Question bank — Stokes' theorem — statement, curl-circulation connection

False. $S$ can be any piecewise-smooth oriented surface — a dome, a saddle, a crumpled sheet. Flatness is only convenient for computing; the theorem never asks for it.

True. The right side integrates the zero vector, giving $0$, so the circulation must also be $0$. This is exactly the "locally conservative" special case (see Conservative Fields & Potential Functions).

False. One vanishing loop can happen by coincidence or symmetry. Conservative means every closed loop gives $0$, which forces $\nabla \times \mathbf{F}=0$ everywhere (on a simply-connected region).

True, provided $\nabla \times \mathbf{F}$ is defined everywhere between them. Both equal the same ${\oint }_C\mathbf{F}\cdot d\mathbf{r}$, so they must equal each other — this is the "swap to the easy surface" trick.

False. It needs only continuous first partial derivatives ($C^1$). Conservative fields are the boring case where both sides are zero.

False. In 3D the curl is a vector with three components; Green's scalar $Q_x-P_y$ is only its $\hat{\mathbf{k}}$-component (the piece pointing out of the plane).

False. Reversing traversal flips the sign of ${\oint }_C\mathbf{F}\cdot d\mathbf{r}$. To keep the equality true you must also flip $\hat{\mathbf{n}}$, which flips the flux — the right-hand rule keeps them locked together (see Orientation & the Right-Hand Rule).

True. Take $S$ flat in the $xy$-plane with $\hat{\mathbf{n}}=\hat{\mathbf{k}}$; then $(\nabla \times \mathbf{F})\cdot \hat{\mathbf{k}}=Q_x-P_y$ and Stokes collapses to exactly Green's Theorem.

False. The theorem needs $\mathbf{F}$ to be $C^1$ on all of $S$. A singularity punches a hole in the hypotheses; you must exclude that point (e.g. cut it out and add its little boundary loop).

True, but then $C$ is all its boundary — both the outer rim and the inner rim, each oriented by the right-hand rule. Circulation means summing over every boundary curve.

The dot with $d\mathbf{S}$ is missing. Curl is a vector; you must project it onto the surface normal: $(\nabla \times \mathbf{F})\cdot d\mathbf{S}=(\nabla \times \mathbf{F})\cdot \hat{\mathbf{n}}dA$. Dropping the dot silently assumes the curl points straight through the surface.

The sign is determined, not free. Once you fix the traversal of $C$, the right-hand rule fixes $\hat{\mathbf{n}}$; picking the other sign contradicts the orientation you already chose and gives a wrong answer.

Symmetry isn't conservativeness. Its curl is $(0,0,2)\ne 0$, so it circulates — the unit-circle loop gives $2\pi$, not $0$.

Stokes relates circulation to the flux of the curl $\nabla \times \mathbf{F}$, not the flux of $\mathbf{F}$ itself. Flux of $\mathbf{F}$ is what the Divergence Theorem (Gauss) deals with — a different theorem.

Only the curl-flux is forced equal. Ordinary flux of $\mathbf{F}$ through two surfaces can differ by whatever $\iiint \nabla \cdot \mathbf{F}dV$ sits between them (divergence theorem), which need not be zero.

The vector area element $(1,1,1)dA$ already bakes in the $\sqrt{3}$ stretch of the tilted graph; $(1,1,1)$ has length $\sqrt{3}$, not $1$. The unit normal is $\frac{1}{\sqrt{3}}(1,1,1)$ — don't confuse the two.

Every interior edge is shared by two neighboring patches and gets walked once each way; those opposite traversals cancel. Only boundary edges have no neighbor to cancel against, so they survive as $C$.

For a tiny loop of area $\Delta A$ with normal $\hat{\mathbf{n}}$, the circulation is $\approx (\nabla \times \mathbf{F})\cdot \hat{\mathbf{n}}\Delta A$. Dividing by $\Delta A$ and shrinking the loop gives the curl component — literally circulation density.

Only the spin aligned with the surface's own axis pokes through it; spin sideways to the surface contributes nothing to that surface's flux. The dot product $\cos \theta$ factor keeps exactly the aligned part.

Both sides of Stokes flip sign under a reversal (traversal for the left, normal for the right). Locking them by one consistent rule keeps the equality an equality instead of an "equals up to a sign."

The right side of Stokes depends on $S$ only through its edge $C$; two surfaces sharing $C$ share the same ${\oint }_C$, hence the same curl-flux. Pick whichever surface makes the integral simplest.

A conservative field is $\mathbf{F}=\nabla f$, and its curl is $\nabla \times \nabla f=0$ identically. Stokes then makes every loop's circulation the flux of $0$, which is $0$.

Because it is Stokes applied to $\mathbf{E}$: the circulation of $\mathbf{E}$ around a loop equals the flux of $\nabla \times \mathbf{E}$ through it, which Faraday sets equal to the changing magnetic flux (see Maxwell's Equations).

Zero. There is no length to integrate over, and the surface it bounds has zero area, so both sides collapse to $0$ consistently.

Its boundary $C$ is empty, so ${\oint }_C\mathbf{F}\cdot d\mathbf{r}=0$. Therefore ${\iint }_S(\nabla \times \mathbf{F})\cdot d\mathbf{S}=0$: the curl-flux through any closed surface is always zero.

Zero, because $(\nabla \times \mathbf{F})\cdot \hat{\mathbf{n}}=0$ at every point — the spin runs parallel to the surface and none of it pokes through.

No, only its boundary (via Stokes) or equivalently its projected vector area matters. Two dented surfaces with the same rim give the same flux of a constant curl.

Then $\nabla \times \mathbf{F}=0$ no longer forces zero circulation, because you cannot fill in a surface avoiding the hole. Loops encircling the hole can have nonzero circulation.

Zero. A constant field has $\nabla \times \mathbf{F}=0$, so its curl-flux is $0$; equivalently, walking a closed loop returns you home with the pushes summing to nothing.