Integral Theorems of Vector Calculus: Green, Stokes and Gauss

The integral theorems of vector calculus relate an integral over a region to an integral over its boundary: Green’s theorem links a double integral to a line integral around a plane curve, Stokes’ theorem links a surface integral of the curl to a line integral around the edge, and Gauss’s divergence theorem links a triple integral of the divergence to the flux through the closed boundary surface. In brief: for F = (−y, x) and the unit disc, ∮ F · dr = 2π and ∬ (∂Q/∂x − ∂P/∂y) dA = ∬ 2 dA = 2π. Both sides agree, which is Green’s theorem.

All three theorems are higher-dimensional versions of the Newton-Leibniz formula ∫_a^b f′ dx = f(b) − f(a): the integral of a derivative over a region equals the values on its boundary. This guide defines divergence and curl, states each theorem with its conditions, and checks them in worked examples, with a summary table.

What are divergence and curl?

For F = (P, Q, R), the divergence is div F = ∂P/∂x + ∂Q/∂y + ∂R/∂z, a scalar measuring how much the field spreads out from a point (a source if positive). The curl is curl F = (R_y − Q_z, P_z − R_x, Q_x − P_y), a vector measuring local rotation. In the plane, the scalar curl is Q_x − P_y. A field with zero curl on a simply connected region is conservative.

Example 1: computing both

F = (x²y, yz, xz). div F = 2xy + z + x. curl F = (0 − y, 0 − z, 0 − x²) = (−y, −z, −x²).

What does Green’s theorem say?

Let D be a region in the plane bounded by a piecewise smooth, simple closed curve C traversed counterclockwise (the region stays on the left). If P and Q have continuous partial derivatives, then ∮_C P dx + Q dy = ∬_D (∂Q/∂x − ∂P/∂y) dA.

Example 2: verification on a square

F = (−y, x) on the unit square [0, 1]². Right side: ∬ 2 dA = 2. Left side: the bottom edge (y = 0, dx) gives ∫ −0 dx = 0; the right edge (x = 1, dy) gives ∫₀¹ 1 dy = 1; the top edge (y = 1, x from 1 to 0) gives ∫₁⁰ −1 dx = 1; the left edge (x = 0, dy downward) gives 0. Total 2 ✓.

Example 3: area by a line integral

Taking P = −y/2 and Q = x/2 gives Q_x − P_y = 1, so the area is A = (1/2) ∮ (x dy − y dx). For the ellipse x = a cos t, y = b sin t: x dy − y dx = ab dt, so A = (1/2) · ab · 2π = πab ✓.

Example 4: a hard-looking line integral

∮ (x² + y) dx + (3x − y²) dy around the circle x² + y² = 4. Q_x − P_y = 3 − 1 = 2. The integral equals 2 · area = 2 · 4π = 8π.

Example 5: a triangle

∮ y² dx + x² dy around the triangle (0, 0), (1, 0), (0, 1), counterclockwise. Q_x − P_y = 2x − 2y. ∬ (2x − 2y) dA = ∫₀¹ ∫₀^1−x (2x − 2y) dy dx = ∫₀¹ [2x(1 − x) − (1 − x)²] dx = (1 − 2/3) − 1/3 = 0, so the integral is 0.

What does Stokes’ theorem say?

Let S be an oriented, piecewise smooth surface whose boundary is the closed curve C, oriented consistently with S (right-hand rule: with the thumb along n, the fingers curl along C). Then ∮_C F · dr = ∬_S (curl F) · n dS. Green’s theorem is the special case of a flat surface in the xy-plane. A consequence is that the flux of curl F through a closed surface is zero.

Example 6: a hemisphere

F = (−y, x, 0), S the upper unit hemisphere with the outward normal, C the unit circle in the xy-plane. curl F = (0, 0, 2). On the circle, ∮ F · dr = ∫₀^2π 1 dt = 2π. On the surface, ∬ (0, 0, 2) · n dS = 2 ∬ n_z dS, which equals 2 times the area of the projection onto the xy-plane, 2π. The two sides agree ✓.

Example 7: choosing an easier surface

Only the boundary matters. Instead of the hemisphere, you can use the flat disc spanning the same circle, which gives ∬ 2 dA = 2π directly.

Example 8: a line integral through the curl

F = (z, x, y), C the unit circle x² + y² = 1 in the plane z = 0. curl F = (1, 1, 1). With the flat disc and n = (0, 0, 1): ∬ 1 dA = π. Direct check: on C, z = 0 so F · dr = x dy = cos t · cos t dt, with the integral π ✓.

What does Gauss’s divergence theorem say?

Let E be a solid whose boundary S is a closed, piecewise smooth surface with the outward normal. Then ∯_S F · n dS = ∭_E div F dV. The flux out of a closed surface equals the total source strength inside.

Example 9: the sphere again

F = (x, y, z), div F = 3. The triple integral over the ball of radius R is 3 · (4πR³/3) = 4πR³.

Example 10: a box

F = (x², y², z²) over [0, 1]³. div F = 2(x + y + z). ∭ = 2 · 3 · (1/2) = 3. Direct check: on each of the three faces with coordinate 1, the flux is ∬ 1 dA = 1, and the three faces with coordinate 0 give 0, so the total is 3 ✓.

Example 11: a cylinder

F = (x, y, z²) over the solid cylinder x² + y² ≤ 1, 0 ≤ z ≤ 2. div F = 1 + 1 + 2z = 2 + 2z. ∭ (2 + 2z) dV = ∫₀^2π ∫₀¹ ∫₀² (2 + 2z) r dz dr dθ = 2π · (1/2) · (4 + 4) = 8π.

Example 12: a hole in the domain

For the field F = r/|r|³ (the inverse-square field), div F = 0 away from the origin, yet the flux through any sphere around the origin is 4π. The theorem cannot be applied across the origin, where F is not defined; the nonzero flux shows a point source there.

Table of the three theorems

Who discovered these theorems?

George Green published his theorem in an 1828 essay on electricity and magnetism. The divergence theorem was found in special cases by Lagrange (1762), Gauss (1813) and Mikhail Ostrogradsky (1826), and is called the Gauss-Ostrogradsky theorem in several countries. Stokes’ theorem is named after George Gabriel Stokes, who posed it as an exam problem in 1854; a version had been communicated to him by William Thomson (Lord Kelvin). The general modern form for differential forms is often called the generalized Stokes theorem.

Where are they used?

Gauss’s theorem leads to the continuity equation for fluids and to Gauss’s law in electrostatics. Stokes’ theorem gives Faraday’s law and Ampère’s law in integral form. Green’s theorem underlies planimeters, instruments that measure areas by tracing a boundary, and the shoelace formula for polygons.

Which mistakes are common?

Wrong orientation

Green’s theorem needs the counterclockwise direction. For Stokes, apply the right-hand rule. For Gauss, the normal points outward.

Applying the theorem to a region with holes in the field

F must be smooth on the whole region, including the interior. Singularities (Example 12) invalidate the direct use of the theorem.

Forgetting boundary pieces

For a closed solid, the flux integral covers every face. If the boundary has several components (a region with a hole), add the inner boundary with the opposite orientation.

Mixing up the quantities

Gauss uses the divergence (a scalar), Stokes the curl (a vector). Taking the divergence in Stokes’ theorem is a common slip.