# Triple Integral: Convert from Cartesian to Cylindrical Coordinates

1. Nov 10, 2008

### daveyman

1. The problem statement, all variables and given/known data
This is my last question about triple integrals in cylindrical coordinates.

Evaluate the integral by changing to cylindrical coordinates:

$$\int _{-3}^3\int _0^{\sqrt{9-x^2}}\int _0^{9-x^2-y^2}\sqrt{x^2+y^2}dzdydx$$

2. Relevant equations
In cylindrical coordinates, $$x^2+y^2=r^2$$ and $$x=r\cos{\theta}$$.

3. The attempt at a solution
My converted integral looks like this:

$$\int _0^{\pi }\int _0^{\sqrt{\frac{18}{1+\text{Cos}[\theta ]^2}}}\int _0^{9-r^2}r^2dzdrd\theta$$

This isn't quite right. Any ideas?

2. Nov 10, 2008

### tiny-tim

Hi daveyman!

Hint: If |x| ≤ 3 and |y|2 ≤ 9 - x2, then … r what, and θ what?

3. Nov 10, 2008

### daveyman

$$r=\sqrt{18-x^2}$$ which, in cylindrical coordinates is $$r=\sqrt{18-r^2\cos^2 \theta}$$.

So would the new integral be

$$\int _0^{\pi }\int _0^{\sqrt{18-r^2*\text{Cos}[\theta ]^2}}\int _0^{9-r^2}r^2dzdrd\theta$$?

4. Nov 10, 2008

### tiny-tim

No!

Get some graph paper and shade in all the points for which |x| ≤ 3 and |y|² ≤ 9 - x². ​

5. Nov 10, 2008

### daveyman

It is a circle at the origin with radius of three. So the integral will be

$$\int _0^{\pi }\int _0^3\int _0^{9-r^2}r^2dzdrd\theta$$

Thank you!

6. Nov 10, 2008

### tiny-tim

Yup!

The moral of this … always draw the region first …

it's almost impossible to work out the limits without a diagram

(though of course, you must then prove them without a diagram … but that's much easier once the diagram has told you the answer! )​

7. Jul 11, 2011

### feldy90

Hey- I know this is quite an old thread... But just wondering how you know that the dθ part is between 0 and pi??
Cheers