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Calculus III (Multivariable) Unit 8: Surface integrals, Stokes & divergence

Surface integrals measure flux through a surface. Stokes' and the divergence theorem connect them to line and volume integrals.

Covers: Parametric surfaces, flux, Stokes' theorem, divergence theorem

Study guide

Surfaces can be parametrized as r(u, v) = ⟨x(u, v), y(u, v), z(u, v)⟩, such as a sphere in spherical angles or a graph z = f(x, y) using x and y as parameters. The vectors r_u and r_v are tangent to the surface, and their cross product r_u × r_v is a normal vector. Surface area is ∬ |r_u × r_v| dA.

A surface integral of a scalar function, ∬_S f dS, uses dS = |r_u × r_v| du dv. The flux of a vector field through an oriented surface, ∬_S F·dS = ∬ F·(r_u × r_v) du dv, measures how much of the field flows through it. Orientation matters: choose the normal (outward for closed surfaces) consistently.

The divergence of F, ∇·F = ∂P/∂x + ∂Q/∂y + ∂R/∂z, measures how much the field spreads out from a point. The curl, ∇ × F, measures rotation. A conservative field has zero curl.

Stokes' theorem relates the circulation around a closed curve to the curl through any surface it bounds: ∮_C F·dr = ∬_S (curl F)·dS, with the orientation of C and S linked by the right-hand rule. The divergence theorem relates the outward flux through a closed surface to the divergence inside: ∯_S F·dS = ∭_E div F dV. Both generalize Green's theorem and the Fundamental Theorem of Calculus: the integral of a derivative over a region equals a sum over its boundary.

For a surface z = f(x, y) with an upward normal, dS = ⟨−f_x, −f_y, 1⟩ dA, so the flux is ∬ (−P f_x − Q f_y + R) dA over the surface's shadow in the xy-plane. curl F = ⟨R_y − Q_z, P_z − R_x, Q_x − P_y⟩, the determinant with rows (i, j, k), (∂/∂x, ∂/∂y, ∂/∂z) and (P, Q, R).

In Stokes' theorem you may use any surface with the given boundary curve, so pick the simplest one, often a flat disc. Orient it so the curve runs counterclockwise when viewed from the tip of the normal. To find the flux through an open surface, you can add a simple lid to make a closed surface, apply the divergence theorem, and subtract the lid's flux.

Key ideas

Formulas and rules

dS = |r_u × r_v| du dv
div F = ∇·F, curl F = ∇ × F
Stokes: ∮_C F·dr = ∬_S (curl F)·dS
Divergence: ∯_S F·dS = ∭_E div F dV

Common mistakes

Words to know

Divergence of F
∇·F
Curl of F
∇ × F
Stokes' theorem
∮_C F·dr = ∬_S (curl F)·dS
Divergence theorem
Outward flux through a closed surface = ∭ div F dV
Normal vector for a parametrized surface
r_u × r_v
Parametric surface
r(u, v) = ⟨x(u, v), y(u, v), z(u, v)⟩
Surface area (parametric)
∬ |r_u × r_v| dA
Surface integral
∬_S f dS, with dS = |r_u × r_v| du dv.
Flux
∬_S F·dS: how much of F passes through S.
Orientation
The chosen normal direction of a surface (outward for closed surfaces).
Curl of a conservative field
0

Sample practice questions

17 more questions like these, from level 1 to 5, are in the practice set.

Question 1 (level 1)

div ⟨x, y, z⟩

  1. 0
  2. 1
  3. 3
  4. ⟨1, 1, 1⟩
Show the answer and steps

Answer: C) 3

  1. div ⟨x, y, z⟩ = ∂x/∂x + ∂y/∂y + ∂z/∂z.
  2. = 1 + 1 + 1 = 3.
Question 2 (level 1)

Find div ⟨4x, −5y, −2z⟩.

  1. ⟨4, −5, −2⟩
  2. −3
  3. 0
  4. 40
Show the answer and steps

Answer: B) −3

  1. div F = ∂P/∂x + ∂Q/∂y + ∂R/∂z.
  2. 4 + (−5) + (−2) = −3.
Question 3 (level 3)

Outward flux of ⟨x, y, z⟩ through the unit sphere.

  1. 0
  2. 4π
  3. 4π/3
  4. 12π
Show the answer and steps

Answer: B) 4π

  1. Divergence theorem: flux = ∭ div F dV.
  2. div F = 3.
  3. 3 × (4/3)π = 4π.