last edited: Dec 19, 2005