Integration in d-dimensional spherical coordinates

Question

Can someone please tell me why $$\left(\int_0^\infty dr\, e^{-r^2}\right)^d=\int_0^\infty dr\,r^{d-1}S_d e^{-r^2}?$$

Why doesn't $d$ end up joining the exponential?

Answer 1

You can replace your potential $(\int ...\text dr)^d$ by a multi dimensional integral $$\left(\int_0^\infty\text dr\, e^{-r^2}\right)^d = \int_{R^d}\text d^dr e^{-\left(r_1^2+...+r_d^2\right)} = \int_0^\infty\text dr\,r^{d-1}S_d e^{-r^2}$$