1. triple integrals

the integral from 0 to 2pi, the integral from 2 to 3, the integral from sqrt(4-r^2) to sqrt(9-r^2)

of (1)/(r^2+z^2)^pi dz dr dtheta

this is in cylindrical coordinates.
how do i finish this integral by switching to other coordinates? should i switch to spherical or polar?? thanks!

2. Originally Posted by holly123
the integral from 0 to 2pi, the integral from 2 to 3, the integral from sqrt(4-r^2) to sqrt(9-r^2)

of (1)/(r^2+z^2)^pi dz dr dtheta

this is in cylindrical coordinates.
how do i finish this integral by switching to other coordinates? should i switch to spherical or polar?? thanks!
Is this what you're asking for help on?

$\int_0^{2\pi}\int_2^3\int_{\sqrt{4-r^2}}^{\sqrt{9-r^2}}\frac{1}{(r^2+z^2)^{\pi}}\,dz\,dr\,d\theta$

If so, I'd go spherical.

Hint 1:

$r\to\rho\cos\phi$
$z\to\rho\sin\phi$
$\theta\to\theta$

You have to multiply by the determinant of the Jacobian:

$J(r,z,\theta)=\left|\begin{array}{ccc}\cos\phi&-\rho\sin\phi&0\\\sin\phi&\rho\cos\phi&0\\0&0&1\end {array}\right|=\rho$

Hint 2:
Spoiler:
Since $r^2+z^2=\rho^2\cos^2\phi+\rho^2\sin^2\phi=\rho^2$, the new integral is:

$\int\int\int\frac{J(r,z,\theta)}{(\rho^2)^{\pi}}\, d\rho\,d\phi\,d\theta=
\int\int\int\frac{1}{\rho^{2\pi-1}}\,d\rho\,d\phi\,d\theta$

Figure out the new bounds and integrate.

3. thank you so much! are the bounds
phi is between 0 and pi
theta is between 0 and 2 pi
rho is between 0 and pi/4

4. Originally Posted by holly123
thank you so much! are the bounds
phi is between 0 and pi
theta is between 0 and 2 pi
rho is between 0 and pi/4
$\rho$ shouldn't be an angle. It's the radius of the sphere. In the original integral, $z=\sqrt{4-r^2}\implies z^2+r^2=4\implies \rho^2=4\implies \rho=2$ Similarly, the upper bound is $\rho=3$.

I believe the others are right.