Real Analysis 579532.The condition curl
−→
F =0 suggests the use of Stokes’ theorem:
∫∫Scurl−→
F ·−→ndS=∮
∂C−→
F ·d−→
R.
We expect the answer to the question to be no. All we need is to find a surfaceSwhose
boundary lies in thexy-plane and such that the integral of
−→
G(x, y)on∂Sis nonzero.
A simple example that comes to mind is the interiorSof the ellipsex^2 + 4 y^2 =4.
Parametrize the ellipse asx=2 cosθ,y=sinθ,θ∈[ 0 , 2 π). Then
∮∂S−→
G·d−→
R=
∫ 2 π0(
−sinθ
4,
2 cosθ
4, 0
)
·(−2 sinθ,cosθ, 0 )dθ=∫ 2 π01
2
dθ=π.By Stokes’ theorem this should be equal to the integral of the curl of
−→
F over the interior
of the ellipse. The curl of
−→
F is zero except at the origin, but we can fix that by adding
a smooth tiny upward bump at the origin, which does not alter too much the above
computation. The integral should on the one hand be close to 0, and on the other hand
close toπ, which is impossible. This proves that such a vector field
−→
F cannot exist.
(48th W.L. Putnam Mathematical Competition, 1987, solution from K. Kedlaya,
B. Poonen, R. Vakil,The William Lowell Putnam Mathematical Competition1985–2000,
MAA, 2002)
533.LetD=[a 1 ,b 1 ]×[a 2 ,b 2 ]be a rectangle in the plane, anda, b∈R,a<b.We
consider the three-dimensional parallelepipedV=D×[a, b]. Denote by−→nthe outward
normal vector field on the boundary∂VofV(which is defined everywhere except on the
edges). By the Leibniz–Newton fundamental theorem of calculus,
∫bad
dt∫∫
DG(x, y, t)dxdydt=∫ba∫∫
D∂
∂tG(x, y, t)dxdydt=
∫∫
D∫ba∂
∂tG(x, y, t)dtdxdy=
∫∫
DG(x, y, b)dxdy−∫∫
DG(x, y, a)dxdy=∫
D×{b}G(x,y,t)−→
k ·d−→n+∫
D×{a}G(x,y,t)−→
k ·d−→n,where
−→
k denotes the unit vector that points in thez-direction. With this in mind, we
compute
0 =
∫ba(
d
dt∫∫
DG(x, y, t)dxdy+∮
C−→
F ·d−→
R
)
dt