Triple intergration

b0bd0d

Oh yeah, doing triple integration in rectangle, spherical, and cylindrical coordinates. Help. Me. Please. I need help with converting back and forth. I'm having a hard time seeing what is going on with the limits. I'm searching the Googles, anybody got anything else for me?

Dance Commander

You are going to need to be a lot more specific.

Raiden333

Are you remembering to triangulate the polar matrices to offset the bound limits?

Taximes

Are you making the connection between the triple integral and integrating over a volume?

For example, if you want to integrate over a cube (let's say it's 3x4x5), you would find the volume of that by calculating x * y * z, right? To integrate over the volume, it's just ∫∫∫(something) dx dy dz, where the limits are x=0 to 3, y=0 to 4, z=0 to 5.

Cylindrical/spherical are the same, except that volume is computed differently, so it's, for example, r drdθdz in cylindrical (r dθ is the arc length, times width dr, times height dz).

b0bd0d

Okay example problems:
http://www.math.siu.edu/classes/math251-1-086/38-Practice%20Exam%203.pdf

Boutros

Like for problem 4 you have a cone topped by part of a sphere, so your limits on the phi, the angle from the z axis are from 0 to half the base angle of the cone, limits on r are 0 to the radius of the sphere, and limits on theta are 0 to 2*pi.

In cylindrical it is a lot tougher, I would split the integral into two parts, the easy cone part and the sphere section, and the limits of integration for r in the sphere section would be some kind of nasty function of z (or you could have z as a nasty function of r and have simpler limits on r).

You are usually only going to use spherical or cylindrical coordinates when the symmetry of the problem makes sense to do so, so don't find you limits in cartesian and convert, think about the geometry of the system in the coordinate system you are integrating in.