Triple integrals

1. Mar 24, 2005

physicsss

Find the mass of a ball B given by "x^2+y^2+z^2≤a^2" if the density at any point is proportional to its distance from the z-axis using cylindrical coordinates So is the density equal to K*sqrt(x^2+y^2), or K*r?

Using triple integral of f(rcosθ, rsinθ, z)*r*dz*dr*dθ) I got the following: the triple integral of K*r^2dz*dr*dθ, and the limits for the integral w/ respect to z being from -sqrt(a^2-x^2-y^2) to sqrt(a^2-x^2-y^2), which becomes -sqrt(a^2-r^2) to sqrt(a^2-r^2), limits w/ respect to r being from 0 to a and w/ respect to theta from 0 to 2*pi.

Doing it I find the integral very hard to integrate because I can't do u-subsititon with two r^2's.

Am I doing anythign wrong? Thanks in advance.

Last edited: Mar 24, 2005
2. Mar 24, 2005

dextercioby

U can very well use spherical coordinates.I'm sure the integration is immediate.

Daniel.

3. Mar 24, 2005

physicsss

I can't. the stupid book asks us to use cylindrical coordinates.

Is my setup right though?

4. Mar 24, 2005

dextercioby

Why?What's the point...?Why is it hard to integrate...

$$M=k\int_{0}^{2\pi}d\varphi \int_{0}^{a}r^{2} \ dr \int_{-\sqrt{a^{2}-r^{2}}}^{+\sqrt{a^{2}-r^{2}}} \ dz$$

I think it's trivial.

Daniel.

Last edited: Mar 24, 2005
5. Mar 24, 2005

physicsss

You forgot a r...in cylindrical coordinates you there's an extra r.

6. Mar 24, 2005

dextercioby

You're right,i edited.Then do it...It's not difficult.You can make a "sin/cos" substitution and solve it easily.

Daniel.