Triple integral in spherical coordinates

hitemup
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Homework Statement



Evaluate
\int \int \int _R (x^2+y^2+z^2)dV

where R is the cylinder
0\leq x^2+y^2\leq a^2,
0\leq z\leq h

Homework Equations


[/B]
x = Rsin\phi cos\theta
y = Rsin\phi sin\theta
z = Rcos\phi

The Attempt at a Solution


[/B]
2*\int_{0}^{\pi/2}d\phi \int_{0}^{2\pi}d\theta \int_{0}^{h/cos\phi}dR R^4sin\phi

Are the bounds correct?
 
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Do you have to use spherical co-ordinates, why not cylindrical ones ?
 
hitemup said:

Homework Statement



Evaluate
\int \int \int _R (x^2+y^2+z^2)dV

where R is the cylinder
0\leq x^2+y^2\leq a^2,
0\leq z\leq h

Homework Equations


[/B]
x = Rsin\phi cos\theta
y = Rsin\phi sin\theta
z = Rcos\phi

The Attempt at a Solution


[/B]
2*\int_{0}^{\pi/2}d\phi \int_{0}^{2\pi}d\theta \int_{0}^{h/cos\phi}dR R^4sin\phi

Are the bounds correct?
This is not set properly if you insist on using spherical coordinates. You would have to break it up into two integrals, one where ##\rho## goes from ##0## to the side of the cylinder and another where it goes from ##0## to the top. A much better choice is cylindrical coordinates. Try setting it up that way and check back with us.
 
Yes I realized at one time it would be easier if I used cylindrical coordinates, since the region itself is a cylinder. But I just wanted to know how to proceed with spherical coordinates.
 
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hitemup said:
Yes I realized at one time it would be easier if I used cylindrical coordinates, since the region itself is a cylinder. But I just wanted to know how to proceed with spherical coordinates.

OK, so give it a go. First set up$$
\int_{\theta \text{ limits}}\int_{\phi \text{ limits}}\int_0^{\rho \text{ on the sides in terms of } \phi} \text{Integrand }
\rho^2\sin\phi~d\rho d\phi d\theta$$then add$$
\int_{\theta \text{ limits}}\int_{\phi \text{ limits}}\int_0^{\rho \text{ on the top in terms of } \phi} \text{Integrand }\rho^2\sin\phi~d\rho d\phi d\theta$$
 
EDIT : Sorry, i thaught they were cylindrical
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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