Surface Integrals and Flux
Parameterise a surface, build its area element from a cross product, and integrate a field through it to measure flux.
What a learner can do afterwards
- Parameterise a surface and compute its area element
- Choose an orientation and say what the sign of the flux means
- Evaluate the flux of a field through a plane or a sphere
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A surface integral adds something over a curved surface in space. The scalar version multiplies a function by the area element dS, which gives things like the mass of a shell from its density. The vector version is flux: you dot the field with the unit normal and integrate, which measures the net flow through the surface. Think of flux as the volume of fluid crossing the skin per unit time.
Everything rests on a parametrisation r(u, v), which maps a flat parameter domain onto the surface. The tangent vectors r sub u and r sub v span a small parallelogram, so dS is the length of r sub u cross r sub v, times du dv. The cross product itself is a normal vector. For a graph z = g(x, y), dS becomes the square root of 1 plus gx squared plus gy squared, times dx dy.
Parametrise the plane z = 2x + y as r(u, v) = (u, v, 2u + v). Then r sub u = (1, 0, 2) and r sub v = (0, 1, 1), and their cross product is (-2, -1, 1), of length root 6. So dS = root 6 du dv. The graph formula agrees: 1 + 2 squared + 1 squared = 6.
Orientation is chosen, never given. The normal has two possible directions, and picking one fixes the sign of every flux number you compute. Field lined up with the normal pours through and counts positive, field against it counts negative, and field sliding along the surface contributes nothing. For closed surfaces the convention is outward, and you must keep the normal consistent or patches will cancel by mistake.
Two flux checks. The field (0, 0, 3) through the 2 by 5 rectangle in the plane z = 0, with normal (0, 0, 1): the dot is 3 everywhere, so the flux is 3 times 10, which is 30. On a sphere of radius R centred at the origin, the outward unit normal at (x, y, z) is (x, y, z) divided by R. For the field (x, y, z) on the unit sphere the dot is x squared + y squared + z squared = 1, so the flux equals the sphere area, 4 pi.
Check a new setup with integrand 1 in the scalar version: it must return the plain area. A sphere of radius R has area 4 pi R squared. If the check fails, look at the cross product and the parameter domain first.
Parametrise the skin, dot the field with your chosen normal, and integrate over the flat parameter domain.
The lesson and quiz items were written for this dataset. The lesson and quiz items q1, q3 and q5 to q8 were later revised.
Extra items x1 to x5 were written for this dataset. The rewritten reading was written for this dataset. Other reading entries summarise the Khan Academy articles Surface integrals and Surface area integrals. Videos are from Khan Academy.
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