Calculate the following integral over the unit ball Given $a_1a_2+b_1b_2+c_1c_2=0$ calculate:
$$\int\int\int_{x^2+y^2+z^2<1}(a_1x+b_1y+c_1z)(a_2x+b_2y+c_2z)\,\mathrm dx\,\mathrm dy\,\mathrm dz.$$
My intuition is that it's $0$ and I need to find some change of basis to flip the sign of the integral, but I'm struggling to find it.
Any help?
 A: Your intuition is right.
The function you integrate can be represented as $(r*a)(r*b)$, where $a$ and $b$ are vectors.
These two vectors are perpendicular to each other, so you can now turn the basis so that one axis is parallel to vector $a$, and another one - parallel to vector $b$. In this basis the function you need to integrate looks much more simple: $(r*a)(r*b) = A*x*B*y$. ($A$ and $B$ are the lengths of the vectors $a$ and $b$). So, you need to calculate:
$\int\int\int_{x^2+y^2+z^2<1}(A*x*B*y)dxdydz$
Now just note, that for each small volume around some point (x, y, z) there exists a corresponding small volume around (-x, y, z) where the function has opposite value. So the integral is 0.
A: Another approach:
Let's call the integrand function $u$. by hypothesis, $u$ is harmonic (that is, $\Delta u = 0$, where $\Delta$ denotes the laplacian operator). It is known that the average value of a harmonic function over a ball is given by the value of the function at the center of the ball:
$$
\frac{1}{|B_{1}|} \int_{B_1} u = u(0, 0, 0) \ ,
$$
where $|B_1|$ denotes the volume of the unit ball. Since $u(0,0,0)=0$, we conclude.
