Triple Integral in Cylindrical Coordinates

[daveyman]

Revised question is below.

 

[daveyman]

Homework Statement

Evaluate \int\int\int_E{\sqrt{x^2+y^2} dV where E is the region that lies inside the cylinder x^2+y^2=16 and between the planes z=-5 and z=4.

Homework Equations

For cylindrical coordinates, r^2=x^2+y^2.

The Attempt at a Solution

The inside of the integral becomes \sqrt{r^2}=r. Then I integrated using the following bounds: \int _0^{2*\pi }\int _{-5}^4\int _0^4r drdzd\theta

However, this gives me an answer of 144\pi. I've tried several things in Mathematica and I finally tried \int _0^{2*\pi }\int _{-5}^4\int _0^4r^2drdzd\theta which actually gave me the right answer of 384\pi. However, integrating over r^2 makes no sense.

Any ideas?

 

[gabbagabbahey]

The infinitesimal volume element in cylindrical coordinates is dV=rdrd\theta dz not just drd\theta dz

 

[daveyman]

Thank you for your quick response! I have one more question, though.

Doesn't the dr take care of the radial component of the volume? Why does it need to be r dr?

I realize this is a naive question, but I really appreciate your help.

 

[gabbagabbahey]

dr does take care of the radial component, but the tangential component is r d\theta not d \theta...remember that the arc length subtended by an angle \theta is r \theta...the same is true for the infinitesimal change in arc length.

 

[daveyman]

Oh I get it. Thank you so much!