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$ n \in \mathbb{N},b >0$

Define $S_n$ as the n-dimensional Sphere in $\mathbb{R}^n$.

I cannot figure out the appropiate transformation to use the transformation theorem such that the following integral transforms into an integral having the following area of integration

$\int_{||x||_2^2 \leq b} e^{-\frac{1}{2} ||x||_2^2} \lambda_n(dx) \to \int_0^\sqrt{b}\int_{S_{n-1}} .... $

Does anyone have an idea?

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For $f \in L^1(\mathbb{R}^n)$ the formula $$\int_{B_R(0)} f(x) \ dx = \int_0^R \int_{\partial B_\rho(0) } f(\xi) \ dS(\xi) \ d \rho = \int_0^R \int_{\partial B_1(0)} f(\xi)\ dS(\xi)\ \rho^{n-1} \ d\rho $$ holds. In your case you can apply this for $R = \sqrt{b}$.

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  • $\begingroup$ Your argument seems reasonable, thank you. But which measure do you mean by $dS(\cdot)$? $\endgroup$ – Slyder Feb 2 at 16:32
  • $\begingroup$ How do you do you derive $p^{n-1}?$ I assume you applied the transformation theorem to: $\Psi: \partial B_1(0) \to \partial B_p(0),\ \Psi(x) = p\cdot x$ which, if I am not mistaking yields the integral: $\int_0^R \int_{B_1(0)}f(\psi(y))p^n d\lambda^n(y) d\rho$, as the Jacobian of $\Psi$ is a diagonalmatrix with $p$ on the diagonal. What am I missing? $\endgroup$ – Slyder Feb 2 at 16:38
  • $\begingroup$ With $dS$ I mean the "surface measure" $dS = \sqrt{g} d\lambda^{n - 1}$ where $\sqrt{g}$ is the gramian determinant. $\endgroup$ – eddie Feb 2 at 16:51
  • $\begingroup$ Okay, I think I figured it out. We know $M_r := \partial B_{r}(0) $is a n-1 dimensional submanifold. Suppose, $\{ \phi: V_{\alpha} \to T_\alpha\ \}_{\alpha \in \mathbb{N}} $ is an atlas of $M_\rho$. Therefore, $\phi_{\alpha}^r: r^{-1}T_\alpha \to r^{-1} V_\alpha $, $x\mapsto r^{-1}\phi(rx)$ is a local map of $B_1(0)$, which is well defined (right?) Accordingly, we derive the corresponding atlas. Then we use the above mentioned transformation between $\Psi: \phi^{-1} (r^{-1}V_\alpha \cap M_1) \to \phi^{-1} (r^{-1}V_\alpha \cap M_r)$. Which has the determinant $r^{n-1}$. $\endgroup$ – Slyder Feb 4 at 17:17

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