# Visualising surface integrals

For a current problem I am working on, I have run into angular surface integrals, i.e. the differential solid angle $\text{d}\Omega$. Specifically the surface integrals are defined by $$A_{\mu\nu} = \int\int_{\rm S} \text{d}\Omega~ \frac{x_\mu x_\nu}{r^2},$$ where $x_\mu=\{x,y,z\}$ for $\mu=1,2,3$ and $r = \sqrt{x_1^2+x_2^2+x_3^2}$ and the differential solid angle is the usual definition $\text{d}\Omega=\sin(\theta)d\theta d\phi$. It is easy to show, by evaluating the integrals analytically that $$A_{\mu\nu} = \frac{4\pi}{3}\delta_{\mu\nu}.$$ Here $\delta_{\mu\nu}$ is the Kronecker delta. What if I wanted to visualise these surface integrals? I tried plotting the angular functions which are being integrated i.e. for $A_{22}$ $$A_{22} = \int\int_{\rm S} \text{d}\Omega~ \frac{x_2 x_2}{r^2} = \int\int_{\rm S} \text{d}\Omega~ \left(\sin(\theta)\sin(\phi)\right)^2$$ I plotted the angular function $f(\theta,\phi)=\left(\sin(\theta)\sin(\phi)\right)^2$, and get the figure below

Now for $A_{23}$, which equals zero, I get the following $$A_{23} = \int\int_{\rm S} \text{d}\Omega~ \frac{x_2 x_3}{r^2} = \int\int_{\rm S} \text{d}\Omega~ \left(\sin(\theta)\sin(\phi)\cos(\theta)\right)$$ which when plotted looks like

I can obviously see the symmetries and anti-symmetries for two different surfaces although I can't figure out why the second one has a surface integral of zero. To me it should be the first, as they cancel. Am I plotting these surface integrals correctly? Can someone explain the difference between the two surfaces, and why the second is equal to zero and not the first?

Thanks.