Vector Calculus

Why only the endpoints matter: along any path $\mathbf r(t)$, $\mathbf F\cdot\mathbf r' = \nabla f\cdot\mathbf r' = \tfrac{d}{dt}f(\mathbf r(t))$, so the line integral is the total change of $f$ (proof below). The test comes from mixed partials: if $P = f_x$ and $Q = f_y$, then $P_y = f_{xy} = f_{yx} = Q_x$. For $\mathbf F = (y, x) = \nabla(xy)$, the work from $(0, 0)$ to $(2, 1)$ is $xy$ at the end minus $xy$ at the start, $2 - 0 = 2$, whether you go straight or turn a corner at $(2, 0)$.

Why: the double integral of $Q_x - P_y$ adds up the tiny circulations of the cells of the region. Along every edge shared by two cells, the two circulations run in opposite directions and cancel, so only the outer boundary survives (proof sketch below). For the triangle with corners $(0, 0)$, $(4, 0)$, $(0, 3)$, the area formula $\tfrac12\oint(x\,dy - y\,dx)$ along the three straight edges gives $\tfrac12(0 + 12 + 0) = 6$, the area $\tfrac12\cdot4\cdot3$; the shoelace formula is exactly this computation.

Why these derivatives: the divergence $P_x + Q_y$ compares what flows out of a tiny box with what flows in — the $P_x$ term measures how much faster the flow leaves through the right face than it enters through the left. The curl's $z$-component $Q_x - P_y$ compares the push along opposite edges of the box; it is the circulation per unit area, which is why it is the integrand in Green's theorem. In the figure, $(x, y)$ points straight out everywhere (pure source), while $(-y, x)$ circles the origin (pure rotation).

Why: over a short time step the point moves by $d\mathbf r \approx \mathbf r'(t)\,dt$, and the force does work $\mathbf F\cdot d\mathbf r$ on that step; adding the steps gives the first integral. For $ds$, the step's length $|\mathbf r'(t)|\,dt$ is what gets counted, and a length has no direction, so reversing the path leaves $\int_C f\,ds$ unchanged.

Along the diameter, $\mathbf r(t) = (1 - 2t, 0)$ for $0 \le t \le 1$: then $\mathbf F = (0, 1 - 2t)$ and $\mathbf r' = (-2, 0)$, so $\mathbf F\cdot\mathbf r' = 0$ and the integral is $0$. The same endpoints give two different values, so $\mathbf F$ is not a gradient; the test agrees, since $Q_x - P_y = 1 - (-1) = 2 \ne 0$. On the arc the field pushes along the motion the whole way; on the diameter it pushes sideways and does no work.

The mass is $\int_0^1 7t\cdot5\,dt = \tfrac{35}{2}$. Check: the density grows evenly from $0$ to $7$ along a wire of length $5$, so its average is $\tfrac72$ and the mass is $5\cdot\tfrac72 = \tfrac{35}{2}$.

Why the orientation matters: the proof tiles $R$ by tiny cells and lets the shared edges cancel, which works only if every cell is traversed the same way, with its inside on the left. Reversing the boundary flips the sign of the line integral, while the double integral does not change, so Green's theorem is only true for the counterclockwise direction.

By Green's theorem the integral is $\iint_R 2\,dA = 2\cdot4\pi = 8\pi$. The terms $\sin x$ and $e^{y^2}$, which would be hopeless along the circle, never mattered: $\sin x$ in $P$ has $\partial_y = 0$ and $e^{y^2}$ in $Q$ has $\partial_x = 0$. Each is a gradient piece, and gradients have zero loop integrals.

The left side, edge by edge. Bottom ($y = 0$, $x$ from $0$ to $3$): $\int_0^3 x^2\,dx = 9$. Right ($x = 3$, $y$ from $0$ to $2$): $\int_0^2 (3 + y^2)\,dy = 6 + \tfrac83$. Top ($y = 2$, $x$ from $3$ back to $0$): $\int_3^0 (x^2 - 2)\,dx = -3$. Left ($x = 0$, $y$ from $2$ down to $0$): $\int_2^0 y^2\,dy = -\tfrac83$. The total is $9 + 6 + \tfrac83 - 3 - \tfrac83 = 12$, matching.

The area is $\tfrac12\int_0^{2\pi}6\,dt = 6\pi$, which is $\pi ab$ with $a = 3$, $b = 2$ — the same answer the Jacobian topic reached by stretching a disk. A planimeter does this integral mechanically as its arm traces the boundary.

By the chain rule the integrand is exactly $\tfrac{d}{dt}f(\mathbf r(t))$, so the integral is $f(\mathbf r(b)) - f(\mathbf r(a))$ by the ordinary fundamental theorem of calculus. Only the endpoints survive. $\blacksquare$

A general region is tiled by small rectangles: interior edges are traversed twice in opposite directions and cancel, leaving only the outer boundary. $\blacksquare$

If they agree (on a region without holes), CONSTRUCT the potential: integrate $P$ in $x$ to get $f = \int P\,dx + g(y)$ with an unknown function of $y$ alone; differentiate this $f$ in $y$, set it equal to $Q$, and solve for $g'(y)$; integrate once more. Example: $\mathbf F = (2xy,\; x^2 + 3y^2)$. Check: $P_y = 2x = Q_x$ ✓. Integrate: $f = x^2y + g(y)$. Match: $f_y = x^2 + g'(y) \stackrel{!}{=} x^2 + 3y^2$, so $g'(y) = 3y^2$ and $g = y^3$. Potential: $f = x^2y + y^3$, and every line integral of $\mathbf F$ is now just $f$ at the endpoints. The hole caveat is real: $\left(\tfrac{-y}{x^2+y^2}, \tfrac{x}{x^2+y^2}\right)$ passes the derivative test everywhere it is defined yet has integral $2\pi$ around the origin — the missing point matters.

Step 1 — parametrize: $\mathbf r(t) = (\cos t, \sin t)$, $0 \le t \le 2\pi$, so $dx = -\sin t\,dt$, $dy = \cos t\,dt$.

Step 2 — substitute: $\oint(-y\,dx + x\,dy) = \int_0^{2\pi}\left[(-\sin t)(-\sin t) + (\cos t)(\cos t)\right]dt = \int_0^{2\pi}(\sin^2 t + \cos^2 t)\,dt$.

Step 3 — the identity collapses the integrand to $1$: the integral is $2\pi$. RIGHT SIDE (the double integral): $Q_x - P_y = 1 - (-1) = 2$, and $\iint_{\text{disk}} 2\,dA = 2\pi$. The two sides agree — and dividing by $2$ gives the area formula $\tfrac12\oint(x\,dy - y\,dx) = \pi$, the area of the unit disk, computed purely from its boundary.

CURL: the determinant recipe gives $\nabla\times\mathbf F = \left(\partial_y(z^2) - \partial_z(xy),\; \partial_z(x^2) - \partial_x(z^2),\; \partial_x(xy) - \partial_y(x^2)\right) = (0 - 0,\; 0 - 0,\; y - 0) = (0, 0, y)$. Nonzero curl means $\mathbf F$ is NOT conservative — no potential exists, and loop integrals of $\mathbf F$ can be nonzero — even though two of the three components look innocent. One derivative test settles what no amount of staring at the formula would.