# Triple integral in spherical coordinates

1. Feb 29, 2012

### tintin2006

I want to check if I'm doing this problem correctly.
1. The problem statement, all variables and given/known data
Region bounded by $x^2+y^2=4$ and bounded by the surfaces z = 0, and $z=\sqrt{9-x^2-y^2}$.
Set up triple integrals which represent the volume of the solid using spherical coordinates.

2. Relevant equations
$\int\int\int_{V}\rho^2sin\phi \; d\rho d\phi d\theta$

3. The attempt at a solution
The shape is a cylinder with a round top.
It seems that I have to break this into two integrals:
1:
From $x^2+y^2=4$ and $z=\sqrt{9-x^2-y^2}$:
$z=\sqrt{9-4}=\sqrt{5}$
Since $z=\rho cos\phi$, $\rho cos\phi = \sqrt{5} \Rightarrow \rho = \frac{\sqrt{5}}{cos\phi}$
2:
From $z=\sqrt{9-x^2-y^2}$, it is a sphere of radius 3 so ρ=3
The sphere and cylinder meets at point (y,z)=(2, √5) and the radius makes an angle with the point:
$sin\phi = \frac{2}{3}$
$\int_{0}^{2\pi}\int_{0}^{arcsin(\frac{2}{3})}\int_{0}^{3}\rho^2sin\phi \; d\rho d\phi d\theta + \int_{0}^{2\pi}\int_{arcsin(\frac{2}{3})}^{\frac{ \pi }{2}}\int_{0}^{\frac{\sqrt{5}}{cos\phi}}\rho^2 sin \phi \; d\rho d\phi d\theta$
Thank you.

2. Feb 29, 2012

### SammyS

Staff Emeritus
Hello tintin2006. Welcome to PF !

That all looks fine to me.

3. Feb 29, 2012

Thanks :)