How to compute volume of this using spherical coordinates?

Question

I am trying to compute the volume of the object bounded by $z=4-x^2-y^2$ tp $z=0$.

I convert to spherical coordinates, and I end up with the following integral:

$$x = r\cos\theta\sin\phi \\ y= r\sin\theta\sin\phi \\ z = r\cos\phi$$

$$\int_0^{2} \int_0^{\pi/2} \int_0^{2\pi} r^2 sin (\theta) d\theta d \phi dr$$

But, the integral is giving me 0 obviously, because the integral of sine from 0 to 2 pi is 0.

So, I try to switch the order, and get $r^2 $ in terms of $\phi$ , since r is a function of $\phi$.

$$\int_0^{2} \int_0^{2\pi} \int_0^{\pi/2} \frac{4- cos \phi}{2* sin^2 \phi} d \phi d\theta dr$$

However, this integral ends up being unbounded.

What am I doing wrong? Thanks.

Answer 1

What you are doing wrong:

1. The surface $z=4-x^2-y^2$ is not part of a sphere, it is a paraboloid. (The sphere would be $z^2=4-x^2-y^2$, not just $z$.) It means that the spherical coordinates are inappropriate here, you won't get independent integration limits.
2. If it were a sphere, the integral is not zero anyway because it must be $\sin\phi$ in the Jacobian determinant, (not $\sin\theta$), and the interval for $\phi$ is $[0,\pi/2]$.

Answer 2

The best way to solve this is to use Cartesian coordinates. At $z$ the radius of the circle is $\sqrt{4 - z}$, hence we have

$$V = \int_0^4 \pi (4 - z) dz = 8\pi$$

If we really want to use spherical coordinates, note that the surface of the solid is governed by $rcos \phi = 4 - r^2sin^2 \phi$ from which it is very difficult to express either $r$ in terms of $\phi$ or $\phi$ in terms of $r$. Say if we can put $r = f(\phi)$, then

$$V = \int_{\phi = 0}^{\frac{\pi}{2}} \int_{r = 0}^{f(\phi)} \int_{\theta = 0}^{2\pi} r^2 sin\phi d\theta dr d\phi $$