Evaluate the triple integral of: $f(x,y,z)=z(x^2+y^2+z^2)^{-3/2}$
Over the part of the ball: $x^2+y^2+z^2\le 16$ with $z\ge 2$
So I converted to spherical coordinates and got: $$f(\rho,\phi,\theta)=\rho cos(\phi)\rho^{-3/2}$$ $$f(\rho,\phi,\theta)=cos(\phi)\rho^{-1/2}$$
Then for the bounds: $$\rho^2\le 16$$ $$\rho\le 4$$ and $$\rho cos(\phi)\ge 2$$ $$2/cos(\phi)\le \rho$$
So: $$2/cos(\phi)\le \rho\le 4$$
Then since the plane crosses the sphere at $z=2$ where $\rho=4$, and $z=\rho cos(\phi)$: $$4cos(\phi)=2$$ $$\phi = \pi/3$$
So: $$0\le\phi\le\pi/3$$
Since it is a sphere I have: $$0\le\theta\le 2\pi$$
So my bounds are:
$$2/cos(\phi)\le \rho\le 4$$
$$2/cos(\phi)\le \rho\le 4$$
$$0\le\theta\le 2\pi$$
Over: $$f(\rho,\phi,\theta)=cos(\phi)\rho^{-1/2}$$
With $dV=\rho^2 sin(\phi)d\rho d\phi d\theta$, I write the integral as:
$$\int_0^{2\pi}\int_0^{\pi/3}\int_{2/cos(\phi)}^4 \rho^{-1/2} cos(\phi) \rho^2 sin(\phi) d\rho d\phi d\theta$$
$$\int_0^{2\pi}\int_0^{\pi/3}\int_{2/cos(\phi)}^4 \rho^{3/2} cos(\phi) sin(\phi) d\rho d\phi d\theta$$
Is this all correct?
Best Answer
Alternatively, you could set this up in cylindrical coordinates. The integral becomes
$$2 \pi \int_2^4 dz \, z \: \int_0^{\sqrt{16-z^2}} d\rho \, \rho\, (z^2+\rho^2)^{-3/2}$$
The result I get is $\pi$.