# Prove that the volume of a cylinder is pi*a^2*h using triple integration and spherical coordinates

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(a/h) and the second region is the one defined by phi to range from arctan(a/h) till 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 asecθ and r= 0 till bcscθ but my question exactly is how to drive them ?

• – Lord Shark the Unknown Nov 13 '18 at 21:03
• The equations for vertical and horizontal lines in polar coordinate are $r=a \sec \theta$ and $r=b\csc \theta.$ You can adapt these readily. – B. Goddard Nov 13 '18 at 21:07
• @LordSharktheUnknown Ok – John adams Nov 13 '18 at 21:08
• @B.Goddard Please explain how do you drive them – John adams Nov 13 '18 at 21:08
• Start with $y=a$ is the same as $r\sin \theta = a.$ – B. Goddard Nov 13 '18 at 21:27

I would say that to find the volume of a cylinder use cylindrical coordinates.

But you want to use spherical.

$$x^2 + y^2 = a^2\\ z = 0\\ z = h$$

$$x = \rho\cos\theta\sin\phi\\ y = \rho\sin\theta\sin\phi\\ z = \rho\cos \phi$$

$$\rho^2\cos^2 \theta\sin^2\phi + \rho^2\sin^2 \theta\sin^2\phi = a^2\\ \rho^2\sin^2\phi = a^2\\ \rho = a\csc\phi$$

$$\rho\cos \phi = h\\ \rho = h\sec \phi$$

$$a\csc\phi = h\sec \phi\\ \tan\phi = \frac ah$$

$$\int_\limits0^{2\pi}\int_\limits0^{\arctan \frac {a}{h}}\int_\limits0^{h\sec \phi} \rho^2\sin \phi \ d\rho\ d\phi\ d\theta\\ + \int_\limits0^{2\pi}\int_\limits{\arctan \frac {a}{h}}^\frac {\pi}{2}\int_\limits0^{a\csc \phi} \rho^2\sin \phi \ d\rho\ d\phi\ d\theta$$

• How did you drive the limits of upper limits of the radius for each region ? – John adams Nov 13 '18 at 21:17
• @Johnadams Did you ask this question before I completed my last edit? Is it all clear now? – Doug M Nov 13 '18 at 21:29
• Yes , Thank you very much. – John adams Nov 13 '18 at 21:30