Module 44: Surface area, scalar surface integrals, and flux

A. Prerequisites and observable goals#

B. The central idea: a surface needs two parameters and a normal#

A curve uses one parameter; a surface patch uses two: $\vv r(u,v)$. The two derivative vectors describe tangent directions. Their cross product gives a normal and an area scale. A scalar surface integral weights actual curved area. Flux additionally selects the component of a vector field passing through the oriented surface. Choosing an orientation is part of the problem, not an optional sign adjustment afterward.

C. Area elements and orientation#

For a smooth regular patch with $\vv r_u\times\vv r_v\ne\vv0$ and a parameter region that covers the surface once except permissible boundary seams,

For the orientation of $\vv r_u\times\vv r_v$,

The cross product already contains the area scale. Do not multiply its dot product by its norm again. Reversing orientation negates flux but does not change scalar surface area. [O3Surface]

For a graph $z=g(x,y)$, upward vector area is $\langle-g_x,-g_y,1\rangle\dd x\dd y$, and scalar area is $\sqrt{1+g_x^2+g_y^2}\dd x\dd y$. Downward orientation reverses the vector. A closed surface encloses a solid and has a conventional outward orientation; an open patch has a boundary and must be assigned one of its normal choices explicitly.

D. Why cross products appear again#

A small parameter rectangle has tangent side vectors approximately $\vv r_u\Delta u$ and $\vv r_v\Delta v$. Its mapped parallelogram has area $\|\vv r_u\times\vv r_v\|\Delta u\Delta v$. Dividing the cross product by its length yields a unit normal, so multiplying $\vv F\cdot\vv n$ by scalar area cancels the normalization denominator and gives the vector-area formula.

The same geometric distinction appeared in line integrals: density times length used a norm, while work along a direction used a dot product. Here density times area uses a cross-product norm, and flow through an orientation uses a dot product with the cross product.

E. A safe surface-integral workflow#

Identify the surface equation, its boundary, and orientation. Choose a graph or a two-parameter description with a simple parameter domain. Differentiate both parameter directions, compute their cross product, and test its direction at an easy point. Reverse it if necessary. Substitute the field and integrate with either the norm or the oriented vector, depending on the requested quantity. For a closed surface made of several patches, account for every patch with outward normals; a missing disk can change flux even when it looks visually small.

F. Worked examples#

A cylinder’s side. The radius-$a$ cylinder is $\vv r(\theta,z)=\langle a\cos\theta,a\sin\theta,z\rangle$. The cross product $\vv r_\theta\times\vv r_z=\langle a\cos\theta,a\sin\theta,0\rangle$ points outward and has norm $a$. Therefore lateral area over height $H$ is $\int_0^H\int_0^{2\pi}a\dd\theta\dd z=2\pi aH$. The top and bottom disks are not included.

A sphere’s orientation. With fixed radius $R$, use $\vv r(\phi,\theta)=R\langle\sin\phi\cos\theta,\sin\phi\sin\theta,\cos\phi\rangle$. The order $\vv r_\phi\times\vv r_\theta$ points outward and has magnitude $R^2\sin\phi$. Swapping parameter derivative order makes the normal inward. The pole singularities are coordinate degeneracies handled by patching or limits, not geometric tears in the sphere.

G. Independent practice#

F foundation C standard course work T transfer S optional stretch L runnable lab or project. Attempt the complete set before opening its solutions.

H. Complete solutions: Module 44#

Compare the first step where your work differs. A corrected answer is not yet an independent success.

Open every solution

I. Retention and readiness#

Retrieve M42.04 and M41.07. Explain when to use a cross-product norm and when to use its oriented vector directly. Advance when you can identify every patch of a closed surface and test each normal’s direction geometrically.