We have a cylinder of radius $a$ and height $h.$ We need to prove its volume to be equal to $\pi a^2 h$ using triple integral and spherical coordinates.
The best way to solve this problem is to divide the cylinder to two volume regions, the first region is the one defined by $\phi$ to range from $0$ to $\arctan(\frac{a}{h})$ and the second region is the one defined by $\phi$ to range from $\arctan(\frac{a}{h})$ till $\frac{\pi}{2}$.
Surley, for both regions $\theta$ will vary from $0$ till $2\pi$. However, I tried many times to find the proper limits for the radius for each region but I failed. So how do I properly define the radius for each region ?
Edit : I know the limits of each region for the radius $r= 0$ till $a\secθ$ and $r= 0$ till $b\cscθ$ but my question exactly is how to drive them ?
I would say that to find the volume of a cylinder use cylindrical coordinates.
But you want to use spherical.
Your boundaries
$x^2 + y^2 = a^2\\ z = 0\\ z = h$
$x = \rho\cos\theta\sin\phi\\ y = \rho\sin\theta\sin\phi\\ z = \rho\cos \phi$
$\rho^2\cos^2 \theta\sin^2\phi + \rho^2\sin^2 \theta\sin^2\phi = a^2\\ \rho^2\sin^2\phi = a^2\\ \rho = a\csc\phi$
$\rho\cos \phi = h\\ \rho = h\sec \phi$
$a\csc\phi = h\sec \phi\\ \tan\phi = \frac ah$
$\int_\limits0^{2\pi}\int_\limits0^{\arctan \frac {a}{h}}\int_\limits0^{h\sec \phi} \rho^2\sin \phi \ d\rho\ d\phi\ d\theta\\ + \int_\limits0^{2\pi}\int_\limits{\arctan \frac {a}{h}}^\frac {\pi}{2}\int_\limits0^{a\csc \phi} \rho^2\sin \phi \ d\rho\ d\phi\ d\theta$