# Thread: triple integral in cylindrical coordinates

1. ## triple integral in cylindrical coordinates

Convert and evaluate the integral of (from -1 to 1) the integral of (0 to sqrt(1-y^2) ) the integral of (from x^2+y^2 to sqrt(x^2 +y^2) ) xyz dz dx dy into an integral in cylindrical coordinates.

Please do show me the answer of this question if possible. Thank you very much.

2. Originally Posted by kittycat
Convert and evaluate the integral of (from -1 to 1) the integral of (0 to sqrt(1-y^2) ) the integral of (from x^2+y^2 to sqrt(x^2 +y^2) ) xyz dz dx dy into an integral in cylindrical coordinates.

Please do show me the answer of this question if possible. Thank you very much.
$\int_{-1}^1 \int_0^{\sqrt{1-y^2}} \int_{x^2+y^2}^{\sqrt{x^2+y^2}}xyz \ dz \ dx \ dy$.

Look at the area over which we are integrating expressed by $0\leq x\leq \sqrt{1-y^2} \mbox{ and }-1\leq y \leq 1$.
You should realize that this is a semi-circle in the right half plane. Which is expressed in polar form as $-\frac{\pi}{2}\leq \theta \leq \frac{\pi}{2} \mbox{ and }0\leq r\leq 1$.

Using $r=\sqrt{x^2+y^2}$ the integral becomes,
$\int_{-\pi/2}^{\pi/2} \int_0^1 \int_{r^2}^r (r\cos \theta)(r\sin \theta) z r \ dz \ dr \ d\theta$
Use double angle identity,
$\frac{1}{2}\int_{-\pi/2}^{\pi/2}\int_0^1 \int_{r^2}^r r^3 \sin 2\theta z \ dz \ dr \ d\theta$.

3. Hi perfecthacker,

Thank you very much.