Total Mass: Calculating the Mass of a Lamina Using Triple Integrals

[withthemotive]

Homework Statement

A lamina occupies the part of the disk x^2 + y^2 ≤ 16 in the first quadrant and the density at each point is given by the function ρ(x,y) = 2(x^2+y^2) .

What is the total mass? Where is the center of mass? (Once I solve total mass I can solve the center by myself.)

The Attempt at a Solution

Total Mass:

I thought it might be easier to solve if I translate this into cylindrical coordinates, so therefore {0≤ r≤ 4, 0≤theta≤ pi/2, 0≤ z≤ 16-r^2}.

I solved this through triple integrals in the order of (rdzdthetadr) and ended up with an answer of (1024/3)*pi, but I'm being told this is incorrect.

 

[gabbagabbahey]

This seems more like a 2D problem than a 3D problem to me...why are you saying that z goes from zero to 16-r^2?

 

[HallsofIvy]

Homework Statement

A lamina occupies the part of the disk x^2 + y^2 ≤ 16 in the first quadrant and the density at each point is given by the function ρ(x,y) = 2(x^2+y^2) .

What is the total mass? Where is the center of mass? (Once I solve total mass I can solve the center by myself.)

The Attempt at a Solution

Total Mass:

I thought it might be easier to solve if I translate this into cylindrical coordinates, so therefore {0≤ r≤ 4, 0≤theta≤ pi/2, 0≤ z≤ 16-r^2}.

No, this problem is two dimensional. There is no "z". Just use polar coordinates!

I solved this through triple integrals in the order of (rdzdthetadr) and ended up with an answer of (1024/3)*pi, but I'm being told this is incorrect.

 

[withthemotive]

No, this problem is two dimensional. There is no "z". Just use polar coordinates!

Thanks. I debated on doing it only in 2-D, but turned the idea down. Thanks for clearing that up.