# Ngô Quốc Anh

## December 6, 2007

### Surface integral: The symmetric property

Filed under: Các Bài Tập Nhỏ, Giải Tích 5 — Ngô Quốc Anh @ 4:59

Why is $\iint_S(x^2+y^2-2z^2)dA=0$ for $S$ being the unit sphere.

$\displaystyle \int_S x^2\,dA=\int_S y^2\,dA=\int_S z^2\,dA=C$,

by rotational symmetry of the sphere. The integral is

$\displaystyle C+C-2C=0$.

And the easiest way to find $C$ (should you want to) is again by symmetry

$\displaystyle 3C=\iint_S(x^2+y^2+z^2)\,dS=\iint_S 1\,dS=4\pi$,

hence $C=\frac{4\pi}{3}$.