Surface Integrals: Area Elements, Flux and Orientation

A surface integral sums a quantity over a curved surface S: the scalar surface integral ∬_S f dS uses the area element dS = |r_u × r_v| du dv, and the flux integral ∬_S F · n dS measures how much of a vector field passes through S in the direction of the unit normal n. In brief: for the surface z = x² + y² the area element is dS = √(1 + 4x² + 4y²) dx dy. The flux of the constant field F = (0, 0, 1) through the unit square [0, 1] × [0, 1] in the xy-plane with the upward normal is exactly 1.

Surface integrals are the two-dimensional counterpart of line integrals: instead of a curve, one integrates over a surface. This guide explains parametrized surfaces, the area element, scalar integrals, orientation and flux, with worked examples for planes, cylinders, spheres and graphs and a table of standard area elements.

How do you parametrize a surface?

A surface is described by r(u, v) = (x(u, v), y(u, v), z(u, v)) for (u, v) in a parameter region D. Common parametrizations:

• a graph z = g(x, y): r(x, y) = (x, y, g(x, y));
• a cylinder of radius R: r(θ, z) = (R cos θ, R sin θ, z);
• a sphere of radius R: r(φ, θ) = (R sin φ cos θ, R sin φ sin θ, R cos φ), with 0 ≤ φ ≤ π, 0 ≤ θ ≤ 2π, using the angles;
• a plane through a point with two direction vectors: r(u, v) = r₀ + u a + v b.

What is the area element dS?

The vectors r_u and r_v are tangent to the surface. They span a small parallelogram of area |r_u × r_v| du dv. So dS = |r_u × r_v| du dv, and the surface area is A(S) = ∬_D |r_u × r_v| du dv. For a graph z = g(x, y): dS = √(1 + g_x² + g_y²) dx dy.

Example 1: the area of a sphere

r_φ × r_θ has the length R² sin φ. A = ∫₀^2π ∫₀^π R² sin φ dφ dθ = 2π · 2R² = 4πR².

Example 2: the area of a cylinder

r_θ = (−R sin θ, R cos θ, 0), r_z = (0, 0, 1); the cross product has the length R. The lateral area for height h: ∫₀^2π ∫₀^h R dz dθ = 2πRh ✓.

Example 3: a paraboloid

The part of z = x² + y² below z = 1. In polar coordinates, dS = √(1 + 4r²) r dr dθ. A = 2π ∫₀¹ r √(1 + 4r²) dr = 2π · (1/12)(5√5 − 1) = (π/6)(5√5 − 1) ≈ 5.33.

Example 4: a tilted plane

The part of the plane x + y + z = 1 in the first octant has g = 1 − x − y, so dS = √3 dx dy. The projection is a triangle with area 1/2, so the area is √3/2 ≈ 0.866.

How do you compute a scalar surface integral?

∬_S f dS = ∬_D f(r(u, v)) |r_u × r_v| du dv. With f = surface density, the result is the mass of a thin shell.

Example 5: a function on a sphere

∬_S z² dS over the unit sphere = ∫₀^2π ∫₀^π cos² φ sin φ dφ dθ = 2π · 2/3 = 4π/3. By symmetry, ∬ x² dS is also 4π/3, and the three sum to the area 4π ✓.

Example 6: the mass of a shell

The upper hemisphere of radius 2 with density σ = z. dS = 4 sin φ dφ dθ and z = 2 cos φ. Mass = ∫₀^2π ∫₀^π/2 8 cos φ sin φ dφ dθ = 2π · 8 · (1/2) = 8π.

What is orientation of a surface?

A surface is orientable if a continuous choice of unit normal n exists. For a parametrized surface, n = (r_u × r_v)/|r_u × r_v|. Swapping the roles of u and v reverses it. A closed surface (a sphere, a cube surface) is by convention given the outward normal. The Möbius strip is a classic one-sided surface that is not orientable, so flux through it cannot be defined consistently.

How do you compute flux?

The flux of F through the oriented surface S is ∬_S F · n dS = ∬_D F(r(u, v)) · (r_u × r_v) du dv. The cross product already contains both n and dS, so no square root is needed. For a graph z = g(x, y) with the upward normal, r_x × r_y = (−g_x, −g_y, 1), and the flux is ∬ (−P g_x − Q g_y + R) dx dy for F = (P, Q, R).

Example 7: flux through a sphere

F = (x, y, z) through the sphere of radius R with the outward normal. On the sphere F · n = R, so the flux = R · 4πR² = 4πR³. This equals three times the volume of the ball (the divergence of F is 3).

Example 8: flux through a graph

F = (0, 0, z) through z = 1 − x − y in the first octant, upward. Here g_x = g_y = −1; the integrand is R = z = 1 − x − y. Flux = ∫₀¹ ∫₀^1−x (1 − x − y) dy dx = 1/6.

Example 9: flux through a cylinder

F = (x, y, 0) through the lateral surface of the cylinder x² + y² = 4, 0 ≤ z ≤ 3, outward. On the cylinder, F · n = (2 cos θ, 2 sin θ, 0) · (cos θ, sin θ, 0) = 2, so the flux = 2 · (2π · 2 · 3) = 24π.

Example 10: a closed cube

F = (x, y, z) through the surface of the cube [0, 1]³, outward. On the face x = 1, F · n = 1 and the area is 1, giving 1. On the face x = 0, F · n = −x = 0. The three pairs give 3 in total, the volume times div F = 3 ✓.

Table of area elements

Where do these ideas come from?

Archimedes found the area of a sphere (4πR²) in antiquity. The area of a parametrized surface by a double integral was developed in the eighteenth century by Euler and Lagrange. The flux concept grew out of the study of fluid flow and of electric and magnetic fields in the nineteenth century, in the works of Gauss, Green and Ostrogradsky.

Which mistakes are common?

Forgetting the area factor

Writing dx dy instead of dS on a tilted or curved surface underestimates the area. The factor √(1 + g_x² + g_y²) is required.

Wrong sign of the normal

The flux changes sign if the orientation is reversed. State whether the normal is upward, outward or inward, and check the order of u and v.

Using the square root twice

In the flux formula with r_u × r_v, the length is already included. Do not divide by |r_u × r_v| and multiply by it again unless you use the n dS form.

Forgetting the end caps

A closed solid such as a cylinder has the lateral surface plus the top and bottom. Include each piece when the problem asks about the whole boundary.