Exercises — Stokes' theorem — statement, curl-circulation connection

The left side is the circulation, a line integral around the loop $C$: ${\oint }_C\mathbf{F}\cdot d\mathbf{r}.$ The right side is the flux of the curl through the disk $S$ bounded by $C$: ${\iint }_S(\nabla \times \mathbf{F})\cdot d\mathbf{S}.$ Nothing is computed yet — the point is only to recognise which object lives where: a loop on the left, a surface on the right, and the curl appears only on the right.

Using the determinant layout with $P=2y,Q=-3x,R=z$: $\nabla \times \mathbf{F}=({\partial }_{y}R-{\partial }_{z}Q,{\partial }_{z}P-{\partial }_{x}R,{\partial }_{x}Q-{\partial }_{y}P).$ Term by term:

• ${\partial }_{y}R-{\partial }_{z}Q={\partial }_y(z)-{\partial }_z(-3x)=0-0=0$
• ${\partial }_{z}P-{\partial }_{x}R={\partial }_z(2y)-{\partial }_x(z)=0-0=0$
• ${\partial }_{x}Q-{\partial }_{y}P={\partial }_x(-3x)-{\partial }_y(2y)=-3-2=-5$

So $\nabla \times \mathbf{F}=(0,0,-5)$. It's a vector, constant, pointing in $-\hat{\mathbf{k}}$.

Point your right thumb up along $\hat{\mathbf{n}}=\hat{\mathbf{k}}$. Your fingers curl counter-clockwise as seen from above (from the $+z$ side). So $C$ is traversed counter-clockwise in the $xy$-plane. That is the amber arrow in the figure. If instead $\hat{\mathbf{n}}=-\hat{\mathbf{k}}$, everything flips to clockwise.

From L1.2, $\nabla \times \mathbf{F}=(0,0,-5)$. The surface is the unit disk in the $xy$-plane with $\hat{\mathbf{n}}=\hat{\mathbf{k}}$, so $d\mathbf{S}=(0,0,1)dA$. What we do: dot the curl with the normal — this extracts the piece of swirl that actually threads through the surface. $(\nabla \times \mathbf{F})\cdot \hat{\mathbf{n}}=(0,0,-5)\cdot (0,0,1)=-5.$ The integrand is constant, so the flux is just $-5\times (\text{area of disk})$: ${\iint }_S(\nabla \times \mathbf{F})\cdot d\mathbf{S}=-5\cdot \pi (1{)}^2=-5\pi .$ Therefore ${\oint }_C\mathbf{F}\cdot d\mathbf{r}=-5\pi$.

Parametrise the unit circle: $\mathbf{r}(t)=(\cos t,\sin t,0)$, $t:0\to 2\pi$, so $d\mathbf{r}=(-\sin t,\cos t,0)dt$. On the curve $\mathbf{F}=(2\sin t,-3\cos t,0)$. Dot them: $\mathbf{F}\cdot d\mathbf{r}=\left(2\sin t\right)(-\sin t)+(-3\cos t)(\cos t)=-2{\sin }^{2}t-3{\cos }^{2}t.$ Simplify — one algebra step. Use ${\sin }^{2}t=1-{\cos }^{2}t$, so $-2{\sin }^{2}t=-2(1-{\cos }^{2}t)=-2+2{\cos }^{2}t$. Then $-2{\sin }^{2}t-3{\cos }^{2}t=(-2+2{\cos }^{2}t)-3{\cos }^{2}t=-2-{\cos }^{2}t.$ Integrate $0\to 2\pi$. Since ${\int }_0^{2\pi }{\cos }^{2}tdt=\pi$: ${\int }_0^{2\pi }(-2-{\cos }^{2}t)dt=-2(2\pi )-\pi =-4\pi -\pi =-5\pi .✓$ Matches L2.1 exactly — the theorem holds.

Curl:

• ${\partial }_y(2z)-{\partial }_z(x)=0$
• ${\partial }_z(y)-{\partial }_x(2z)=0$
• ${\partial }_x(x)-{\partial }_y(y)=1-1=0$

So $\nabla \times \mathbf{F}=(0,0,0)$ everywhere. By Stokes, the flux of a zero curl through any surface is $0$, hence ${\oint }_C\mathbf{F}\cdot d\mathbf{r}=0$ for every closed loop. This is exactly the conservative-field situation: zero curl $\Rightarrow$ locally a gradient $\Rightarrow$ closed-loop circulation vanishes.

Curl:

• ${\partial }_y(y^2)-{\partial }_z(x^2)=2y-0=2y$
• ${\partial }_z(z^2)-{\partial }_x(y^2)=2z-0=2z$
• ${\partial }_x(x^2)-{\partial }_y(z^2)=2x-0=2x$

So $\nabla \times \mathbf{F}=(2y,2z,2x)$.

Surface — where does $d\mathbf{S}=(-g_x,-g_y,1)d{A}_{xy}$ come from? Write the triangle as a graph $z=g(x,y)=1-x-y$. Picture a tiny rectangle in the flat $xy$-shadow with sides $dx$ (along $x$) and $dy$ (along $y$). Lift its two edges up onto the slanted surface: the $dx$-edge becomes the little 3D vector $\mathbf{a}=(1,0,g_x)dx$ (move one step in $x$, the height changes by $g_x$), and the $dy$-edge becomes $\mathbf{b}=(0,1,g_y)dy$. The true patch on the surface is the parallelogram these two span, and the vector that is perpendicular to the patch with length equal to the patch's area is their cross product: $\mathbf{a}\times \mathbf{b}=(-g_x,-g_y,1)dxdy.$ That single object $(-g_x,-g_y,1)d{A}_{xy}$ is $d\mathbf{S}$: its direction gives the orientation (which way is "up/out") and its magnitude gives the tilted area. Here $g_x=g_y=-1$, so $d\mathbf{S}=(1,1,1)d{A}_{xy}.$ The $+1$ in the third slot means it points to the $+z$ side, i.e. away from the origin — the required orientation. Good.

Dot product: $(2y,2z,2x)\cdot (1,1,1)=2y+2z+2x=2(x+y+z).$ The collapse: on the plane $x+y+z=1$, this is just $2(1)=2$ — a constant!

Integrate over the shadow: projecting onto the $xy$-plane gives the triangle $x\ge 0,y\ge 0,x+y\le 1$, whose area is $\frac{1}{2}$. ${\oint }_C\mathbf{F}\cdot d\mathbf{r}={\iint }_{\text{tri}}2d{A}_{xy}=2\cdot \frac{1}{2}=1.$

Curl: ${\partial }_x(x)-{\partial }_y(-y)=1-(-1)=2$, other components $0$, so $\nabla \times \mathbf{F}=(0,0,2)$.

Orientation first — both lids must use the upward normal. The boundary $C$ is walked CCW seen from above. By the right-hand rule (thumb along $\hat{\mathbf{n}}$, fingers along $C$), the compatible normal points upward ($+z$ component positive) on whichever lid we cap with. So:

• flat disk → $\hat{\mathbf{n}}=+\hat{\mathbf{k}}$;
• hemisphere → outward-and-upward normal $\hat{\mathbf{n}}=(x,y,z)$ (which has $z\ge 0$, so its third slot is $\ge 0$: upward-facing, as required).

Both choices have a non-negative $\hat{\mathbf{k}}$-component, so both are the same orientation induced by the CCW rim — this is what lets us compare them.

(a) Flat disk: $\hat{\mathbf{n}}=\hat{\mathbf{k}}$, integrand $(0,0,2)\cdot (0,0,1)=2$, area $\pi$: ${\iint }_{\text{disk}}2dA=2\pi .$

(b) Hemisphere — why the same answer, seen geometrically. The curl is the constant vertical vector $(0,0,2)$. With the upward normal fixed above, $(0,0,2)\cdot d\mathbf{S}=2(d{S}_z)$, where $d{S}_z$ is the signed area of the patch's horizontal shadow — signed positive precisely because the normal points upward. Now tile the hemisphere into tiny patches: a steep patch near the equator is large, but it casts a thin shadow; a flat patch near the top is small but casts a shadow almost equal to itself. Add up all the (positively-signed) shadows and you exactly re-cover the flat unit disk once — no overlaps, no gaps. So ${\iint }_{\text{hemisphere}}(0,0,2)\cdot d\mathbf{S}=2\cdot (\text{total shadow area})=2\cdot (\text{area of unit disk})=2\pi .$ Had we (wrongly) taken the downward hemisphere normal, every $d{S}_z$ would flip sign and we'd get $-2\pi$ — disagreeing with the rim, a signal we'd broken the right-hand rule.

They must agree — both surfaces share the boundary circle $C$ with the same induced orientation, and Stokes says the flux of a curl depends on $S$ only through its boundary. This is the superpower from Example 3 of the parent note: pick whichever lid is easiest.

Curl (away from the axis): with $P=\frac{-y}{x^2+y^2},Q=\frac{x}{x^2+y^2}$, the $\hat{\mathbf{k}}$-component is $Q_x-P_y=\frac{y^2-x^2}{(x^2+y^2{)}^2}-\frac{y^2-x^2}{(x^2+y^2{)}^2}=0,$ and the other components vanish because $R=0$ and $P,Q$ don't depend on $z$. So $\nabla \times \mathbf{F}=0$ everywhere it is defined.

Line integral: on $\mathbf{r}(t)=(\cos t,\sin t,0)$, $x^2+y^2=1$, so $\mathbf{F}=(-\sin t,\cos t,0)$ and $d\mathbf{r}=(-\sin t,\cos t,0)dt$: $\mathbf{F}\cdot d\mathbf{r}={\sin }^{2}t+{\cos }^{2}t=1,{\int }_0^{2\pi }1dt=2\pi .$

The resolution: Stokes requires $\mathbf{F}$ to be $C^1$ (continuous first derivatives — see the definition box at the top) on a surface whose entire interior is filled. But $\mathbf{F}$ is undefined on the whole $z$-axis ($x=y=0$), and every disk capping the unit circle must pass through that axis. There is no valid surface, so Stokes simply does not apply — no contradiction. This is precisely the parent note's warning about punctured surfaces.

By Stokes, ${\oint }_C\mathbf{F}\cdot d\mathbf{r}={\iint }_{\text{disk}}(0,0,x^2+y^2)\cdot (0,0,1)dA={\iint }_{\text{disk}}(x^2+y^2)dA.$

Switch to polar coordinates ($x^2+y^2=r^2$, $dA=rdrd\theta$) because the integrand and region are both circular — polar makes the limits clean: ${\int }_0^{2\pi }{\int }_0^1{r}^2\cdot rdrd\theta ={\int }_0^{2\pi }d\theta {\int }_0^1{r}^{3}dr.$ Evaluate each piece: ${\int }_0^{2\pi }d\theta =2\pi$ and ${\int }_0^1{r}^{3}dr={\left[\frac{r^4}{4}\right]}_0^1=\frac{1}{4}$. Multiply: ${\oint }_C\mathbf{F}\cdot d\mathbf{r}=2\pi \cdot \frac{1}{4}=\frac{\pi }{2}.$ So the circulation is $\frac{\pi }{2}$.

Curl: as before, $\nabla \times \mathbf{F}=(0,0,2)$.

Wall flux: the side wall's outward normal $\hat{\mathbf{n}}=(x,y,0)$ is purely horizontal, so $(\nabla \times \mathbf{F})\cdot \hat{\mathbf{n}}=(0,0,2)\cdot (x,y,0)=0.$ The flux through the wall is $0$, hence ${\oint }_{\partial S}\mathbf{F}\cdot d\mathbf{r}=0$.

Why the two rims run opposite ways — from the right-hand rule. Stand on the outside of the wall so the outward normal $\hat{\mathbf{n}}=(x,y,0)$ pokes into your chest. The right-hand rule says: with the surface on your left, you walk the boundary keeping that normal consistent. Do this at one spot on the wall and trace where "forward" leads:

• Along the top rim ($z=2$) the induced direction comes out clockwise seen from above.
• Along the bottom rim ($z=0$) it comes out counter-clockwise seen from above.

They are opposite because the wall lies between them and the surface must stay on the same side of your body as you slide from bottom edge to top edge — flipping the sense of circulation. Concretely, each single circle gives $\pm 2\pi$ (from L3.2), so: ${\oint }_{\partial S}\mathbf{F}\cdot d\mathbf{r}=\underset{\text{bottom, CCW}}{\underbrace{(+2\pi )}}+\underset{\text{top, CW}}{\underbrace{(-2\pi )}}=0,$ in agreement with the zero wall-flux. The vanishing flux is precisely telling us the two rim circulations are equal and opposite.

Here $\nabla f=(f_x,f_y,f_z)$, so we set $P=f_x,Q=f_y,R=f_z$ and compute the curl:

• ${\partial }_{y}R-{\partial }_{z}Q=f_{zy}-f_{yz}$
• ${\partial }_{z}P-{\partial }_{x}R=f_{xz}-f_{zx}$
• ${\partial }_{x}Q-{\partial }_{y}P=f_{yx}-f_{xy}$

By Clairaut's theorem (for a $C^2$ function — continuous second partials — the mixed partials are equal, $f_{xy}=f_{yx}$ etc.), each bracket is $0$. Hence $\nabla \times (\nabla f)=0$.

Stokes interpretation: feed a gradient field $\mathbf{F}=\nabla f$ into Stokes. The right-hand side becomes the flux of $\nabla \times (\nabla f)=0$, which is $0$ through any surface. Therefore the left-hand side vanishes too: ${\oint }_C\nabla f\cdot d\mathbf{r}={\iint }_S(\nabla \times \nabla f)\cdot d\mathbf{S}={\iint }_{S}0\cdot d\mathbf{S}=0$ around every closed loop $C$. This is exactly the statement that potential fields do zero net work around any closed path — the "what goes up comes back down" of energy. Stokes turns the algebraic identity $\nabla \times \nabla f=0$ into the physical fact that gradient fields are conservative.

With $\hat{\mathbf{n}}=\hat{\mathbf{k}}$ and $\mathbf{F}=(P(x,y),Q(x,y),0)$, only the $\hat{\mathbf{k}}$-slot of the curl matters: $(\nabla \times \mathbf{F})\cdot \hat{\mathbf{k}}={\partial }_{x}Q-{\partial }_{y}P.$ Then Stokes reads ${\oint }_C(Pdx+Qdy)={\iint }_S(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y})dA,$ which is Green's theorem verbatim. So Green is the flat, 2D shadow of Stokes.

Apply Stokes to the left side, turning the circulation of $\mathbf{E}$ into a curl flux: ${\oint }_C\mathbf{E}\cdot d\mathbf{r}={\iint }_S(\nabla \times \mathbf{E})\cdot d\mathbf{S}.$ Equate with the right side, pulling the time derivative inside (fixed surface): ${\iint }_S(\nabla \times \mathbf{E})\cdot d\mathbf{S}=-{\iint }_S\frac{\partial \mathbf{B}}{\partial t}\cdot d\mathbf{S}.$ Since this holds for every surface $S$, the integrands must match pointwise: $\boxed{\nabla \times \mathbf{E}=-\frac{\partial \mathbf{B}}{\partial t}}$ the differential form of Faraday's law — one of Maxwell's equations. Stokes is the bridge between the integral (loop/flux) and local (curl) statements.

L2.1 $:::-5\pi$ L2.2 $:::-5\pi$ (matches) L2.3 $:::0$ L3.1 $:::1$ L3.2 $:::2\pi$ for both lids L4.1 $:::2\pi$ (Stokes inapplicable — singular axis) L4.2 $:::\pi /2$ L4.3 $:::0$