$\iiint_Ez(x^2+y^2+z^2)^{-3/2}dV$ where $E=\{(x,y,z):x^2+y^2+z^2\leq16, z\geq2\}$

Question

Convert to spherical coordinates and evaluate: $$\iiint_Ez(x^2+y^2+z^2)^{-3/2}dV$$ where $E$ is the region satisfying the following inequalities: $x^2+y^2+z^2\leq16$, $z\geq2$.

When I drew out the graph. I got the boundaries to be $2\leq\rho\leq4$, $0\leq\theta\leq2\pi$, $0\leq\phi\leq \frac{\pi}{3}$.

But apparently, my $\rho$ is wrong. it's supposed to be $\frac{2}{\cos\phi}\leq\rho\leq4$.

I can't seem to see why. How exactly was it chosen? From what i gather it's suppose to be the range of values that the parameters can take.

Answer 1

The region $E$ is a spherical cap. If you draw a ray from the origin through $E$, it will enter $E$ through the plane $z=2$ and exit through the sphere $x^2 + y^2 + z^2 = 16$. That is, for a fixed $\theta$ and $\phi$, the lower limit of $\rho$ comes from the plane and the upper limit of $\rho$ comes from the sphere.

You are right that the upper limit is $\rho = 4$; that describes the sphere. But a lower limit of $2$ would describe a smaller sphere. Instead, convert the equation $z=2$ into spherical coordinates: $\rho \cos\phi = 2$. Therefore the lower limit of $\rho$ is $\frac{2}{\cos\phi}$.

Answer 2

It can be hard to do these without drawing a picture and I can't figure out how to do that here.

Your bounds on $\theta$ and $\phi$ are correct.

When you integrate with respect to the radius $\rho$, then the upper limit is indeed $4$ (the outside of the sphere $x^2 + y^2 + z^2 = 16$), but the lower limit will depend on $\phi$.

IF you draw a picture of the sphere $x^2 + y^2 + z^2 = 16$ and the plane $z=2$, you need to integrate from the plane to the surface of the sphere. SAo you have a right angle triangle with side $2$ and angle $\phi$. Then the hypotenuse is $2 / \cos(\phi)$ so this will be your lower limit.