Intuition for a Ingenious Integral Substitution In the class Measure Theory and Integration, the official solution gave an ingenious solution to the following computation:
For $a \in \mathbb{R}^3$ and $r>0$ compute $\int_{\|x \|_2 \leq r} \langle a,x \rangle^2 \ d \lambda^3(x)$.
I'm wondering about the motivation for the chosen substitution! What motivates the author to think of that?
Sketch. Let $Q \in O_3$ be the orthogonal $3 \times 3$ matrix that maps $Qa = \|a \|_2 e_1$. Then $\Phi : \xi \mapsto rQ^T \xi$ is a $C^1$-diffeomorphism that maps
$$ \{\xi \in \mathbb{R}^3: \|\xi \|_2 \leq 1 \} \to \{x \in \mathbb{R}^3: \|x \|_2 \leq r \} , \ |\det(D \Phi(\xi))| = r^3.$$
Change of variables along with some computation yields
$$ \int_{\|x \|_2 \leq r} \langle a,x \rangle^2 \ d \lambda^3(x) = r^5 \|a \|_2^2 \int_{\| \xi \|_2 \leq 1} \xi_1^2 \ d \lambda^3(\xi). $$
We finish by integrating along $\xi_1$-intersections of the unit ball $B^3$ due to Tonelli (I'm not sure if it's well known that the next step is allowed. It was my first time seeing it and took me a bit to prove it's validity.)
$$ \int_{B^3} \xi_1^2 \ d\lambda^3(\xi) = \int_{-1}^1 \xi_1^2 \lambda^2 \left(\sqrt{1-\xi_1^2} B^2 \right) \ d \lambda^1(\xi_1) = \frac{4 \pi}{15}. $$
So! Where did that $\Phi$ come from? Is it geometrically clear that one should seek integration along intersections in the end?
 A: The idea is that, since the ball of radius $r$ centred at the origin is symmetric, and in particular, invariant with respect to any unitary transformation, integrating $\langle a,x\rangle^2$ over that ball, is the same as integrating $\langle b,x\rangle^2$, for every other vector $b$, such that $\|a\|=\|b\|$. Clearly, $b=\|a\|e_1$ is a very convenient choice.
Next step, turn the integration over a ball of radius $r$, to an integration over a ball of radius 1. That's precisely what $\Phi$ does. It takes care of both steps, in order to simplify the integration.
A: Also note that the same symmetry argument can give you
$$
\int_{B^3} \xi_1^2 \ d\lambda^3(\xi) =\int_{B^3} \xi_2^2 \ d\lambda^3(\xi) =\int_{B^3} \xi_3^2 \ d\lambda^3(\xi) .
$$ Thus the last integral is equal to
$$\begin{eqnarray}
\int_{B^3} \xi_1^2 \ d\lambda^3(\xi)&=&\frac{1}{3}\int_{B^3} (\xi_1^2+\xi_2^2+\xi_3^2) \ d\lambda^3(\xi)\\
&=&\frac{1}{3}\int_{\Bbb S^2}\int_0^1 r^4 dr d\sigma_x\tag{polar coordinate change}\\& =&\frac{|\Bbb S^2|}{15}=\frac{4\pi}{15}.
\end{eqnarray}$$
