Parametric surfaces and scalar surface integrals

Try

A roof is the graph z=x+y over the unit square. Its horizontal projection has area 1, but the sloping roof has larger area. Parameterize it by r(u,v)=(u,v,u+v), with 0≤u,v≤1. The parameter directions become tangent vectors r_u=(1,0,1) and r_v=(0,1,1). Their parallelogram gives the local surface patch produced by a small parameter rectangle.

Predict

Predict whether the roof’s area factor is 1, 2 or √3. The cross product r_u×r_v=(−1,−1,1) has magnitude √3, so total area is √3. Adding the lengths of the tangent vectors would not give area; multiplying their lengths would ignore the angle between them. The cross-product magnitude includes both lengths and the sine of their included angle.

Experiment

Now use the curved graph z=u²+v² over the same square. The tangent vectors are (1,0,2u) and (0,1,2v), so the area factor is √(1+4u²+4v²). It changes across the patch. The setup is exact even though the remaining double integral is less convenient than the plane example.

At (u,v)=(0,0), the factor is 1; at (1,1), it is 3. Compare these local values and explain why multiplying one arbitrarily selected factor by the projected area cannot generally give the exact total surface area.

Understand

For a regular parametrized surface r(u,v), the vectors r_u,r_v are independent and their cross product is nonzero. The scalar surface element is dS=|r_u×r_v|du dv. A scalar surface integral is ∫∫S f dS=∫∫D f(r(u,v))|r_u×r_v|du dv, with suitable continuity and one-to-one coverage apart from boundary exceptions. Repeated coverage counts repeated area unless explicitly corrected.

For a graph z=g(x,y), the map (x,y)↦(x,y,g) gives area factor √(1+g_x²+g_y²). The projected differential dx dy is not itself the sloping surface area. A surface density with mass-per-area units multiplies dS to produce mass. Reversing the order of the two parameters changes the cross-product sign but not its magnitude.

Apply

Parameterize a sphere of radius 3 by r(φ,θ)=(3sin φcos θ,3sin φsin θ,3cos φ), with 0≤φ≤π and 0≤θ≤2π. The cross-product magnitude is 9sin φ. Its integral is 36π. The poles make this particular parametrization nonregular, and the seam repeats boundary coordinates, but those exceptional sets have zero surface area. The sphere itself is smooth there and can be covered by other regular patches.

Repair “a zero parameter cross product proves the geometric surface is singular.” It proves the chosen parametrization is nonregular at that parameter point. As the sphere poles show, another chart may describe the same surface regularly. Conversely, one must not ignore a degeneracy without examining what set it represents.

Continue

Compare two parametrizations of the same patch and explain which factors change while the final area remains invariant.

Optional depth connects the area factor with the first fundamental form: if E=r_u·r_u, F=r_u·r_v and G=r_v·r_v, then |r_u×r_v|=√(EG−F²). This identity measures how the two parameter directions change lengths and angles, and leads toward differential geometry.

Create a surface-integral setup that lists parameter domain, tangent vectors, cross product, magnitude and composed density. Label whether a drawing is an exact planar projection or a schematic perspective. Next, retain the cross product’s direction instead of discarding it: a signed normal field will convert the same surface geometry into flux.