# Triple integral of a sphere being cut by a plane

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?

• You have made a mistake when converting $f$ to spherical coords. $x^2+y^2+z^2\not=\rho$, instead this is $\rho^2$. – Ruslan Apr 6 '13 at 22:44
• Wow, that makes using spherical coordinates so much easier. – MathMan08 Apr 6 '13 at 23:09

$$2 \pi \int_2^4 dz \, z \: \int_0^{\sqrt{16-z^2}} d\rho \, \rho\, (z^2+\rho^2)^{-3/2}$$
The result I get is $\pi$.
• I did solve it. Work right to left. The inner integral may be done by substituting $u=\rho^2$. The outer integral is then very simple. – Ron Gordon Apr 6 '13 at 22:57
• The volume of the sphere before being cut is $4/3 \pi r^3=4/3 \pi *4^3=256 \pi/3$. So we know the top half of the sphere has a volume of half that which is $128 \pi/3$. So $\pi/2$ for that section of the top half seems very small... – MathMan08 Apr 6 '13 at 23:01