Vector Calculus

Study Sheet

Vector Calculus

Line integrals, Green's theorem, div and curl

Fields, Circulation, and Flux

Tip
Conservative fields have potentials

If F=∇f\mathbf F = \nabla f then ∫CF⋅dr=f(end)−f(start)\int_C \mathbf F\cdot d\mathbf r = f(\text{end}) - f(\text{start}) — the fundamental theorem for line integrals: path-independent, and zero around every loop. Test: in the plane, F=(P,Q)\mathbf F = (P, Q) is conservative when Qx=PyQ_x = P_y.

Tip
Green's theorem trades a loop for a region

∮CP dx+Q dy=∬R(Qx−Py) dA\oint_C P\,dx + Q\,dy = \iint_R (Q_x - P_y)\,dA for a counterclockwise loop. The famous special case 12∮(−y dx+x dy)=area\tfrac12\oint(-y\,dx + x\,dy) = \text{area} powers planimeters and the shoelace formula alike.

Concept
Divergence and curl, in one breath

∇⋅F\nabla\cdot\mathbf F measures how much the field SPREADS from a point (source strength); ∇×F\nabla\times\mathbf F measures how much it ROTATES. Conservative fields are curl-free; incompressible flows are divergence-free — and the Stokes/divergence theorems extend Green to surfaces and solids.

Side note
Significance: conservation laws live here

Green's theorem and its 3D siblings are the mathematical form of 'what is created inside crosses the boundary.' Fluid flow, heat, electric charge — every conservation law in physics is one of these theorems applied to the right field. The planimeter (a 19th-century gadget that measures area by tracing a boundary) is Green's theorem built in brass.

Try it
Try it: spot the conservative field

Is F=(2xy,x2)\mathbf F = (2xy, x^2) conservative? Work: Py=2xP_y = 2x and Qx=2xQ_x = 2x — equal, so yes; potential f=x2yf = x^2y (check ∇f=(2xy,x2)\nabla f = (2xy, x^2) ✓). So ∫F⋅dr\int\mathbf F\cdot d\mathbf r from (0,0)(0,0) to (3,2)(3,2) along ANY path is f(3,2)=18f(3,2) = 18.

Proofs & Why It Matters

Tip
Proof: the fundamental theorem for line integrals

If F=∇f\mathbf F = \nabla f and CC is parametrized by r(t)\mathbf r(t), a≤t≤ba \le t \le b, then ∫CF⋅dr=∫ab∇f(r(t))⋅r′(t) dt\int_C \mathbf F\cdot d\mathbf r = \int_a^b \nabla f(\mathbf r(t))\cdot\mathbf r'(t)\,dt.

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

Tip
Proof sketch: Green's theorem on a rectangle

On [a,b]×[c,d][a,b]\times[c,d]: ∬Qx dA=∫cd[Q(b,y)−Q(a,y)]dy\iint Q_x\,dA = \int_c^d\left[Q(b,y) - Q(a,y)\right]dy, which is the line integral of Q dyQ\,dy up the right side and down the left. Similarly −∬Py dA-\iint P_y\,dA produces P dxP\,dx along the bottom and top. Adding gives the counterclockwise loop integral.

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

Tip
Proof: the ½∮(x dy − y dx) area formula

Apply Green's theorem with P=−y2P = -\tfrac{y}{2}, Q=x2Q = \tfrac{x}{2}: then Qx−Py=12+12=1Q_x - P_y = \tfrac12 + \tfrac12 = 1, so the loop integral equals ∬1 dA\iint 1\,dA — the enclosed area. Choosing polygon vertices as the path gives the shoelace formula as a corollary. ■\blacksquare

Going Deeper: Explanations & Worked Problems

Concept
Conservative or not: the full workflow

Given F=(P,Q)\mathbf F = (P, Q), the question "does a potential exist?" has a mechanical answer. Test: compute QxQ_x and PyP_y; if they differ, no potential, stop.

If they agree (on a region without holes), CONSTRUCT the potential: integrate PP in xx to get f=∫P dx+g(y)f = \int P\,dx + g(y) with an unknown function of yy alone; differentiate this ff in yy, set it equal to QQ, and solve for g′(y)g'(y); integrate once more. Example: F=(2xy,;x2+3y2)\mathbf F = (2xy,; x^2 + 3y^2). Check: Py=2x=QxP_y = 2x = Q_x ✓. Integrate: f=x2y+g(y)f = x^2y + g(y). Match: fy=x2+g′(y)=!x2+3y2f_y = x^2 + g'(y) \stackrel{!}{=} x^2 + 3y^2, so g′(y)=3y2g'(y) = 3y^2 and g=y3g = y^3. Potential: f=x2y+y3f = x^2y + y^3, and every line integral of F\mathbf F is now just ff at the endpoints. The hole caveat is real: (−yx2+y2,xx2+y2)\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 2π2\pi around the origin — the missing point matters.

Example
Worked: verifying Green's theorem on the unit circle

Take F=(−y,x)\mathbf F = (-y, x) around the unit circle, counterclockwise. LEFT SIDE (the line integral, done honestly):

Step 1 — parametrize: r(t)=(cos⁡t,sin⁡t)\mathbf r(t) = (\cos t, \sin t), 0≤t≤2π0 \le t \le 2\pi, so dx=−sin⁡t dtdx = -\sin t\,dt, dy=cos⁡t dtdy = \cos t\,dt.

Step 2 — substitute: ∮(−y dx+x dy)=∫02π[(−sin⁡t)(−sin⁡t)+(cos⁡t)(cos⁡t)]dt=∫02π(sin⁡2t+cos⁡2t) dt\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 11: the integral is 2π2\pi. RIGHT SIDE (the double integral): Qx−Py=1−(−1)=2Q_x - P_y = 1 - (-1) = 2, and ∬disk2 dA=2π\iint_{\text{disk}} 2\,dA = 2\pi. The two sides agree — and dividing by 22 gives the area formula 12∮(x dy−y dx)=π\tfrac12\oint(x\,dy - y\,dx) = \pi, the area of the unit disk, computed purely from its boundary.

Example
Worked: divergence and curl side by side

Let F=(x2,;xy,;z2)\mathbf F = (x^2,; xy,; z^2). DIVERGENCE: differentiate each component by ITS OWN variable and add: ∂x(x2)+∂y(xy)+∂z(z2)=2x+x+2z=3x+2z\partial_x(x^2) + \partial_y(xy) + \partial_z(z^2) = 2x + x + 2z = 3x + 2z — at (1,2,3)(1,2,3) this is 9>09 > 0, so the field is expanding there (a net source).

CURL: the determinant recipe gives ∇×F=(∂y(z2)−∂z(xy),;∂z(x2)−∂x(z2),;∂x(xy)−∂y(x2))=(0−0,;0−0,;y−0)=(0,0,y)\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 F\mathbf F is NOT conservative — no potential exists, and loop integrals of F\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.