Explanation of Surface Integral problem

Question

The above image is a proposed solution to finding a surface integral over a sphere of radius 4 for the vector field function $F = <x^2,y^2,z^2>$. I understand the explanation, however, I don't understand why the integral over the surface of $x^3$ would be zero. In other words, after you've found the normal what is the reasoning (using symmetry) that the surface integral is $0$? I'm sorry if this is confusing, but please click the link above to view the problem and the explanation....I don't understand how "by symmetry" the integral is evaluated to zero.

Answer 1

In general, given an integral $\int_A f$, where $A$ enjoys a certain symmetry $\mathcal{S}$, ie $\mathcal{S}(A) = A$:

• if $f$ is odd with respect to $\mathcal{S}$, ie $\mathcal{S}(f) = -f$, then $\int_A f = 0$;
• if $f$ is even with respect to $\mathcal{S}$, ie $\mathcal{S}(f) = f$, then $\int_A f = 2\int_{A_{\text{sym}}} f$.

Specifically, $A = \left\{x^2+y^2+z^2=4\right\}$ and $f = x^3+y^3+z^3$; since $A$ enjoys symmetry $\mathcal{S}(x,\,y,\,z) = (-x,\,-y,\,-z)$ and $f$ is odd with respect to $\mathcal{S}$, then it follows tha $\int_A f = 0$.