# Triple Integration of a Sphere in Cylindrical Coordinates

1. Apr 2, 2009

### Zarlucicil

1. The problem statement, all variables and given/known data

The problem was to find the volume enclosed by a sphere of radius "a" centered on the origin by crafting a triple integral and solving for it using cylindrical coordinates.

2. Relevant equations

$$x^{2}+y^{2}+z^{2}=a^{2}$$ : Equation for a sphere of radius "a" centered on the origin.

$$\iiint\limits_E dV$$ : Triple integral for finding volume of a region $$E$$.

3. The attempt at a solution

I solved the triple integral (but I don't think it's right) and got this: $$\frac{4}{3}a^{2}\pi$$ ---> Actually, I think I solved the integral right, but I think my bounds are incorrect.

I used the following as my bounds and subsequent iterated integral:

$$E=\{ \ (r,\theta,z) \ | \ 0\leq r\leq a, \ 0\leq \theta\leq 2\pi, \ -\sqrt{a^2-r^2}\leq z\leq \sqrt{a^2-r^2} \ \}$$

$$\int^{2\pi}_{0}\int^{a}_{0}\int^{\sqrt{a^2-r^2}}_{\sqrt{a^2-r^2}} dz dr d\theta$$

If my proposed answer isn't right could the problem lie within my bounds? I'm not really great at determining the bounds for iterated integrals yet >.<'

Thanks :D

2. Apr 2, 2009

### tiny-tim

Hi Zarlucicil!

(have a pi: π and a theta: θ and try using the X2 tag just above the Reply box )

erm … it isn't dz dr dθ, it's … ?

3. Apr 2, 2009

### xepma

Well, your bounds are in fact correct. What's missing is a volume element, namely, you need to have $$r drd\theta dz$$. But maybe you just missed it, since without it you probably won't get the final answer you mentioned (which isn't correct by the way - you need $$4\pi a^3/3$$).

You can also write $$r$$ in terms of $$z$$, which is in my opinion a bit more intuitive. In that case you get:

$$\int^{2\pi}_{0}\int^{-a}_{a}\int^{0}_{\sqrt{a^2-z^2}} r dr dz d\theta$$