Integrating Non-Constant Charge Density in a Volume: Tips and Tricks

AI Thread Summary
The discussion focuses on integrating a non-constant charge density function, ρ_v = 10z^2ρ^{-0.1x}sin(yπ), over a specified volume. The main challenge arises from the non-constant nature of ρ, complicating the integration process. Participants suggest that the integration should be approached by treating the variables independently and integrating one at a time while holding others constant. Clarification is provided that ρ is not a variable but rather a constant, simplifying the integration. Ultimately, the integral can be managed without excessive complication if approached correctly.
stargoo
Messages
3
Reaction score
0

Homework Statement


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 dxdydz, but \rho^{-.1x}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 \rhoto \sqrt{x^2+y^2}, but that doesn't help either. Any help would be appreciated.
 
Physics news on Phys.org
got this from google [integral] b^{x} dx = b^{x} / ln(b) + C

Simple: look in integral tables :P
 
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.
 
oops, I couldn't read properly the question..

It seems a real big mess to me ><
 
thanks for trying, though!
 
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!
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top