Calculate the integral of $\phi(x,y,z)$ over the surface of the area of the unit sphere

Question

Given a scalar field through $\phi(x,y,z) = xyz $. Calculate the integral of $\phi(x,y,z)$ over the surface of the area of the unit sphere : $ I = \int_{Sus} \phi dA$ such that Sus is the surface area of the unit sphere. I want to calculate the integral in spherical coordinates.

Answer 1

$I=0$ simply considering the simetry for $\phi$. It's an odd function, and for every value $x_0yz$ we have another changing the sign of $x_0$: $-x_0yz$ which cancels each other in the integral. Anyway,

We need to perform three changes. First, express $\phi$ in polar coordinates,

$$\begin{cases} x=r\sin\theta\cos\varphi\\ y=r\sin\theta\sin\varphi\\ z=r\cos\theta \end{cases}$$

$\phi(x,y,z)=xyz=r\sin\theta\cos\varphi\,r\sin\theta\sin\varphi\,r\cos\theta=r^3\sin^2\theta\cos\theta\sin\varphi\cos\varphi$, but $r=1$, so, $\phi(x,y,z)=\sin^2\theta\cos\theta\sin\varphi\cos\varphi$

Second, stablish the integration limits: $0\leq\theta\lt\pi\;;0\leq\varphi\lt 2\pi$ and third, set the surface element $dS=r^2\sin\theta\,d\theta\,d\varphi$, but again $r=1$, so $dS=\sin\theta\,d\theta\,d\varphi$

$$I = \int_{Sus} \phi dA=\int_0^{2\pi}\int_0^\pi\sin^2\theta\cos\theta\sin\varphi\cos\varphi\,d\theta\,d\varphi=0$$