# Evaluate the integral by changing to cylindrical coordinates.

1. Jun 3, 2010

### meeklobraca

I wish I knew how to type this out with the proper symbols but here it goes. It says to change the following to cylindrical coordinates and evaluate

(x^2 + y^2)^(1/2) dz dy dx where -3<=x<=3, 0<=y<=(9-9x^2)^1/2, 0<=z<=9-x^2-y^2

2. Relevant equations

3. The attempt at a solution

I got 162pi/5

Would anybody mind confirming?

Cheers!

2. Jun 3, 2010

### HallsofIvy

That upper limit on the y-integral can't be "$\sqrt{9- 9x^2}$" since that would give an ellipse with semi-axis on the x-axis 1- x can't go from 0 to 3 inside that!

I will assume this is actually
$$\int_{x=0}^3\int_{y=0}^{\sqrt{9- x^2}} \int_{z= 0}^{9- x^2- y^2}dzdydx$$

That, now, is an integration over the eighth of a sphere, in the first quadrant, with center at (0, 0, 0) and radius 3.

In cylindrical coordinates, r will go from 0 to 3 while $\theta$ goes from 0 to $\pi/2$. z goes from 0 to $9- x^2- y^2= 9- r^2$ and the differential of volume is $r dr d\theta$.

To see the LaTex code for formulas in a post, just click on the formula.

3. Jun 3, 2010

### meeklobraca

Yes that is the correct integral. IM curious as to why its from 0 to pi/2. If the x value is from -3 to 3, isnt that a half a circle, hence 0 to pi?

4. Jun 4, 2010

### HallsofIvy

You are right- I misread the problem. If you look at my first integral, you will see I have x running from 0 to 3 rather than -3 to 3.

Yes, $\theta$ should go from 0 to $\pi$, not $\pi/2$.