More on Multivariable Calculus

This post covers some material that a first course in vector calculus often skips:

• computing areas with Green’s theorem;
• an example of the generalised Stokes theorem;
• divergence, curl and the Laplacian in orthogonal curvilinear coordinates.

Areas from Green’s theorem

As an example, we show that an ellipse with semi-major axis \(a\) and semi-minor axis \(b\) has area \(\pi ab\).

There are many ways to do this. ProofWiki gives two methods, but both are rather laborious. Green’s theorem, the two-dimensional case of Stokes’ theorem, gives a much shorter route, and the idea is worth knowing in its own right.

Green’s theorem states that for a region \(D\) in the plane with boundary curve \(C\), traversed anticlockwise,

\[\oint_C(P \, d x+Q \, d y)=\iint_D\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right) d x\, d y .\]

If we choose \(P\) and \(Q\) so that \(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}=1\), the right-hand side is just the area of \(D\). Taking \(P=-\frac{1}{2}y\) and \(Q=\frac{1}{2}x\) gives

\[A=\frac{1}{2} \oint_C(x \, dy-y \, dx).\]

With the standard parametrisation of the ellipse, \((x,y)=(a\cos t,\,b\sin t)\) for \(0\leq t\leq 2\pi\),

\[A =\frac{1}{2} \int_0^{2 \pi}\left[(a \cos t)(b \cos t)-(b \sin t)(-a \sin t)\right] d t =\frac{a b}{2} \int_0^{2 \pi} d t=\pi a b .\]

As an exercise (*), show that

\[A = \oint_C x \, dy = -\oint_C y \, dx = \frac{1}{2} \oint_C \left(x \, dy - y \, dx\right).\]

Exercises

• Show that the area of a circle of radius \(r\) is \(\pi r^2\). (Secondary school)
• Show that the area between one arch of the cycloid \(x = r(t - \sin t)\), \(y = r(1 - \cos t)\) and the \(x\)-axis is \(3 \pi r^2\). (Undergraduate, years 1–2)
• Prove the exercise marked (). Why might one of these forms be quicker for the cycloid? (Undergraduate, years 1–2*)

The generalised Stokes theorem

The generalised Stokes theorem states that for a differential form \(\omega\) on an oriented manifold \(\Sigma\) with boundary \(\partial\Sigma\),

\[\int_{\Sigma} d \omega=\int_{\partial \Sigma} \omega .\]

Let’s check it on the unit ball \(\Sigma\) in \(\mathbb{R}^3\), whose boundary is the unit sphere, with the 2-form

\[\omega=x^1\, d x^2 \wedge d x^3 .\]

Boundary integral. Parametrise the unit sphere by

\[x^1=\sin\theta\cos\phi, \qquad x^2=\sin\theta\sin\phi, \qquad x^3=\cos\theta .\]

Then \(dx^2 \wedge dx^3 = \sin^2\theta\cos\phi \; d\theta\wedge d\phi\), so

\[\omega=\sin ^3 \theta \cos ^2 \phi \; d \theta \wedge d \phi,\]

and

\[\int_{\partial\Sigma} \omega=\int_{0}^\pi \int_{0}^{2 \pi} \sin ^3 \theta \cos ^2 \phi \; d \phi\, d \theta=\frac{4}{3}\cdot \pi = \frac{4 \pi}{3} .\]

Volume integral. On the other hand \(d \omega=d x^1 \wedge d x^2 \wedge d x^3\), the volume form, so

\[\int_{\Sigma} d \omega= \int_{0}^{2\pi} \int_{0}^{ \pi} \int_{0}^{1} r^2 \sin \theta \; dr \, d\theta\, d\phi = \frac{4\pi}{3},\]

the volume of the unit ball. The two sides agree, as Stokes’ theorem predicts.

Exercises

• Given \(\omega = x^1\, d x^2 \wedge d x^3\), verify that \(d \omega = d x^1 \wedge d x^2 \wedge d x^3\). (Undergraduate, year 3)
• Verify the expression for \(dx^2 \wedge dx^3\) on the sphere. (Undergraduate, year 3)

Green’s theorem from Stokes’ theorem

We can recover one of the area formulas above using differential forms. Let \(\alpha = y\, dx\), so that \(d\alpha = dy \wedge dx = -\,dx\wedge dy\). Then

\[\text{Area of } D=\iint_D d x \wedge d y=-\iint_D d\alpha=-\oint_{\partial D} \alpha=-\oint_{\partial D} y\, dx .\]

More generally, for the 1-form \(P\, dx + Q\, dy\), Stokes’ theorem gives

\[\oint_{\partial D} P\, d x+Q\, d y=\iint_{D}dP \wedge dx+dQ \wedge dy .\]

Since \(dP=\frac{\partial P}{\partial x} dx + \frac{\partial P}{\partial y} dy\) and \(dx\wedge dx = 0\),

\[\iint_{D}\left(\frac{\partial P}{\partial y}\, d y \wedge dx+\frac{\partial Q}{\partial x}\, dx \wedge dy\right) = \iint_{D}\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)dx\,dy,\]

which is Green’s theorem: Stokes’ theorem in two dimensions. Another proof is in Toby Lam’s post.

Orthogonal curvilinear coordinates

This section is based on Dr B. J. Pendleton’s lecture notes for Electromagnetism and Relativity at the University of Edinburgh.

As a warm-up, compute the scale factors and unit vectors of spherical polar coordinates. The position vector is

\[\mathbf{r} = r \sin \theta \cos \phi \, \mathbf{e}_x + r \sin \theta \sin \phi \, \mathbf{e}_y + r \cos \theta \, \mathbf{e}_z,\]

so

\[\begin{aligned} \frac{\partial \mathbf{r}}{\partial r} &= \sin \theta \cos \phi \, \mathbf{e}_x + \sin \theta \sin \phi \, \mathbf{e}_y + \cos \theta \, \mathbf{e}_z &&\Rightarrow\quad h_r = 1, \\ \frac{\partial \mathbf{r}}{\partial \theta} &= r \cos \theta \cos \phi \, \mathbf{e}_x + r \cos \theta \sin \phi \, \mathbf{e}_y - r \sin \theta \, \mathbf{e}_z &&\Rightarrow\quad h_\theta = r, \\ \frac{\partial \mathbf{r}}{\partial \phi} &= -r \sin \theta \sin \phi \, \mathbf{e}_x + r \sin \theta \cos \phi \, \mathbf{e}_y &&\Rightarrow\quad h_\phi = r \sin \theta, \end{aligned}\]

where \(h_i = \lvert\partial\mathbf{r}/\partial u_i\rvert\). Dividing by the scale factors gives the unit vectors:

\[\begin{aligned} \mathbf{e}_r&=\sin \theta \cos \phi \, \mathbf{e}_x+\sin \theta \sin \phi \, \mathbf{e}_y+\cos \theta \, \mathbf{e}_z,\\ \mathbf{e}_\theta&=\cos \theta \cos \phi \, \mathbf{e}_x+\cos \theta \sin \phi \, \mathbf{e}_y-\sin \theta \, \mathbf{e}_z, \\ \mathbf{e}_\phi&=-\sin \phi \, \mathbf{e}_x+\cos \phi \, \mathbf{e}_y . \end{aligned}\]

Divergence

In general orthogonal coordinates \((u_1, u_2, u_3)\) with scale factors \(h_1, h_2, h_3\),

\[\nabla \cdot \mathbf{a}=\frac{1}{h_1 h_2 h_3}\left[\frac{\partial}{\partial u_1}\left(a_1 h_2 h_3\right)+\frac{\partial}{\partial u_2}\left(a_2 h_3 h_1\right)+\frac{\partial}{\partial u_3}\left(a_3 h_1 h_2\right)\right].\]

In spherical polar coordinates this becomes

\[\begin{aligned} \nabla \cdot \mathbf{a} & =\frac{1}{r^2 \sin \theta}\left[\frac{\partial}{\partial r}\left(r^2 \sin \theta\, a_r\right)+\frac{\partial}{\partial \theta}\left(r \sin \theta\, a_\theta\right)+\frac{\partial}{\partial \phi}\left(r a_\phi\right)\right] \\ & =\frac{1}{r^2} \frac{\partial}{\partial r}\left(r^2 a_r\right)+\frac{1}{r \sin \theta}\left[\frac{\partial}{\partial \theta}\left(\sin \theta\, a_\theta\right)+\frac{\partial a_\phi}{\partial \phi}\right]. \end{aligned}\]

Curl

\[\nabla \times \mathbf{a}=\frac{1}{h_1 h_2 h_3}\begin{vmatrix} h_1 \mathbf{e}_1 & h_2 \mathbf{e}_2 & h_3 \mathbf{e}_3 \\ \frac{\partial}{\partial u_1} & \frac{\partial}{\partial u_2} & \frac{\partial}{\partial u_3} \\ h_1 a_1 & h_2 a_2 & h_3 a_3 \end{vmatrix},\]

which in spherical polar coordinates reads

\[\nabla \times \mathbf{a}=\frac{1}{r^2 \sin \theta}\begin{vmatrix} \mathbf{e}_r & r \mathbf{e}_\theta & r \sin \theta \, \mathbf{e}_\phi \\ \frac{\partial}{\partial r} & \frac{\partial}{\partial \theta} & \frac{\partial}{\partial \phi} \\ a_r & r a_\theta & r \sin \theta \, a_\phi \end{vmatrix} .\]

Laplacian

For a scalar field \(f\),

\[\nabla^2 f=\frac{1}{h_1 h_2 h_3}\left[\frac{\partial}{\partial u_1}\left(\frac{h_2 h_3}{h_1} \frac{\partial f}{\partial u_1}\right)+\frac{\partial}{\partial u_2}\left(\frac{h_3 h_1}{h_2} \frac{\partial f}{\partial u_2}\right)+\frac{\partial}{\partial u_3}\left(\frac{h_1 h_2}{h_3} \frac{\partial f}{\partial u_3}\right)\right],\]

and in spherical polar coordinates

\[\begin{aligned} \nabla^2 f & =\frac{1}{r^2 \sin \theta}\left[\frac{\partial}{\partial r}\left(r^2 \sin \theta \frac{\partial f}{\partial r}\right)+\frac{\partial}{\partial \theta}\left(\sin \theta \frac{\partial f}{\partial \theta}\right)+\frac{\partial}{\partial \phi}\left(\frac{1}{\sin \theta} \frac{\partial f}{\partial \phi}\right)\right] \\ & =\frac{1}{r^2} \frac{\partial}{\partial r}\left(r^2 \frac{\partial f}{\partial r}\right)+\frac{1}{r^2 \sin \theta}\frac{\partial}{\partial \theta}\left(\sin \theta \frac{\partial f}{\partial \theta}\right)+\frac{1}{r^2 \sin ^2 \theta}\frac{\partial^2 f}{\partial \phi^2} . \end{aligned}\]

Exercise

• Write down all of these formulas for cylindrical polar coordinates as well. (Undergraduate, year 2)

Changes on 25 October 2026

What was wrong. In the Stokes example, \(\alpha\) stood both for the differential form and for an angle on the sphere. A link to lecture notes no longer worked.

What changed. The form is now \(\omega\) and the angle \(\theta\), the dead link is removed and the typesetting is tidied.