# Why $r^{2}$ instead of $r^3$ in this change of variable?

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!

• Its a surface integral, so only 2 dimensions are involved. Dec 18, 2019 at 21:55
• @Spencer could you explain why that changes the Jacobian somehow? Dec 18, 2019 at 22:02

What you have is a surface integral. In a surface integral you deal with a parametrization (in the case of your sphere) $$\sigma(\theta,\phi)=(x_1+r\cos\theta\sin\phi,x_2+r\sin\theta\sin\phi,x_3+r\cos\phi),\ \ 0\leq\theta\leq2\pi,\ \ 0\leq\phi\leq\pi.$$ And your integral is calculated by $$\tag1 \int_{\partial B(x,r)}f(\sigma)\,d\sigma=\int_0^\pi\int_0^{2\pi}f(\sigma(\theta,\phi))\,\|\sigma_\theta\times\sigma_\phi\|\,d\theta\,d\phi.$$ You can rewrite $$\sigma$$ as $$\sigma(\theta,\phi)=x+r\,\omega(\theta,\phi),$$ where $$\omega$$ is the parametrization of the unit ball centered at the origin. If you calculate $$\omega_\theta\times\omega_\phi$$, you'll see that $$\sigma_\theta\times\sigma_\phi=r^2\,\omega_\theta\times\omega_\phi.$$ So $$(1)$$ becomes \begin{align} \int_0^\pi\int_0^{2\pi}f(\sigma(\theta,\phi))\,\|\sigma_\theta\times\sigma_\phi\|\,d\theta\,d\phi&=\int_0^\pi\int_0^{2\pi}f(x+r\omega(\theta,\phi))\,r^2\,\|\omega_\theta\times\omega_\phi\|\,d\theta\,d\phi\\ \ \\ &=\int_{\partial B(0,1)} f(x+r\omega)\,r^2\,d\omega. \end{align}
$$\omega$$ is confined to $$\partial B(0,1)$$. So the map you define $$T$$ is not from $$\mathbb{R}^3$$ to $$\mathbb{R}^3$$. Rather it is from $$\partial B(0,1)$$ to $$\partial B(x,r)$$. So there will be some $$2\times 2$$ Jacobian, depending on parametrization.
So far, this is just an attempt to explain why $$r^3$$ is not the determinant of the Jacobian.
So why does it actually work out to $$r^2$$? I will try to avoid some paramterization of these surfaces. Instead, recall that a determinant expresses the factor by which volume (or area in 2D, or hyper-volume...) is scaled. If you can picture your map $$T$$, it is uniformly scaling the sphere's surface area by a factor of $$r^2$$. Whatever parameterization you choose, calculating the determinant of the Jacobian needs to simplify down to $$r^2$$ with this map $$T$$.