Surface Area and Scalar Surface Integrals

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Small parameter rectangles

A parametrized surface $\mathbf r(u,v)$ maps a flat parameter region to space. Its tangent vectors $\mathbf r_u$ and $\mathbf r_v$ span a small parallelogram. Their cross-product magnitude gives the area scale:

dS=\|\mathbf r_u\times\mathbf r_v\|\,du\,dv.

Consequently $\iint_S f\,dS=\iint_D f(\mathbf r(u,v))\|\mathbf r_u\times\mathbf r_v\|du\,dv$ . Use a regular parametrization covering the surface once, except possibly along negligible seams.

Visual guide

VISUAL GUIDEA parameter square stretches into a tilted patch

The surface r(u, v) = (u, v, u + v) maps a unit square to this projected parallelogram. Its tangent edges have cross-product magnitude √3, so true area is √3 even though the parameter area is 1. Projection does not preserve area.

A surface given as a graph

For $z=g(x,y)$ , choose $\mathbf r=(x,y,g)$ . Then $dS=\sqrt{1+g_x^2+g_y^2}\,dx\,dy$ . The square root accounts for tilt. Surface area is generally larger than the area of its horizontal projection. Orientation does not affect a scalar surface integral.

Worked example: a tilted sheet

On $z=x+y$ above the unit square, $g_x=g_y=1$ , so the area is $\sqrt3$ . If the surface density is $\delta=z=x+y$ , its mass is

\int_0^1\int_0^1(x+y)\sqrt3\,dy\,dx=\sqrt3.

The numerical equality of mass and area here comes from average density $1$ , not from ignoring density.

PAUSE & THINKA quick check, not a grade

Try it yourself.

Choose an answer and explain your reasoning to yourself. Use a hint if you get stuck.

For a regular parametrized surface r(u,v), what factor converts du dv to surface area?

||rᵤ×rᵥ||

rᵤ·rᵥ

||rᵤ+rᵥ||

Hint 1 · Find a starting point

Two tangent vectors span a small parallelogram.

Hint 2 · Take the next step

Its area is the magnitude of their cross product.

Show the reasoning

Answer: ||rᵤ×rᵥ||

dS=||rᵤ×rᵥ|| du dv measures area without choosing an orientation.

Before moving on: what would make one of the other answers wrong? Saying why is part of understanding.

Worked example: cylindrical wall

For $\mathbf r(\theta,z)=(R\cos\theta,R\sin\theta,z)$ , with $0\le\theta\le2\pi$ and $0\le z\le H$ , the cross-product magnitude is $R$ . Thus lateral area is $2\pi RH$ . The top and bottom disks are separate pieces and must be added if the requested surface is closed.

Regularity and coverage

A zero cross product indicates a degenerate parameter patch. Isolated coordinate singularities can often be handled by another patch or a limiting argument; a map that traces the same area repeatedly instead overcounts it.

Practice

Find the area of $z=2x$ above a unit square.
Find the lateral area of a cylinder with $R=2,H=3$ .
Give the mass of that wall for constant density $5$ .

Show worked solutions

The area scale is $\sqrt{1+4}=\sqrt5$ , so area is $\sqrt5$ .
$2\pi RH=12\pi$ .
Multiply area by density: $60\pi$ . No end caps are included.

Further study

MIT OpenCourseWare: Multivariable Calculus provides a full university course with additional lectures and exercises.

MAKE IT YOURS

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Can you explain how the visualization connects to this lesson’s goal? If a step still feels uncertain, put this lesson on your review list and try the check again another day.

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