# Help with triple integration problem

1. Aug 25, 2005

### mister_okay

hey! i need some help with a triple integration problem using spherical coordinates. it's the volume of a small part of a sphere. rho from 5 to 6, phi from pi/6 to pi/4 and theta from pi/4 to pi/3.

i got an answer of (-91/72) x pi x (sqrt(2)-sqrt(3))...am i right? Thanks!

2. Aug 25, 2005

### arildno

Well, the answer ought to be:
$$\frac{6^{3}-5^{3}}{3}(\frac{\pi}{3}-\frac{\pi}{4})(\cos\frac{\pi}{6}-\cos\frac{\pi}{4})$$
So, if your answer is just a simplification of this, then it is correct.

EDIT:
Yup, it is correct.

Last edited: Aug 25, 2005
3. Aug 25, 2005

### mister_okay

Hey thanks arildno :)

but one question..

shouldnt it be cos(pi/4)-cos(pi/6) instead of cos(pi/6)-cos(pi/4)?

4. Aug 25, 2005

### iNCREDiBLE

No, since:

$$\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} sin\phi d\phi = [-cos\phi]_{\frac{\pi}{6}}^{\frac{\pi}{4}} = -cos(\frac{\pi}{4}) - (-cos(\frac{\pi}{6})) = cos(\frac{\pi}{6}) - cos(\frac{\pi}{4})$$

Last edited: Aug 25, 2005
5. Aug 26, 2005

### mister_okay

oh ok...forgot about the negative. thanks!