# Triple Integral in Cylindrical Coordinates

1. Nov 9, 2008

### daveyman

Revised question is below.

Last edited: Nov 9, 2008
2. Nov 9, 2008

### daveyman

1. The problem statement, all variables and given/known data
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.

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

3. 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?

3. Nov 9, 2008

### gabbagabbahey

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

4. Nov 9, 2008

### 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.

Last edited: Nov 9, 2008
5. Nov 9, 2008

### 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.

6. Nov 9, 2008

### daveyman

Oh I get it. Thank you so much!