Line and Surface Integrals

Ordinary integrals add a quantity up along a straight axis; the MSTE paper occasionally asks you to add one up along a curve or through a boundary instead. This lesson covers the vector-calculus tools the CELE actually tests: the line integral of a scalar field, the line integral of a vector field that gives work and circulation, conservative fields and the path independence that makes them easy, Green theorem for a closed plane curve, the divergence and curl operators, and flux with the divergence theorem. Every rule is carried all the way to a number with units so you can check your own habits against it.

Line integral of a scalar field

A line integral of a scalar field adds up the values of a function f(x, y) along a curve C, weighting each value by the little piece of arc length ds it sits on:

integral over C of f ds.

To evaluate it you parametrize the curve as r(t) = (x(t), y(t)) for t running from a to b, and replace ds by the speed of that parametrization times dt:

ds = sqrt((dx/dt)^2 + (dy/dt)^2) dt.

With f = 1 the integral just returns the length of C. With f equal to a linear density the integral returns the mass of a wire, and dividing by the length returns the average value of f along the curve. The single habit that matters is never to forget the sqrt factor: ds is arc length, not dt and not dx.

Worked example: A thin wire runs in a straight line from (0, 0) to (1, 1), with position in meters, and carries a linear density lambda(x, y) = (x + y) kg/m. Parametrize the segment as r(t) = (t, t) for t from 0 to 1, so dx/dt = 1 and dy/dt = 1 and ds = sqrt(1^2 + 1^2) dt = sqrt(2) dt. Along the wire lambda = t + t = 2t, so the mass is

m = integral over C of lambda ds = integral from 0 to 1 of (2t) sqrt(2) dt = 2 sqrt(2) [t^2 / 2] from 0 to 1 = sqrt(2) = 1.41 kg.

Dropping the sqrt(2) and integrating 2t dt alone would give 1.00 kg, the classic slip of treating ds as dt.

Line integral of a vector field and work

When the field is a vector F = (P, Q) rather than a scalar, the natural thing to add up is the component of F along the direction of travel. That is the line integral of a vector field, written with the tangent step dr = (dx, dy):

integral over C of F dot dr = integral over C of (P dx + Q dy).

Physically this is the work done by a force field F as its point of application moves along C, and for a flow field it is the circulation around C. You evaluate it the same way, by parametrizing: substitute x(t), y(t), dx = (dx/dt) dt, dy = (dy/dt) dt and integrate the resulting one-variable integral in t.

Worked example: Compute the work of F = (y, 2x), with force in newtons and position in meters, along the parabola y = x^2 from (0, 0) to (1, 1). Parametrize as r(t) = (t, t^2) for t from 0 to 1, so dr = (1, 2t) dt and F = (y, 2x) = (t^2, 2t). Then F dot dr = (t^2)(1) + (2t)(2t) = t^2 + 4t^2 = 5t^2, and

W = integral from 0 to 1 of 5t^2 dt = 5 [t^3 / 3] from 0 to 1 = 5/3 = 1.67 J.

Worked example (a different path): Take the same field F = (y, 2x) but travel the straight segment y = x from (0, 0) to (1, 1). Now r(t) = (t, t), dr = (1, 1) dt, and F = (t, 2t), so F dot dr = t + 2t = 3t and W = integral from 0 to 1 of 3t dt = 3/2 = 1.50 J. The two answers differ, 1.67 J versus 1.50 J, which is the whole point: for a general field the work depends on the path, not just on the endpoints.

Conservative fields and path independence

A vector field F is conservative when it is the gradient of some scalar potential f, that is F = grad f = (f_x, f_y). Conservative fields are the pleasant special case where the line integral forgets the path entirely and depends only on the endpoints. That is the fundamental theorem for line integrals:

integral over C of grad f dot dr = f(B) - f(A).

Two consequences follow at once. First, the integral is path independent, so any route from A to B gives the same work. Second, the integral around any closed loop is zero, because B and A are the same point. In two dimensions there is a fast test for whether F = (P, Q) is conservative: on a simply connected region it is conservative exactly when

partial P / partial y = partial Q / partial x.

If that equality holds, you recover the potential f by integrating P in x, then fixing the leftover function of y by matching f_y to Q.

Worked example: Show that F = (2xy, x^2) is conservative and use it to find the work from (1, 0) to (2, 3), positions in meters and force in newtons. Here P = 2xy and Q = x^2, so partial P / partial y = 2x and partial Q / partial x = 2x; they match, so F is conservative. Integrating P = 2xy in x gives f = x^2 y + g(y), and then f_y = x^2 + g'(y) must equal Q = x^2, forcing g'(y) = 0, so f = x^2 y. By the fundamental theorem the work is

W = f(2, 3) - f(1, 0) = (2^2)(3) - (1^2)(0) = 12 - 0 = 12 J,

the same for every path between those two points. Reversing the endpoints would flip the sign to -12 J, the common orientation mistake.

Green theorem

Green theorem is the plane bridge between a line integral once around a closed curve and a double integral over the region it encloses. For a positively (counterclockwise) oriented simple closed curve C bounding a region R,

closed integral over C of (P dx + Q dy) = double integral over R of (partial Q / partial x - partial P / partial y) dA.

A clean special case gives area straight from the boundary. Choosing P = -y and Q = x makes the integrand partial Q / partial x - partial P / partial y = 1 + 1 = 2, so the double integral is twice the area. Rearranged, the area of R is

A = (1/2) closed integral over C of (x dy - y dx).

Worked example (area): Find the area of the ellipse x^2 / a^2 + y^2 / b^2 = 1 with a = 3 m and b = 2 m, using the boundary formula. Parametrize the boundary as x = a cos t, y = b sin t for t from 0 to 2 pi, so dx = -a sin t dt and dy = b cos t dt. Then x dy - y dx = (a cos t)(b cos t) dt - (b sin t)(-a sin t) dt = ab (cos^2 t + sin^2 t) dt = ab dt, and

A = (1/2) integral from 0 to 2 pi of ab dt = (1/2)(ab)(2 pi) = pi ab = pi (3)(2) = 6 pi = 18.85 m^2.

Worked example (line to area): Evaluate the closed integral over C of (x^2 dx + xy dy) where C is the triangle with vertices (0, 0), (1, 0), and (1, 1) traversed counterclockwise. Here P = x^2 and Q = xy, so partial Q / partial x - partial P / partial y = y - 0 = y. By Green theorem the line integral equals the double integral of y over the triangle, where x runs from 0 to 1 and y runs from 0 to x:

double integral over R of y dA = integral from 0 to 1 of [ integral from 0 to x of y dy ] dx = integral from 0 to 1 of (x^2 / 2) dx = (1/2)(1/3) = 1/6 = 0.167.

Doing that line integral segment by segment would take three parametrizations; Green theorem replaces all three with one easy double integral.

Divergence and curl

Two derivative operators summarize how a vector field spreads and how it swirls. For F = (P, Q) in the plane the divergence is a scalar that measures net outflow per unit area,

div F = partial P / partial x + partial Q / partial y,

and the (scalar) curl measures the local rate of rotation,

curl F = partial Q / partial x - partial P / partial y.

The curl is exactly the integrand in Green theorem, which is why Green theorem is read as "circulation around C equals the total curl inside R." A field with zero curl everywhere on a simply connected region is conservative, tying this section back to path independence. In three dimensions F = (P, Q, S) has div F = partial P / partial x + partial Q / partial y + partial S / partial z, and its curl is a full vector, but the board rarely needs more than the divergence in 3D.

Worked example: For the velocity field F = (x^2 y, x y^2), with velocity in m/s and position in m, compute the divergence and curl at the point (1, 2). The divergence is

div F = partial / partial x (x^2 y) + partial / partial y (x y^2) = 2xy + 2xy = 4xy,

so at (1, 2) it is 4(1)(2) = 8 per second (units of 1/s, since it is a velocity per length). The scalar curl is

curl F = partial / partial x (x y^2) - partial / partial y (x^2 y) = y^2 - x^2,

so at (1, 2) it is 2^2 - 1^2 = 4 - 1 = 3 per second. Swapping the two terms and writing x^2 - y^2 would give -3, the sign mistake to watch for.

Flux and the divergence theorem

Flux measures how much of a field crosses a boundary. Across a closed plane curve C with outward unit normal n, the flux is the line integral of the outward component of F,

flux = closed integral over C of F dot n ds,

and the two-dimensional divergence theorem (the flux form of Green theorem) turns that boundary integral into a double integral of the divergence over the region R inside:

closed integral over C of F dot n ds = double integral over R of div F dA.

Worked example: Find the flux of the field F = (x, y), with F in m/s and position in m, outward across the circle of radius R = 3 m. Its divergence is div F = 1 + 1 = 2, a constant, so the double integral is just 2 times the enclosed area:

flux = double integral over R of div F dA = 2 (pi R^2) = 2 pi (3^2) = 18 pi = 56.55 m^2/s.

A direct check confirms it: on the circle the outward normal is n = (x, y) / R, so F dot n = (x^2 + y^2) / R = R^2 / R = R, and multiplying by the circumference 2 pi R gives 2 pi R^2, the same answer.

The idea lifts to three dimensions unchanged. The (3D) divergence theorem says the flux of F out through a closed surface equals the triple integral of the divergence over the solid it bounds:

closed surface integral of F dot n dS = triple integral over V of div F dV.

Worked example: Find the flux of F = (x, y, z), with F in m/s and position in m, outward through the sphere of radius R = 2 m. The divergence is div F = 1 + 1 + 1 = 3, so

flux = triple integral over V of 3 dV = 3 (4/3 pi R^3) = 4 pi R^3 = 4 pi (2^3) = 32 pi = 100.5 m^3/s.

Forgetting the factor of 3 and reporting only the volume 4/3 pi R^3 = 33.5 m^3/s is the usual error.

Line integral	Formula	What it totals
Scalar field along C	integral over C of f ds	length of C (f = 1), mass of a wire, average of f
Vector field along C	integral over C of F dot dr	work done by F, circulation around C
Flux across a plane curve	closed integral over C of F dot n ds	net outflow of F through the boundary

Theorem	Statement	Trades this for that
Fundamental theorem for line integrals	integral over C of grad f dot dr = f(B) - f(A)	a gradient line integral for endpoint values
Green theorem	closed integral (P dx + Q dy) = double integral (Q_x - P_y) dA	a boundary line integral for an area double integral
Divergence theorem (3D)	closed surface integral F dot n dS = triple integral div F dV	a closed-surface flux for a volume triple integral

Exam-day strategy

• Parametrize every line integral first: write r(t) = (x(t), y(t)), then substitute and integrate in t; for a scalar integral never drop the ds = sqrt((dx/dt)^2 + (dy/dt)^2) dt factor, since that is what makes it arc length.
• Match the integrand to the question: f ds for length or mass of a wire, F dot dr for work or circulation, and F dot n ds for flux across a boundary.
• Before grinding out a vector line integral, test whether F is conservative with partial P / partial y = partial Q / partial x; if it holds, find the potential f and use f(B) - f(A), which ignores the path entirely and turns the closed-loop answer into zero.
• Reach for Green theorem whenever the curve is closed: it replaces one line integral around C with a single double integral of (Q_x - P_y) over the region, and the choice P = -y, Q = x gives the area formula A = (1/2) closed integral of (x dy - y dx).
• Keep the operators straight: divergence adds the matching partials (P_x + Q_y) and measures outflow, while the scalar curl subtracts the crossed partials (Q_x - P_y) and measures swirl; a zero-curl field on a simply connected region is conservative.
• For flux across a closed boundary, convert to a divergence integral: in the plane use double integral of div F dA, and in space use triple integral of div F dV, which for a constant divergence is just that divergence times the area or the volume.
• Watch orientation and units: counterclockwise is positive for Green theorem and outward is positive for flux, so reversing either flips the sign; carry the units through so work lands in joules, plane flux in m^2/s, and space flux in m^3/s.