Derivation for the integrating term in line integrals and volume integrals in spherical coordinates

Question

Can anyone refer me to, or respond with, the derivation for the integrating term in line integrals $dl=dr\hat{r}+rd\theta\hat{\theta}+r\sin\theta\ d\phi\hat{\phi}$ and volume integrals $dV=r^2\sin\theta dr d\theta d\phi$. I'm having trouble understanding why the angle $\phi$ needs some form of $r\sin\theta$ in front of it.

Answer 1

Let's connect your notation to geography:

• $\theta$ is latitude with $0$ being North Pole, $\pi/2$ the equator, $\pi$ is South Pole
• $\phi$ is the longitude, perhaps$ 0\le \phi\le 2\pi$ with $0$ being the Greenwich meridian. (This does not really matter, all meridians look the same on the globe).

Unlike meridians, the circles of constant latitude do not look the same on the globe: their length varies with the latitude. At latitude $\theta$, the circle has radius $r\sin\theta$ and its length is $2\pi r \sin \theta$ where $r$ is the radius of the Earth.

Say, you are traveled westward and increased your longitude (value of $\phi$) by some amount $d\phi$. What is the distance you traveled? It's $\frac{d\phi}{2\pi}$ portion of the latitude circle, that is $2\pi r \sin \theta \frac{d\phi}{2\pi} = r\sin\theta \,d\phi$.

With $\theta$ it is easier: you are traveling along a meridian, which is a half-circle of radius $r$. So the displacement is $r\,d\theta$. With $dr$ it's even more straightforward, and you get $dl$ by adding the displacements.

For the volume element $dV$, the standard way is to take the differentials $$x=r\sin\theta \cos\phi \implies dx = \sin\theta \cos\phi\,dr+ r\cos\theta \cos\phi\,d\theta - r\sin\theta \sin\phi d\phi$$ $$y=r\sin\theta \sin\phi \implies dy = \sin\theta \sin\phi\,dr+ r\cos\theta \sin\phi\,d\theta + r\sin\theta \cos\phi d\phi$$ $$z=r\cos\theta \implies dz = \cos\theta \,dr+ -r\sin\theta \,d\theta $$ and put the coefficients on the right (i.e., partial derivatives) into a $3\times 3$ determinant to calculate the volume. It simplifies a good deal.

Answer 2

First, one computes the "coordinate basis vectors." The coordinate basis vectors are not necessarily unit vectors. Rather, given a function $\mathbf r(r, \theta, \phi)$ that maps the coordinates to positions, one obtains the basis vectors as follows:

$$\mathbf e_r = \frac{\partial \mathbf r}{\partial r}, \quad \mathbf e_\theta = \frac{\partial \mathbf r}{\partial \theta}, \quad \mathbf e_\phi = \frac{\partial \mathbf r}{\partial \phi}$$

The most direct, if tedious, way to do this is to break $\mathbf r$ into a cartesian basis. That is, write

$$\mathbf r = \mathbf e_x r \sin \theta \cos \phi + \mathbf e_y r \sin \theta \sin \phi + \mathbf e_z r\cos \theta$$

You should then be able to verify that, taking a partial derivative with respect to $r$, we get

$$\mathbf e_r = \frac{\partial \mathbf r}{\partial r} = \mathbf e_x \sin \theta \cos \phi + \mathbf e_y \sin \theta \sin \phi + \mathbf e_z \cos \theta \implies |\mathbf e_r| = 1$$

A similar process tells you that $|\mathbf e_\theta| = r$ and $|\mathbf e_\phi| = r \sin \theta$.

Now, what is it you do when you integrate a line integral? You might have some function $F(\mathbf r)$ such that you want to integrate

$$\int_\ell F(\mathbf r) \, d\boldsymbol \ell$$

The "dumb" way to do this is to integrate over some parameterized curve $\boldsymbol \ell(t)$. Then $d\boldsymbol \ell$ really means $\boldsymbol \ell'(t) \, dt$.

Suppose that you choose $t = \theta$ for fun, and we take a curve that has $r = R$, some constant, from $\theta = 0$ to $\theta = \pi$. Let's also choose $F = 1$ to get

$$\int_0^\pi 1 \, \ell'(\theta) \, d\theta$$

I'll leave it to you to verify this, but $\ell'(\theta)$ here is $R \hat \theta = \mathbf e_\theta$ everywhere on the curve. That's the convenient property of using $\mathbf e_\theta$ instead of $\hat \theta$: when you hold $r, \phi$ constant and parameterize the curve with $\theta$, the tangent vector is $\mathbf e_\theta$, not $\hat \theta$. It's the same for the other coordinates, and this is why we use $\mathbf e_r, \mathbf e_\theta, \mathbf e_\phi$ instead of $\hat r, \hat \theta, \hat \phi$.

That should adequately explain what's going on with line integrals, but what about volume integrals? This is typically explained as a Jacobian determinant, but a more "geometric" explanation is to use exterior or clifford algebra. As the line integral involves a tangent vector, a volume integral involves a tangent "volume" formed by the three coordinate basis vectors. This is done using a wedge product. So $dV$ is formed by

$$dV \equiv (\mathbf e_r \wedge \mathbf e_\theta \wedge \mathbf e_\phi) (\mathbf e_x \wedge\mathbf e_y \wedge\mathbf e_z)^{-1} \, dr \, d\theta \, d\phi$$

First, why $(\mathbf e_x \wedge\mathbf e_y \wedge\mathbf e_z)^{-1}$? This is an arbitrary choice of orientation for 3d space.

Second, $\mathbf e_r \wedge \mathbf e_\theta \wedge \mathbf e_\phi$ is proportional to $\mathbf e_x \wedge\mathbf e_y \wedge\mathbf e_z$: why? Because there is only one kind of volume in 3d space. Knowing that $|\mathbf e_\theta| = r$ and $|\mathbf e_\phi| = r \sin \theta$ and that all the vectors are orthogonal, we get

$$\mathbf e_r \wedge \mathbf e_\theta \wedge \mathbf e_\phi = r^2 \sin \theta (\hat r \wedge \hat \theta \wedge \hat \phi)$$

The remaining wedge products cancel, and you get your canonical volume element.