Evaluate $$\iint_S|ax+by+cz|dS$$ where S is the unit sphere: $x^2+y^2+z^2=1$; a,b,c are real numbers
My book says I should be able to solve this using symmetry but I'm not sure how to approach this with the absolute value sign. Should I just integrate over a certain portion of the sphere or should I try something else. If anyone could give me an idea or a good starting point that would be greatly apprieciated.
Try prooving this formula (or just use it to calculate the integral):
$$\iint_S f(ax+by+cz){\rm d}S = 2\pi \int_{-1}^1 f(u \sqrt{a^2+b^2+c^2}) {\rm d}u$$
$S$: $x^2+y^2+z^2=1$
Proof:
If we take the argument of $f$:
$$ax+bx+cy = (a,b,c).(x,y,z)$$
you can see that it is just a dot product of two vectors, which can be expressed as:
$$(a,b,c).(x,y,z)= \sqrt{a^2+b^2+c^2}.1.\cos \theta = \sqrt{a^2+b^2+c^2} \cos \theta$$
because $(x,y,z) \in S$, so $|(x,y,z)| = 1$.
If we now express the left side of the formula in spherical coordinates: $$\iint_S f(ax+by+cz){\rm d}S = \int_0^{2\pi}{\rm d}\varphi \int_0^{\pi} \sin \theta f(\sqrt{a^2+b^2+c^2}\cos \theta){\rm d}\theta = $$ $$=2\pi\int_0^{\pi} \sin \theta f(\sqrt{a^2+b^2+c^2}\cos \theta){\rm d}\theta$$ and use substitution $u = \cos \theta$, $du = -\sin \theta$, we get $$2\pi\int_{1}^{-1} f(u\sqrt{a^2+b^2+c^2})(-{\rm d}u) = 2\pi\int_{-1}^{1} f(u\sqrt{a^2+b^2+c^2}){\rm d}u$$
and we have the formula above.
Calculation of your integral:
$$\iint_S |ax+by+cz|{\rm d}S = 2\pi \int_{-1}^1 |\sqrt{a^2+b^2+c^2}u|{\rm d}u = 2\pi \sqrt{a^2+b^2+c^2}\int_{-1}^1 |u|{\rm d}u =$$
$$= 4\pi \sqrt{a^2+b^2+c^2}\int_{0}^1 u{\rm d}u = 2\pi \sqrt{a^2+b^2+c^2}$$