Why doesn't integrating over a sphere with $\phi$ between $0$ and $2\pi$ and $\theta$ between $0$ and $\pi$ work?

Question

One way of integrating over a sphere with $p = 1$ is by integrating over $p$ from $0$ to $1$, $\phi$ from $0$ to $\pi$, and $\theta$ from $0$ to $2 \pi$

Here is a link that can graph parametric surfaces in spherical coordinates: http://www.math.uri.edu/~bkaskosz/flashmo/sphplot/

If you enter in what is above, then you will find that the graph is indeed, a sphere. Furthermore, if you integrate $p^2\sin (\phi)$ over these ranges, you will get $\frac {4\pi}{3}$ which is the expected value.

I noticed that $p$ from $0$ to $1$, $\phi$ from $0$ to $2\pi$, and $\theta$ from $0$ to $\pi$ also seems to create a sphere.
Note the subtle change: $\phi$ is from $0$ to $2\pi$ and $\theta$ is from $0$ to $1\pi$. If you plug this in to the grapher, you find that what you get resembles a sphere.
However, when you integrate $p^2\sin (\phi)$ over $p$ from $0$ to $1$, $\phi$ from $0$ to $2\pi$, and $\theta$ from $0$ to $\pi$, you get $0$. Why is that?

Answer 1

When you change to spherical coordinates (with the convention that you seem to be using), the absolute value of the Jacobian determinant is $|\rho^2 \sin \phi|$, and this does only simplify to $\rho^2 \sin\phi$ if $\sin\phi \ge 0$, which is the case in the interval $0 \le \phi \le \pi$, but not in the interval $\pi < \phi < 2 \pi$. But if you instead integrate $|\rho^2 \sin \phi|$ over the ranges that you suggest, then you indeed get $4\pi/3$.

Answer 2

The key fact is that the sphere $$x^2+y^2+z^2=R^2$$ is parametrized by

• $x=R\cos \theta \sin \phi$
• $y=R\sin \theta \sin \phi$
• $z=R\cos \phi$

with

• $\theta \in [0,2\pi)$
• $\phi \in[0,\pi]$

and at each point $P(x,y,z)$ on the sphere corresponds an unique pair for the parameters $(\theta, \phi)$.

Of course we can also use different range for the parametrization as for example

• $\theta \in [-\pi,\pi)$
• $\phi \in[0,\pi]$

(credits http://mathworld.wolfram.com/SphericalCoordinates.html)