Question bank — Green's theorem — proof sketch, both forms

False. $C$ must also be simple (no self-crossings) and $P,Q$ must have continuous partials on all of $D$; a singularity inside breaks the hypothesis.

False. One loop giving zero is a single data point; conservative means every closed loop gives zero, which needs $Q_x=P_y$ throughout a simply-connected region.

False. Green's theorem is a 2D Fundamental Theorem: the inner $y$-integral in the proof collapses by FTC to boundary values, so the whole result is "boundary remembers the interior" one dimension up.

False. They are the same theorem for two field arrangements; applying circulation to $(\tilde{P},\tilde{Q})=(-Q,P)$ yields the flux form for $(P,Q)$ with no new work.

True — provided the partials are continuous on all of $D$. If there is a hole/singularity, the double integral hypothesis fails and the loop integral can be nonzero.

False. It negates only the line integral side; the double integral over the fixed region $D$ is unchanged. So a clockwise loop equals $-{\iint }_D(Q_x-P_y)dA$.

True. Stokes' relates ${\oint }_C\mathbf{F}\cdot d\mathbf{r}$ to the flux of curl through a surface; taking the surface flat in the $xy$-plane reduces $\text{curl}\cdot \hat{\mathbf{k}}$ to $Q_x-P_y$, recovering the circulation form.

True. Both say total outward flux across a boundary equals the integral of divergence inside; Green's flux form is that statement in the plane.

Error: a punctured disk is not simply connected — a loop around the hole cannot shrink to a point, so curl-free does not force conservative there.

Error: the field is singular at the origin, so Green's hypothesis fails on any $D$ containing it; the actual loop integral is $2\pi$, not $0$.

Error: that is the divergence, used in the flux form. The circulation integrand is the difference $Q_x-P_y$, the scalar curl.

This is actually correct in effect — swapping to $P_y-Q_x$ is the same as prepending a minus for the reversed orientation. Just make sure you flip it once, not twice.

Error: on a vertical segment $x$ is constant so $dx=0$; those pieces contribute nothing to $\oint Pdx$.

Error: some regions (e.g. an annulus, or an L-shape read the wrong way) aren't Type I as a whole; you must chop into Type I/II pieces where internal cuts cancel.

Error: each cut is traversed twice in opposite directions, so those Line Integrals cancel, leaving only the outer boundary.

Because $Q_x-P_y$ is the counterclockwise (positive) scalar curl; to match its sign convention you must walk so the region stays on your left.

If $Q_x-P_y$ is an odd function over a symmetric region (e.g. an integrand odd in $y$ over a disk), the double integral is $0$ by cancellation, so the loop integral is $0$ with no computation.

Rotating the unit tangent $\mathbf{T}=(dx/ds,dy/ds)$ clockwise by 90° points outward when the region is on your left; that rotation sends $(a,b)\mapsto (b,-a)$.

So each piece is a single-valued graph $y=g(x)$; the counterclockwise top curve runs right-to-left, which supplies the minus sign that matches FTC's boundary evaluation.

The theorem needs continuous partials on the whole enclosed region; excising a singularity with a small loop restores the hypothesis, and that inner loop (oriented so the region stays on your left, i.e. clockwise around the hole) enters the boundary sum.

Circulation measures net spin (curl) around the loop; flux measures net outflow (divergence) through it — Green's packages both scalar quantities as boundary-versus-interior statements.

Because $Q_x-P_y=\frac{1}{2}-(-\frac{1}{2})=1$, so the double integral becomes ${\iint }_{D}1dA=\text{Area}(D)$; the field is engineered to make the integrand exactly $1$.

Both sides are $0$ trivially; it is consistent but carries no information — a degenerate sanity check.

The double integral is $0$ and the boundary has zero length, so both sides vanish; the theorem holds vacuously in the limit.

It is not simply connected, so its boundary is two loops; orient the outer counterclockwise and the inner clockwise (region on your left throughout) and sum both line integrals.

The theorem may fail — the proof needs continuous partial derivatives on an open set containing $D$, not just continuity of $P,Q$.

Fine — Green's theorem only requires piecewise-smooth, so finitely many corners are allowed; you just integrate over the smooth arcs and add.

The theorem doesn't apply directly since $C$ isn't simple; split it into simple sub-loops, each with its own consistent orientation, and treat them separately.

Positive: $\nabla \cdot \mathbf{F}=2$, so flux $={\iint }_{D}2dA=2A$ — the radially-outward field always pushes outward.