# Thread: Spherical polar coordinates =p

1. ## Spherical polar coordinates =p

Consider the region: $\displaystyle \Omega:=\{(x,y,z) \in \mathbb{R}^3| y \geq 0, z \geq 0, x^2+y^2 \leq 4 \}$.

Note: $\displaystyle \Omega$ is the half of the upper solid hemisphere of radius 2 which lies to the right of the x-z plane.

Use spherical polar coordinates to evaluate $\displaystyle \int \int \int_ {\Omega} \frac{yze^{-x}}{\sqrt{x^2+y^2}}~dx~dy~dz$.
Methinks I have to integrate $\displaystyle \int_0^{2 \pi} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \int_0^1 r^2 sin (\phi)\left( \frac{rsin (\theta) sin (\phi) r cos (\phi) e^{-r cos(\theta) sin (\phi)}}{\sqrt{(rsin(\theta) cos(\phi))^2+(rsin(\theta) sin (\phi))^2}} \right) dr~d \phi~ d \theta$

$\displaystyle \int_0^{2 \pi} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \int_0^1 r^4 sin(\phi) \left( \frac{sin(\theta) sin(\phi) cos(\phi)e^{-rcos(\theta)sin(\phi)}}{r sin(\theta)} \right)~dr~d \phi~d \theta$

$\displaystyle \int_0^{2 \pi} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \int_0^1 r^3 sin^2 (\phi)cos(\phi)e^{-rcos(\theta) sin(\phi)}~dr~d \phi~d \theta$

Is this right? The integral seems so complicated!!