Converting integral into equivalent in spherical coordinates

[Smily]

I dont know how to solve this problem.

Convert the integral:

. . .[pi/2-integral-pi][0-integral-2][0-integral-(4-r^2)^(1/2)] r^2 cos(theta) dz dr d(theta)

...to a equivalent integral in spherical coordinates.

I know that:

. . .x = p sin(theta) cos(theta)
. . .y = p sin(theta) sin(theta)
. . .z = p cos(theta)
. . .p = 2 cosQ
. . .r = 2 cos(theta)

But I don't know how to convert it. Can somebody help me please? :roll:

 

[galactus]

Use \(\displaystyle r={\rho}sin({\phi})\) to find your \(\displaystyle {\phi}\) limits.

\(\displaystyle 2=2sin({\phi})\Rightarrow{\phi}=\frac{\pi}{2}\)

Of course, you know \(\displaystyle {\rho}=2\)

Theta stays the way it is.


\(\displaystyle \L\\\int_{\frac{\pi}{2}}^{\pi}\int_{0}^{\frac{\pi}{2}}\int_{0}^{2}{\rho}^{3}sin^{2}({\phi})cos({\theta})d{\rho}d{\phi}d{\theta}\)