# Limits of integration spherical coordinates

I have seen a lot of exercises where they solve a triple integral using spherical coordinates. But I'm confused about the limits that one should use. For example when they integrate over a sphere like $x^2 + y^2 + z^2 = 4$ I do understand why the limit are $0 \leq \rho \leq 2$ , $0\leq \theta \leq 2\pi$, but I can't get why this one ends in $\pi$ and not in $2\pi$, $0 \leq \phi \leq \pi$.
Thank You!

$0 \leq \phi \leq \pi$ Otherwhise you would be counting the volume twice!

Indeed, fix $\ \theta$ then, the equations $0 \leq \rho \leq 2$ , $0\leq \phi \leq \pi$ describe a half circle that has "vertical" diameter (see picture below)

By making this half-circle turn around $0\leq \theta \leq 2\pi$ we get the full sphere

• thanks a lot. the half-circle (0 to π) definition for ϕ and turning it around through 360 degrees to cover the entire θ range (0 to 2π) is the best explanation of this problem i've also been suffering with for many many years. – nyxee Feb 21 '18 at 21:05

Stand with your arm held directly above your head. Pretend your arm has radius $\rho=2$. Now swing it through $0\le\phi\le\pi$. You ought to have made a semicircle, and now your arm is resting against your leg. Next, keep swinging your arm through $\phi$, but also turn full circle on the balls of your feet, $0\le\theta\le2\pi$. Your arm ought to have swept out a sphere.

If your arm had initially gone from $0\le\phi\le2\pi$, you would have swept out two spheres in the end.

Using Mathematica, i have noticed that when we interchange the limits, the volume doesn't change.. i.e using { 0≤ϕ≤π and 0≤θ≤2π } OR { 0≤ϕ≤2π and 0≤θ≤π } yielded the same volume

SphericalPlot3D[1, {\[Theta], 0, 3 Pi/2}, {\[Phi], 0, Pi}, AxesLabel -> {x, y}, PlotLabel -> r == 1, AxesEdge -> {{-1, -1}, {1, -1}, {-1, -1}}]
SphericalPlot3D[1, {\[Theta], 0, Pi}, {\[Phi], 0, 3 Pi/2}, AxesLabel -> {x, y}, PlotLabel -> r == 1, AxesEdge -> {{-1, -1}, {1, -1}, {-1, -1}}]