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While I am aware that when integrating over a ball in $\mathbb{R}^n$, we have

$\int_{B(0,R)}f(x)dx=\int_{S^{n-1}}\int_0^Rf(\gamma r)r^{n-1}drd\sigma(\gamma)$

I cannot figure why it is true that

$\int_{S(0,r)}f(x)d\sigma(x)=\int_{S(0,1)}f(y)r^{n-1}d\sigma(y)$

I know this is very trivial but I keep thinking that on changing the variables, the Jacobian should be $r^n$ and not $r^{n-1}$, and that is clearly wrong.

Would be grateful for the help. I only require a vector calculus explanation.

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2 Answers 2

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It is the surface integral, not the volume integral. Surface area is proportional to $r^{n - 1}$. If you want to use Jacobian determinant, I guess you might need to be able to formulate what $d\sigma$ is first. Iterated spherical coordinates (I don't know what it's actually called) should work.

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In $\mathbb R^3$, the area of a sphere is $4 \pi r^2$, so when you change scale from $R$ to $1$ you need to multiply by a factor $R^2$.

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But why a factor of $R^2$. It must be the Jacobian of some transformation. I just want to know what that transformation is. –  Vivek Aug 22 '12 at 22:56

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