English Deutsch Français Italiano Español Português 繁體中文 Bahasa Indonesia Tiếng Việt ภาษาไทย
All categories

Hello,

Currently in calc we're learning about cylindrical and spherical coordinates.

One of the questions that I cannot figure out is finding the volume of a cylinder with height H and radius R using SPHERICAL coordinates not cylindrical. Assuming that phi is the angle in the xy-plane and theta the azimuthal angle from the z-axis. So the format would be ∫∫∫ p^2 sinθ dp dθ dφ . So I think the first integral; the limits of phi should be from 0 to 2π but the next two I don't know.

Any help would be appreciated

thanks

2007-12-03 04:46:59 · 2 answers · asked by Jason 2 in Science & Mathematics Mathematics

2 answers

It is very helpful if you first draw a picture, which you've probably already done. Put the centre of the bottom of the cylinder at the origin.

Next observe that you will need to express the volume as the sum of two integrals. In the first, theta (azimuthal angle) goes from 0 to 2 pi, r goes from 0 to the top surface, and phi (polar angle) goes from 0 to the angle at the "corner", which is arctan R/H. In the second, theta still goes from 0 to 2 pi, r goes from 0 to the side surface, and phi goes from arctan R/H to pi/2.

The top surface, which would be z = H if you were using cylindrical coordinates, is now r cos phi = H (which is r = H sec phi) in spherical coordinates (you have probably been introduced to the conversion factors relating (x,y,z) to (r,theta,phi) in your class). The side surface would be described by r^2 = R^2 + z^2 in cylindrical coordinates, is now, by the same conversion, r^2 = R^2 + r^2 (cos phi)^2. After a little simplification this is expressed as r^2 = R^2 (csc phi)^2, or r = R csc phi.

Now, the first integral is

Int(theta from 0 to 2 pi) Int(phi from 0 to arctan R/H) Int(r from 0 to H sec phi) r^2 sin phi dr dphi dtheta,

and the second is

Int(theta from 0 to 2 pi) Int(phi from arctan R/H to pi/2) Int(r from 0 to R csc phi) r^2 sin phi dr dphi dtheta.

I hope you can take it from there.

2007-12-03 05:26:24 · answer #1 · answered by acafrao341 5 · 1 0

Suggest you just find the volume of 1/4 0f the cylinder and multiply reult by 4. Then integral becomes
4∫∫∫ p^2 dp dθ dφ , where phi goes from 0 to pi/2, rho goes from 0 to sqrt(R^2+H^2), and theta goes from 0 to arcsin(H/R)

2007-12-03 13:13:14 · answer #2 · answered by ironduke8159 7 · 0 1

fedest.com, questions and answers