# Evaluating Volume Integrals and Divergence Theorm

1. Nov 30, 2013

### Thadis

1. The problem statement, all variables and given/known data
Evaluate the integral as either a volume integral of a surface integral, whichever is easier.

$\iiint \nabla .F\,d\tau$ over the region $x^2+y^2+z^2 \leq 25$, where $F=(x^2+y^2+z^2)(x*i+y*j+z*k)$

2. Relevant equations
$\iiint \nabla .F\,d\tau =\iint F.n\,d\sigma$

3. The attempt at a solution
I believe this would be easier to do as a volume integral though I am honestly not sure saying I am not really understand how things like line integrals and surface integrals work in the sense of things like Stokes Theorm and Div. Theorm.

For the volume intregral I did just integrated the Div(F) from -5 to 5 for all three integrals saying that is the range each of the variables will go through. I plugged this into Wolfram Alpha and got an answer of 25000 though I am not sure if I did this question correctly.

If I wanted to do a solve this using the right hand side of the Div. Theorem how would I approach this?

2. Nov 30, 2013

### LCKurtz

If you let all three variables go from -5 to 5 you are describing a region shaped like a rectangular block. You have a sphere.

3. Nov 30, 2013

### vela

Staff Emeritus
That's not what I got from doing what you said you did, which suggests you didn't calculate div F correctly. Show us how you calculated div F.

You may want to switch to spherical coordinates to do the integrals.

4. Nov 30, 2013

### Thadis

Ahh, I knew I was forgetting something. For this case once I converted x,y, and z into sphrerical coordinates wouldnt the limits of integration be r=0 to r=5, θ= 0 to 2pi, and $\phi$=0 to 2pi? Also just to double check. Would taking the volume integral actually easier in this case?

5. Nov 30, 2013

### Thadis

I was just thinking about it but would actually turning it into a surface integral make it easier? Saying you would have the double integral of the force dotted into the normal, which would give out $(x^2+y^2+z^2)^2/5$ since the normal to the sphere is just the normalized position vector? And since this is only going to be on the surface of the sphere we know that $(x^2+y^2+z^2)=r=5$ . So would the answer then be 20pi^2? Since I would be able to draw out a consant 5, and then all that would be left would to do the integral that has no terms left inside with the limits of the two intergals both being from 0 to 2pi?

6. Nov 30, 2013

### vela

Staff Emeritus
The answer is $12500\pi$.

7. Nov 30, 2013

### vela

Staff Emeritus
You're saying the integral is equal to that, or F dotted into the normal? Where did the 5 come from?

$r = \sqrt{x^2+y^2+z^2}$

Those limits aren't correct. The polar angle doesn't go from 0 to $2\pi$.

8. Dec 1, 2013

### Thadis

Sorry, was typing that in a hurry saying I was having to leave right then.

What I was trying to say, since the normal to the sphere is the normalized position vector, which would be ${x,y,z}/5$ since we know the distance to the surface is 5. If we dot this into the force that would create, $(x^2+y^2+z^2)*{x,y,z}.{x,y,z}/5$ after doing the dot product turns into $(x^2+y^2+z^2)^2/5$.

After this I did mess up and say (x^2+y^2+z^2)=5 where I should of said it is equal to 25. So that means the integrad is equal to 25^2/5?

And since the azimuthal angle is rotating completely around that means the polar angle only actually needs to go from 0 to pi correct?

I must be making some sort of mistake somewhere saying the answer I am getting is only like 250pi. Is there anything that you can see that I am doing wrong? Or is my entire train of thought forgetting a crucial point?

9. Dec 1, 2013

### vela

Staff Emeritus
What did you use for $d\sigma$? Remember it's proportional to $r^2$.

10. Dec 1, 2013

### Thadis

Ahh $d\sigma=r^2sin(\theta) d\theta d\phi$. Forgot that it was not just dd/theta d\phi So that means that it would be $\iint r^6/5*sin(\theta)*d\theta*d\phi$ if the limits are \theta=0 to pi and \phi=0 to 2pi?

11. Dec 1, 2013

### vela

Staff Emeritus
Yup.

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