Use spherical coords to find the volume of the solid:

hank

Junior Member
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Sep 13, 2006
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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:

02ππ6π202aρ2sinϕ dρdϕdθ\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?
 
Inb spherical coordinates the sphere and the plane z=a are ρ=2a\displaystyle {\rho}=2a and ρ=asec(ϕ)\displaystyle {\rho}=a\cdot{sec({\phi})}, respectively.

They intersect at ϕ=π3\displaystyle {\phi}=\frac{\pi}{3}

02π0π30asecϕρ2sinϕdρdϕdθ+02ππ3π202aρ2sinϕdρdϕdθ\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:
 
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