Evaluating surface integral using symmetry

Question

Compute $$\iint_S(x^2 + y^2 − 2z^2)dA$$ where $S$ is the unit sphere. Can you find the answer by symmetry considerations without doing any calculations?

For computation using symmetry, the only way I know is showing that the integrand is odd and hence the surface integral vanishes by the symmetry of the surface.

But I don't know how to show (if ?) the integrand is odd about any plane.

Answer 1

Write $x^2+y^2-2z^2$ as the sum $(x^2-z^2)+(y^2-z^2)$. Then $x^2-z^2$ is odd about the $x=z$ plane, since a reflection about $x=z$ is the map $(x,y,z)\mapsto (z,y,x)$. Similarly $y^2-z^2$ is odd about the $y=z$ plane.

Thus $$\iint_S x^2+y^2-2z^2\,dA = \iint_S x^2-z^2 \,dA + \iint_S y^2-z^2\, dA = 0 + 0 = 0.$$

Alternatively, as suggested by Ted Shifrin's comment, by symmetry, $$ \iint_S x^2\,dA = \iint_S y^2\,dA = \iint_S z^2\,dA, $$ so $$ \iint_S x^2+y^2-2z^2\,dA = (1+1-2)\iint_S x^2\,dA=0. $$

Answer 2

$S$ is the unit sphere implies $x^2+y^2+z^2=1$, therefore $$\int\int_Sx^2+y^2-2z^2~dA=\int\int_S1-3z^2~dA$$