H hank Junior Member Joined Sep 13, 2006 Messages 209 Apr 28, 2008 #1 The solid enclosed by the sphere x^2 + y^2 + z^2 = 4a^2 and the planes z=0 and z = a. Here is my setup: \(\displaystyle \int^{2\pi}_{0} \int^{\frac{\pi}{2}}_{\frac{\pi}{6}} \int^{2a}_{0} \rho^2\sin\phi \ d\rho d\phi d\theta\) Is this correct?
The solid enclosed by the sphere x^2 + y^2 + z^2 = 4a^2 and the planes z=0 and z = a. Here is my setup: \(\displaystyle \int^{2\pi}_{0} \int^{\frac{\pi}{2}}_{\frac{\pi}{6}} \int^{2a}_{0} \rho^2\sin\phi \ d\rho d\phi d\theta\) Is this correct?
G galactus Super Moderator Staff member Joined Sep 28, 2005 Messages 7,203 Apr 28, 2008 #2 Inb spherical coordinates the sphere and the plane z=a are \(\displaystyle {\rho}=2a\) and \(\displaystyle {\rho}=a\cdot{sec({\phi})}\), respectively. They intersect at \(\displaystyle {\phi}=\frac{\pi}{3}\) \(\displaystyle \int_{0}^{2\pi}\int_{0}^{\frac{\pi}{3}}\int_{0}^{a\cdot{sec{\phi}}}{\rho}^{2}sin{\phi}d{\rho}d{\phi}d{\theta}+\int_{0}^{2\pi}\int_{\frac{\pi}{3}}^{\frac{\pi}{2}}\int_{0}^{2a}{\rho}^{2}sin{\phi}d{\rho}d{\phi}d{\theta}\) There.....now integrate away :wink:
Inb spherical coordinates the sphere and the plane z=a are \(\displaystyle {\rho}=2a\) and \(\displaystyle {\rho}=a\cdot{sec({\phi})}\), respectively. They intersect at \(\displaystyle {\phi}=\frac{\pi}{3}\) \(\displaystyle \int_{0}^{2\pi}\int_{0}^{\frac{\pi}{3}}\int_{0}^{a\cdot{sec{\phi}}}{\rho}^{2}sin{\phi}d{\rho}d{\phi}d{\theta}+\int_{0}^{2\pi}\int_{\frac{\pi}{3}}^{\frac{\pi}{2}}\int_{0}^{2a}{\rho}^{2}sin{\phi}d{\rho}d{\phi}d{\theta}\) There.....now integrate away :wink:
