Multivariable Calculus cylindrical coordinates

[math040809]

Could someone help me with this "multivariable Calculus cylindrical coordinates" problem:
A spherical solid has radius a, and the density at P(x,y,z) is directly proportional to the distance from P to a fixed line L through the center of the solid. Find its mass.

I don't know how to solve the problem. I know that I'm suppose to set up a triple integral that uses cylindrical coordinates.
I also know that the rectangular coordinates (x,y,z) and the cylindrical coordinates (r,theta, z) of a pt P are related as follows: x=r cos theta, y=r sin theta, tan theta = y/x, r^2=x^2+y^2, z= z. Also, a triple integral w/ cylindrical coordinates is evaluated as: [integral from alpha to beta, integral from g1(theta) to g2(theta) integral from k1(r,theta) to k2(r, theta) of integrand (f(r,theta,z)r) dz, dr, d(theta)].

My attempt: I've tried setting up the triple integral as: [integral from 0 to 2pi, integral from 0 to a, integral from 0 to sqrt(a^2-r^2) of integrand (kz)r dz, dr, d(theta)]. I got the answer as [k(a^4)(pi)]/4, BUT this is the wrong answer, therefore my triple integral is set up incorrectly.
Thanks for any help.

 

[DrMike]

Could someone help me with this "multivariable Calculus cylindrical coordinates" problem:
A spherical solid has radius a, and the density at P(x,y,z) is directly proportional to the distance from P to a fixed line L through the center of the solid. Find its mass.

I don't know how to solve the problem. I know that I'm suppose to set up a triple integral that uses cylindrical coordinates.
I also know that the rectangular coordinates (x,y,z) and the cylindrical coordinates (r,theta, z) of a pt P are related as follows: x=r cos theta, y=r sin theta, tan theta = y/x, r^2=x^2+y^2, z= z. Also, a triple integral w/ cylindrical coordinates is evaluated as: [integral from alpha to beta, integral from g1(theta) to g2(theta) integral from k1(r,theta) to k2(r, theta) of integrand (f(r,theta,z)r) dz, dr, d(theta)].

My attempt: I've tried setting up the triple integral as: [integral from 0 to 2pi, integral from 0 to a, integral from 0 to sqrt(a^2-r^2) of integrand (kz)r dz, dr, d(theta)]. I got the answer as [k(a^4)(pi)]/4, BUT this is the wrong answer, therefore my triple integral is set up incorrectly.
Thanks for any help.

So, if I understand you correctly, you have

\(\displaystyle \int_0^{2\pi}\int_0^a\int_0^{\sqrt{a^2-r^2}} kz r \, dz dr d\theta\)

I can see problems here. Have you drawn yourself a picture of the solid?

First of all, the density is proportional to the distance from a line through the centre, not the distance to a plane through the centre. Your integrand kz is proportional to the distance to the xy-plane. If you want the distance to the z-axis, this will be r, so integrate kr instead.

Second, your bounds only cover the top half of the sphere.