Surface Integrals — Mathematics Knowledge System

Overview

A surface integral extends the double integral from a flat 2D region to a curved surface in 3D space. A scalar surface integral ∬_S f dS sums a quantity (like mass density) over a surface's area; a vector (flux) surface integral ∬_S F·dS measures how much a vector field flows through the surface — the total flux, essential for the divergence theorem and Stokes' theorem.

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Formal Definition

Definition

For a surface S parametrized by r(u,v), (u,v) ∈ D:

\iint_S f\, dS = \iint_D f(\mathbf r(u,v))\, |\mathbf r_u \times \mathbf r_v|\, dA

|r_u × r_v| dA is the surface area element

Scalar surface integral

\iint_S \mathbf F \cdot d\mathbf S = \iint_D \mathbf F(\mathbf r(u,v)) \cdot (\mathbf r_u \times \mathbf r_v)\, dA

dS = n dS where n is the unit normal; orientation is chosen by the sign of r_u × r_v

Flux (vector) surface integral

\iint_S f\, dS = \iint_D f(x,y,g(x,y)) \sqrt{1+g_x^2+g_y^2}\, dA

Special case when S is the graph of a function over region D in the xy-plane

Surface given as a graph z=g(x,y)

Notation

Notation	Meaning
$\iint_S f\, dS$	Scalar surface integral of f over surface S
$\iint_S \mathbf F \cdot d\mathbf S$	Flux of vector field F through oriented surface S
$\mathbf r_u \times \mathbf r_v$	Cross product of partial derivatives of the parametrization, giving a normal vector scaled by the area element

Worked Examples

The surface is a graph z=g(x,y)=4−x−y, with gₓ=−1, g_y=−1.
$\sqrt{1+g_x^2+g_y^2} = \sqrt{1+1+1} = \sqrt3$
Integrate this constant factor over the unit square.
$\iint_{[0,1]^2} \sqrt3\, dA = \sqrt3 \cdot 1 = \sqrt3$

Answer: √3

Practice Problems

Difficulty 7/10

Find the surface area of the plane z = 2x + 2y over the triangle with vertices (0,0), (1,0), (0,1) in the xy-plane.

Difficulty 6/10

Water flows with constant velocity field F = ⟨0, 0, 5⟩ m/s upward. Find the volume flow rate (flux) through a horizontal square patch of area 4 m² with upward unit normal.

Common Mistakes

Common Mistake

Using dA (flat area element) instead of the scaled surface area element |r_u×r_v| dA when integrating over a curved surface.

A curved surface stretches area relative to its flat parameter domain D; you must multiply by |r_u×r_v| (or √(1+gx²+gy²) for a graph) to correctly account for this stretching.

Common Mistake

Forgetting that flux integrals depend on the CHOICE of orientation (which way the normal points).

Reversing a surface's orientation (choosing the opposite unit normal) negates the flux integral ∬_S F·dS — the orientation must be specified or determined from context (e.g. 'outward normal') before evaluating.

Quiz

A surface integral of a vector field (a flux integral) ∬_S F·dS measures:

the length of a curvehow much of the field flows through the surfacethe volume under a surfacethe slope of the surface

The factor √(1 + g_x² + g_y²) in a surface-area integral accounts for:

the color of the surfacehow much the tilted surface stretches relative to its flat shadow in the xy-planethe surface's temperaturethe number of variables

Summary

Surface integrals generalize double integrals to curved surfaces in 3D: scalar ∬_S f dS sums a quantity over surface area; flux ∬_S F·dS measures a vector field's flow through the surface.
The surface area element is |r_u×r_v| dA for a parametrized surface, or √(1+gx²+gy²) dA for a graph z=g(x,y).
Flux integrals require a chosen orientation (normal direction); reversing it negates the result.
Surface integrals are the key ingredient in both Stokes' theorem and the divergence theorem.

References

WebsiteWikipedia — Surface integral