1

This is a continuum to this question which I seem to not understand the solution.

Integral around unit sphere of inner product

I don't get how the diagonal is done, I get that (for example, for $n=3$ that) $$\iiint_B \langle Av, v \rangle dxdydz = \iiint_B a_{11} x_1^2 + a_{22} x_2^2 + a_{33} x_3^2 dx_1dx_2 dx_3 =$$ $$ = (a_{11} + a_{22} + a_{33}) \iint_B x_1^2dx_1dx_2dx_3$$ Which according to the solution is just $\frac{1}{3} Tr(A) \cdot Area(B)$

The part I fail to see is just how is $$\iiint_Bx_1^2dx_1dx_2dx_2=\frac{1}{3}Area(B)$$

Rab
  • 1,196
  • My intuition is that this is due to the fact you are integrating over the unit sphere, so for any point $p$ on the sphere, there is an antipodal point, $-p$ which "cancels" the contribution of the point $p$ to your integral. – mm8511 Jan 12 '18 at 18:05
  • Yes, that part I get, I edited the question to sign the part I don't understand. – Rab Jan 12 '18 at 19:05

1 Answers1

1

By symmetry, $$\iiint_B x_1^2\,dx_1\,dx_2\,dx_3 =\iiint_B x_2^2\,dx_1\,dx_2\,dx_3 =\iiint_B x_3^2\,dx_1\,dx_2\,dx_3 $$ so that \begin{align} \iiint_B(a_{11}x_1^2+a_{22}x_2^2+a_{33}x_3^2)\,dx_1\,dx_2\,dx_3 &=(a_{11}+a_{22}+a_{33})\iiint_B x_1^2\,dx_1\,dx_2\,dx_3\\ &=\frac{a_{11}+a_{22}+a_{33}}3\iiint_B(x_1^2+x_2^2+x_3^2)\,dx_1\,dx_2\,dx_3\\ &=\frac{\text{Tr}(A)}3\iiint_B(x_1^2+x_2^2+x_3^2)\,dx_1\,dx_2\,dx_3. \end{align} Now we can evaluate $$\iiint_B(x_1^2+x_2^2+x_3^2)\,dx_1\,dx_2\,dx_3$$ by spherical polars...

Angina Seng
  • 161,540
  • Yes, for $n=3$ and $n=2$ I can do it, however the solution seems to be for any $n \in \mathbb N$, which is a bit unclear because I can't evaluate for every $n$ while it seemed immediate for the author, how do we do it? – Rab Jan 12 '18 at 19:03
  • Is $$\int_{B_n} \sum_{i=1}^{n} x_i^2 = Vol(B_n)$$ because it seems like it. – Rab Jan 12 '18 at 19:04
  • @RabMakh That's not true for $n=1$. I get $(n/3)\text{Vol}(B_n)$ (spherical polars). – Angina Seng Jan 12 '18 at 19:34