## *screams in anger* ok, divergence theorem problem

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

-_-
:'(

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 Quote by schattenjaeger over the region x^2+y^2+z^2>=25
is that actually a $<=$ ? (you say the region is a sphere...)
 F=((x^2+y^2+z^2)(xi+yj+zk)), so del F would be 3(x^2+y^2+z^2)?
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.

 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)it's be like , 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?

## *screams in anger* ok, divergence theorem problem

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>

 so F=
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...

 Well that was a dumb mistake. Remember kids, sleep is gooood
 So with that in mind how the devil is the answer what it is?
 because using the correct divergence and going through it I'm getting like 12500pi
 schattenjaeger, I hope you're awake after a long . 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.