I'm taking a PDE's course, and several times there has been an integration over a sphere cropping up.
Oftentimes, we change variables to shift the sphere we originally had to the unit sphere, and carry on calculations from there. I'll give an example:
Let $ f: \mathbb{R}^3 \rightarrow \mathbb{R}$. We want to integrate $f$ over the surface of some ball in 3 dimensions, centered at $x$ and with radius $r$; that is, we are after
$$ \int_{\partial B(x, r)} f(\sigma) \ d\sigma.$$
If we make the change of variables $$ \sigma = x + r\omega, $$
then the integral becomes $$ \int_{\partial B(0, 1)} f(x + r\omega)r^2 \ d\omega. $$
I don't understand why this is...I tried to justify it by using the change of variables theorem (the one involving the Jacobian determinant), but I get a different result, and I don't know what I'm misunderstanding here.
Using the theorem, we can think of this change of variables as $$ T(\omega) = x + r \omega $$
for any $\omega \in \mathbb{R}^3.$ Then \begin{align} && \int_{\partial B(x, r)} f(\sigma) \ d\sigma &=\int_{T^{-1}(\partial B(x, r)) = \partial B(0, 1)} f(T(\omega)) |J(\omega)| d \omega && \end{align}
where $|J(\omega)|$ is the Jacobian determinant of $T$. But $$T(\omega) = \langle x_1 + r \omega_1, x_2 + r \omega_2, x_3 + r \omega_3 \rangle$$ so that $$ T'(\omega) = J(\omega) = \frac{\partial }{\omega_j} ( x_i + r \omega_i) = \delta_{ij} \cdot r,$$ or, in other words, a 3 by 3 matrix with only $r$ on the diagonal, and zero everywhere else. Doesn't that mean the Jacobian determinant should be $r^3$ and not $r^2$?
I would hope that by this point I would have grasped this change of variables theorem, but it's very possible I have misunderstood something along the way. I would really appreciate some help!
Thank you!