Home / Courses / Calculus III (Multivariable) / Unit 8
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
- Parametrize a surface with r(u, v); the normal is r_u × r_v.
- Flux ∬_S F·dS measures flow through a surface.
- curl F measures rotation; div F measures outward spreading.
- Stokes' theorem: circulation around a boundary equals flux of the curl through the surface.
- Divergence theorem: outward flux through a closed surface equals the integral of div F over the solid.
Formulas and rules
Common mistakes
- Using the divergence theorem on a surface that isn't closed.
- Mismatching surface orientation and boundary direction in Stokes' theorem.
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.
div ⟨x, y, z⟩
- 0
- 1
- 3
- ⟨1, 1, 1⟩
Show the answer and steps
Answer: C) 3
- div ⟨x, y, z⟩ = ∂x/∂x + ∂y/∂y + ∂z/∂z.
- = 1 + 1 + 1 = 3.
Find div ⟨4x, −5y, −2z⟩.
- ⟨4, −5, −2⟩
- −3
- 0
- 40
Show the answer and steps
Answer: B) −3
- div F = ∂P/∂x + ∂Q/∂y + ∂R/∂z.
- 4 + (−5) + (−2) = −3.
Outward flux of ⟨x, y, z⟩ through the unit sphere.
- 0
- 4π
- 4π/3
- 12π
Show the answer and steps
Answer: B) 4π
- Divergence theorem: flux = ∭ div F dV.
- div F = 3.
- 3 × (4/3)π = 4π.