# Charge density problem

1. May 17, 2008

### stargoo

1. The problem statement, all variables and given/known data
Find the total charge inside the volume indicated:
$$\rho_v=10z^2\rho^{-0.1x}\sin(y\pi) for -1\leq x\leq 2,0\leq y\leq 1,3\leq z \leq3.6$$

I know I have to integrate over the volume [tex]dxdydz[tex], but [tex]\rho^{-.1x}[tex] just keeps giving me a problem. Is there a substitution that I'm missing here. Integrating by parts just gives a more complicated integral. I've tried converting [tex]\rho[tex] to [tex]\sqrt{x^2+y^2}[tex], but that doesn't help either. Any help would be appreciated.

2. May 17, 2008

### rootX

got this from google [integral] b^{x} dx = b^{x} / ln(b) + C

Simple: look in integral tables :P

3. May 17, 2008

### stargoo

that would be all well and good, except that rho is not a constant in this case. The charge density in the stated problem is a function of cartesian AND cylindrical variables. Hence, my problem.

4. May 17, 2008

### rootX

oops, I couldn't read properly the question..

It seems a real big mess to me ><

5. May 17, 2008

### stargoo

thanks for trying, though!

6. May 18, 2008

### dpowd

Are you sure rho is not a constant? The way I read the problem is rho(sub-v) represents some total charge (although that notation would seem a bit unconventional) defined as a function of x, y, z, and some undeclared (but constant) rho. From that point of view, the integral really isn't as ugly as it looks. The triple integral is performed by integrating with respect to one variable at a time, holding the others constant. Your final answer will just be in terms of rho.
If rho were variable, then you have a variable defined as a function of its anti-derivative, which is a differential equation, and you have a mess on your hands.
Good Luck!