Surface Integrals and Flux · seed 1 · A4, ink-friendly. The answer key prints on its own page for grown-ups.
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Adding things up over curved skins

Mathematics · Calculus & Analysis · ages 20-21
Name ______________________   Date ____________

  1. The surface r(u, v) = (u, v, 2u + 2v) has area element dS = 3 du dv. Is this true?

    Circle one:   True   False

  2. A flux computation comes out negative. What does the sign mean?

    • Net flow runs against the chosen normal
    • The surface has no area anywhere at all
    • The field is zero at every single point
    • The surface has to be a closed surface

  3. What does the flux of a field through a surface measure?

    • The net flow crossing the surface
    • The total area of the parameter domain
    • The total length of the boundary curve

  4. What is the area element dS for a parametrised surface?

    • The length of r sub u cross r sub v, times du dv
    • The dot product of r sub u and r sub v, times du dv
    • The length of r sub u plus r sub v, times du dv

  5. The outward flux of F = (x, y, z) through the sphere of radius 2 centred at the origin is k times pi. What is k?

    Answer: ______________

  6. To test a setup, you put integrand 1 into the scalar surface integral over the unit sphere and get 0. What is the likely fault?

    • The unit sphere has no surface area at all
    • A constant integrand is never allowed here
    • The cross product or the domain is wrong

  7. Flipping the chosen normal changes the sign of the flux.

    Circle one:   True   False

  8. The flat unit square 0 <= x <= 1, 0 <= y <= 1 lies in the plane z = 0. The field F = (0, 0, 5) points straight up. What is the upward flux through the square?

    Answer: ______________

  9. Two students parametrise the same open surface with opposite normals and compare flux values. How should the numbers relate?

    • They must be negatives of each other
    • They must be exactly the same number
    • At least one of them must equal zero

  10. A student dots the field with a tangent vector instead of the normal and calls it flux. What did they compute?

    • The correct flux, found by a shortcut
    • The surface area, found by a new rule
    • Not flux: flux needs the normal part
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Answer key

For grown-ups. Fold this page away before handing over the rest.

Adding things up over curved skins W1-mt_b3FIxOVaTO-s1

  1. True · The tangents are (1, 0, 2) and (0, 1, 2). Their cross product is (-2, -2, 1), of length 3, so dS = 3 du dv.
  2. Net flow runs against the chosen normal · The sign reports net flow relative to the chosen normal.
  3. The net flow crossing the surface · Flux dots the field with the normal and totals what pours through.
  4. The length of r sub u cross r sub v, times du dv · The cross product length, times du dv, is the area of the small patch.
  5. 32 · The outward unit normal is (x, y, z)/2, so F dot n = 4/2 = 2 on the sphere. The area is 16 pi, so the flux is 32 pi.
  6. The cross product or the domain is wrong · Integrand 1 must return the plain area, 4 pi for the unit sphere, so a result of 0 points to the cross product or the parameter domain.
  7. True · Every dot product flips sign with the normal, so the total negates.
  8. 5 · Flux is normal component times area: 5 times 1 = 5.
  9. They must be negatives of each other · Orientation is chosen, and opposite choices negate every contribution.
  10. Not flux: flux needs the normal part · Tangents lie in the skin and miss the crossing part that flux totals.
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