Math 21a: Multivariable calculus Oliver Knill, Fall 2019
Lecture 29: Curl, Divergence and Flux
The curl of F~ = [P, Q] is Qx − Py , a scalar field. The curl of F~ = [P, Q, R] is
curl(P, Q, R) = [Ry − Qz , Pz − Rx , Qx − Py ] .
We can write curl(F~ ) = ∇ × F~ . Fields of zero curl are called irrotational.
1 The curl of the vector field [x2 + y 5 , z 2 , x2 + z 2 ] is [−2z, −2x, −5y 4 ].
If you place a “paddle wheel” pointing into the direction
v, its rotation speed F~ · ~v . The direction in which the
wheel turns fastest, is the direction of curl(F~ ). The
angular velocity is the magnitude of the curl.
The divergence of F~ = [P, Q, R] is div([P, Q, R]) = ∇ · F~ = Px + Qy + Rz . The
divergence of F~ = [P, Q] is div(P, Q) = ∇ · F~ = Px + Qy .
The divergence measures the “expansion” of a field. Fields of zero divergence are incompressible.
With ∇ = [∂x , ∂y , ∂z ], we can write curl(F~ ) = ∇ × F~ and div(F~ ) = ∇ · F~ .
∆f = div(grad(f )) = fxx + fyy + fzz .
is the Laplacian of f . One also writes ∆f = ∇2 f because ∇ · (∇f ) = div(grad(f ).
From ∇ · ∇ × F~ = 0 and ∇ × ∇F~ = ~0, we get
div(curl(F~ )) = 0, curl(grad(f )) = ~0.
2 ~ such that F~ = [x + y, z, y 2 ] = curl(G)?
Question: Is there a vector field G ~
Answer: No, because div(F~ ) = 1 is incompatible with div(curl(G)) ~ = 0.
3 Show that in simply connected region, every irrotational and incompressible field can be
written as a vector field F~ = grad(f ) with ∆f = 0. Proof. Since F~ is irrotational, there exists
a function f satisfying F = grad(f ). Now, div(F ) = 0 implies divgrad(f ) = ∆f = 0.
4 Find an example of a field which is both incompressible and irrotational. Solution. Find
f which satisfies the Laplace equation ∆f = 0, like f (x, y) = x3 − 3xy 2 , then look at its
gradient field F~ = ∇f . In that case, this gives
F~ (x, y) = [3x2 − 3y 2 , −6xy] .
We have now all the derivatives together. In dimension d, there are d fundamental derivatives.
1
grad
1 −→ 1
grad curl
1 −→ 2 −→ 1
grad curl div
1 −→ 3 −→ 3 −→ 1
If a surface S is parametrized as ~r(u, v) = [x(u, v), y(u, v), z(u, v)] over a domain R
in the uv-plane and F~ is a vector field, then the flux integral of F~ through S is
Z Z
F~ (~r(u, v)) · (~ru × ~rv ) dudv .
F
1 Compute the flux of F~ (x, y, z) = [0, 1, z 2 ] through the upper half sphere S parametrized by
~r(u, v) = [cos(u) sin(v), sin(u) sin(v), cos(v)] .
Solution. We have ~ru × ~rv = − sin(v)~r and F~ (~r(u, v)) = [0, 1, cos2 (v)] so that
Z 2π Z π
−[0, 1, cos2 (v)] · [cos(u) sin2 (v), sin(u) sin2 (v), cos(v) sin(v)] dudv .
0 0
R 2π R π Rπ
The flux integral is 0 π/2
− sin2 (v) sin(u)−cos3 (v) sin(v) dudv which is − π/2
cos3 v sin(v) dv =
π/2
cos4 (v)/4|0 = −1/4.
2 Calculate the flux of F~ (x, y, z) = [1, 2, 4z] through the paraboloid z = x2 + y 2 lying
above the region x2 + y 2 ≤ 1. Solution: We can parametrize the surface as ~r(r, θ) =
[r cos(θ),R r sin(θ), r2 ]Rwhere ~rr × ~rθ = [−2r2 cos(θ), −2r2 sin(θ), r] and F~ (~r(u, v)) = [1, 2, 4r2 ].
We get S F~ · dS ~ = 2π 1 (−2r2 cos(v) − 4r2 sin(v) + 4r3 ) drdθ = 2π.
R
0 0
3 ~ for F~ (x, y, z) = [xy, yz, zx], where S is the part
RR
Evaluate the flux integral S curl(F ) · dS
2 2
of the paraboloid z = 4 − x − y that lies above the square [0, 1] × [0, 1] and has an
upward orientation. Solution: curl(F ) = [−y, −z, −x]. The parametrization ~r(u, v) =
[u, v, 4 − u2 − vR2 ] gives ru × rv = [2u, 2v, 1] and curl(F )(~r(u, v)) = [−v, u2 + v 2 − 4, −u]. The
1R1
flux integral is 0 0 [−2uv+2v(u2 +v 2 −4)−u] dvdu = −1/2+1/3+1/2−4−1/2 = −25/6.
