Review / Chapter 2 — Vector Calculus

Stokes' Theorem & Curl of E

Chapter 2 — Vector Calculus

Stokes' theorem is to circulation what the divergence theorem is to flux. It turns a surface integral of a curl into a line integral around the boundary:

\[\int_S (\vec{\nabla}\times\vec{F})\cdot d\vec{A} = \oint_{\partial S}\vec{F}\cdot d\vec{r}.\]

Here \(S\) is any smooth open surface, \(\partial S\) is its boundary curve, \(d\vec{A}\) is the area element with orientation, and the loop integral uses the right-hand rule: fingers curl along \(d\vec{r}\), thumb points along \(d\vec{A}\).

Why it works

Tile the surface \(S\) with tiny patches. By the definition of curl, each patch satisfies \(\oint_{\partial P_i}\vec{F}\cdot d\vec{r} \approx (\vec{\nabla}\times\vec{F})\cdot\hat{n}_i A_i\). Now sum over all patches. Along every internal edge shared by two neighboring patches, the line integrals run in opposite directions and cancel. Only the external boundary survives, and it's exactly \(\oint_{\partial S}\vec{F}\cdot d\vec{r}\). Summing the right-hand sides recovers the surface integral of the curl. Take the limit and you're done.

Surface independence

If \(\vec{F}\) is nice enough everywhere, the surface integral of \(\vec{\nabla}\times\vec{F}\) depends only on the boundary curve, not on which surface you fill in. Stretch a rubber sheet however you want — the flux of the curl through it is the same. That's why "flux of a curl" is often called topologically invariant.

The curl of \(\vec{E}\) is zero (electrostatics)

For any electrostatic field,

\[\boxed{\ \vec{\nabla}\times\vec{E} = 0\ }\]

There are two equivalent ways to see this:

  1. From the potential: \(\vec{E} = -\vec{\nabla} V\), and the curl of any gradient is zero: \(\vec{\nabla}\times(\vec{\nabla} V)\equiv 0\).
  2. From path independence: \(\oint_C\vec{E}\cdot d\vec{r} = 0\) for every closed loop (because electrostatic work around a loop is zero). By Stokes, \(\int_S(\vec{\nabla}\times\vec{E})\cdot d\vec{A} = 0\) for every surface \(S\); since the integrand can be isolated via arbitrary small surfaces, \(\vec{\nabla}\times\vec{E} = 0\) pointwise.

This is a diagnostic test: given a proposed \(\vec{E}(\vec{r})\), if its curl is anywhere nonzero, it can't be an electrostatic field — and you can't write it as \(-\vec{\nabla} V\) globally.

Worked examples

(1) Verify Stokes' theorem for \(\vec{F}=(y,0,0)\) around the unit square in the \(xy\)-plane (corners at \((0,0),(1,0),(1,1),(0,1)\) with counterclockwise orientation, \(d\vec{A}=\hat{z}\,dA\)).

Curl side: \(\vec{\nabla}\times\vec{F}=(0,0,\partial_x(0)-\partial_y(y)) = -\hat{z}\). Surface integral: \(\int_S(-\hat{z})\cdot\hat{z}\,dA = -1\).

Line side: bottom edge \(y=0\Rightarrow \vec{F}=0\), contributes \(0\). Right edge \(x=1,\ y:0\to 1\), \(\vec{F}=(y,0,0)\), but \(d\vec{r}=\hat{y}\,dy\), dot product \(0\). Top edge \(y=1,\ x:1\to 0\), \(\vec{F}=(1,0,0)\), \(d\vec{r}=-\hat{x}\,dx\), dot product \(-1\), integrated over \(x\in[0,1]\) gives \(-1\). Left edge: \(0\) similarly. Total: \(-1\). Matches.

(2) The field \(\vec{E}=(cy,cx,0)\) from Chapter 2 lecture notes. Its curl is \(\partial_x(cx)-\partial_y(cy)=c-c=0\) (other components also zero), so it's a legal electrostatic field — and indeed has potential \(V=-cxy\).

Practice Problems

Problem 1easy
Use Stokes' theorem to compute \(\oint_C\vec{F}\cdot d\vec{r}\) for \(\vec{F}=(0,x,0)\) around any closed curve \(C\) bounding a flat region of area \(A\) in the \(xy\)-plane (counterclockwise).
Hint
Find the curl and dot it with \(\hat{z}\).
Solution

\(\vec{\nabla}\times\vec{F} = \hat{z}\). Stokes: \(\oint = \int_S \hat{z}\cdot\hat{z}\,dA = A\).

Answer: equals the enclosed area \(A\). (This is a standard way to compute areas from line integrals.)

Problem 2easy
A proposed electric field is \(\vec{E}=(0,0,\sin x)\). Is it a valid electrostatic field?
Hint
Compute \(\vec{\nabla}\times\vec{E}\).
Solution

\(y\)-component of the curl: \(\partial_z E_x - \partial_x E_z = 0 - \cos x = -\cos x \neq 0\).

Answer: no — curl is nonzero, so no potential exists.

Problem 3medium
If \(\vec{\nabla}\times\vec{E}=0\) in a simply-connected region, show that \(\oint_C\vec{E}\cdot d\vec{r}=0\) for every closed loop \(C\) in that region.
Hint
Fill \(C\) in with a surface \(S\) and use Stokes.
Solution

Simply-connected means any closed loop \(C\) is the boundary of some surface \(S\) entirely in the region. By Stokes, \(\oint_C\vec{E}\cdot d\vec{r} = \int_S(\vec{\nabla}\times\vec{E})\cdot d\vec{A} = \int_S 0\,dA = 0\).

Answer: follows directly from Stokes' theorem.

Problem 4medium
Compute \(\oint_C\vec{F}\cdot d\vec{r}\) directly for \(\vec{F}=(-y,x,0)\) around the unit circle in the \(xy\)-plane (counterclockwise), then verify with Stokes.
Hint
Parametrize \(x=\cos\theta\), \(y=\sin\theta\); separately, curl is a simple constant.
Solution

Direct: \(d\vec{r}=(-\sin\theta,\cos\theta,0)d\theta\), \(\vec{F}\cdot d\vec{r} = \sin^2\theta+\cos^2\theta = 1\); integral over \(\theta\in[0,2\pi]\) gives \(2\pi\). Stokes: \(\vec{\nabla}\times\vec{F} = 2\hat{z}\); flux through unit disk is \(2\pi\). Match.

Answer: \(2\pi\), and Stokes confirms.

Problem 5medium
Does \(\vec{E}=(2xy,\ x^2 + z^2,\ 2yz)\) qualify as an electrostatic field? If yes, find a potential \(V\).
Hint
Check curl; if zero, integrate a partial.
Solution

Curl components: \(x\): \(\partial_y(2yz)-\partial_z(x^2+z^2)=2z-2z=0\). \(y\): \(\partial_z(2xy)-\partial_x(2yz)=0-0=0\). \(z\): \(\partial_x(x^2+z^2)-\partial_y(2xy)=2x-2x=0\). Good. Now find \(V\) with \(-\partial_x V=2xy \Rightarrow V=-x^2 y + g(y,z)\). Then \(-\partial_y V = x^2 - g_y = x^2+z^2\), so \(g_y=-z^2\) and \(g=-yz^2 + h(z)\). Finally \(-\partial_z V = 2yz - h'(z) = 2yz\), so \(h\) constant.

Answer: yes; \(V(x,y,z) = -x^2 y - y z^2\) (up to constant).