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!

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

2 Answers 2


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}

  • $\begingroup$ Thank you, I think this clears it up! $\endgroup$
    – P. Gillich
    Dec 18, 2019 at 22:19

$\omega$ 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 $T$.


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