# Transformation Theorem on n-dimensional Sphere

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

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?

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 $$R = \sqrt{b}$$.

• Your argument seems reasonable, thank you. But which measure do you mean by $dS(\cdot)$? – Slyder Feb 2 at 16:32
• 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? – Slyder Feb 2 at 16:38
• With $dS$ I mean the "surface measure" $dS = \sqrt{g} d\lambda^{n - 1}$ where $\sqrt{g}$ is the gramian determinant. – eddie Feb 2 at 16:51
• 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 $r^{n-1}$. – Slyder Feb 4 at 17:17