18.022 Recitation Handout (with solutions)
8 December 2014
Let D ⊂ R2 be the region enclosed by the curve r = g(θ), for some C1 , non-negative g : R → R such
that g(x + 2π) = g(x) for all x ∈ R.
1. Calculate the length of ∂D, the boundary of D. Express your answer as an integral involving g
and its first derivative.
Solution. We parametrize the curve by r(θ) = g(θ)(cos θ, sin θ) for θ ∈ [0, 2π]. Then
r0 (θ) =
∂g
(θ)(cos θ, sin θ) + g(θ)(− sin θ, cos θ)
∂θ
and
!2
∂g
|r (θ)| =
(θ) + g(θ)2 .
∂θ
0
2
Hence
2π
Z
length(∂D) =
s
0
∂g
(θ)
∂θ
!2
+ g(θ)2 dθ.
2. Let
(x(θ), y(θ)) = (g(θ) cos θ, g(θ) sin θ)
be a parametrization of ∂D. Calculate the length of ∂D again, but this time express the answer as
an integral involving the derivatives of x and y.
Solution. The velocity is now (x0 (θ), y0 (θ)), and so
2π
Z
length(∂D) =
q
x0 (θ)2 + y0 (θ)2 dθ.
0
3. Calculate the length of ∂D for the case that g(θ) = 1 − cos θ.
Solution. Note that g0 (θ) = sin θ. Hence
2π
Z
`(a, b) =
s
0
2π
Z
=
∂g
(θ)
∂θ
!2
+ g(θ)2 dθ
p
sin2 θ + 1 − 2 cos θ + cos2 θ dθ
p
2(1 − cos θ) dθ
0
2π
Z
=
0
= 8.
4. In the remainder of this problem we will prove an important theorem, called the isoperimetric
inequality (this proof is due to E. Schmidt, from 1938): it states that the length of the boundary of
any shape on the plane is at least equal to the square root of 4π times its area.
Let C be the unit circle. Explain why
I
area(D) + π =
I
∂D
(0, x) · ds +
(−y, 0) · ds.
(1)
C
Solution. This follows from Green’s Theorem.
5. Assume henceforth that g(x) ≤ 1 and g(0) = g(π) = 1. Show that

p


1 − g(θ)2 cos2 θ
0≤θ≤π
(g(θ) cos θ, p
(x(θ), w(θ)) = 

(g(θ) cos θ, − 1 − g(θ)2 cos2 θ π ≤ θ ≤ 2π
(2)
is a parametrization of a unit circle.
Solution. Since g(θ) ≤ 1 then w is real, and x(θ)2 + w(θ)2 = 1. Hence all the points (x(θ), w(θ)) lie
on the unit circle. Now, x(0) = 1 and x(π) = −1, since g(0) = g(π) = 1. Since g is continuous x is
continuous, and so x([0, π]) = [−1, 1]. It follows that (x([0, π]), w([0, π])) is the upper half circle. By
a similar argument (x([π, 2π]), w([π, 2π])) is the lower half circle.
6. Let (x(θ), w(θ)) be the parametrization of the unit circle C from (??). Again let
(x(θ), y(θ)) = (g(θ) cos θ, g(θ) sin θ)
be a parametrization of ∂D. Using (??), show that
I
area(D) + π =
2π
(x(θ), −w(θ)) · (y0 (θ), x0 (θ)) dθ.
0
Solution.
I
area(D) + 2π =
=
I
(0, x) · ds +
∂D
I 2π
(−y, 0) · ds
C
2π
I
0
w(θ)x0 (θ) dθ
x(θ)y (θ) dθ −
0
0
2π
I
=
x(θ)y0 (θ) − w(θ)x0 (θ) dθ
0
2π
I
=
(x(θ), −w(θ)) · (y0 (θ), y0 (θ)) dθ.
0
7. Explain why it follows from the previous question that
2π
I
area(D) + π ≤
q
(x(θ)2 + w(θ)2 ) · (x0 (θ)2 + y0 (θ)2 ) dθ.
0
Solution. The follows immediately from the Cauchy-Schwarz-Bunyakowski inequality for vectors
in R2 .
8. Explain why
area(D) + π ≤ length(∂D).
Solution. Since (x(θ), w(θ)) is a parametrization of a unit circle, x(θ)2 + w(θ)2 = 1. The remaining
integral is the length of ∂D.
√
9. Recall the AMGM inequality: for a, b > 0 it holds that ab ≤ (a + b)/2. Use this to show that
p
4π · area(D) ≤ length(∂D).
For which shape are these two quantities equal?
Solution. Since
area(D) + π ≤ length(∂D),
2area(D) + 2π
≤ length(∂D).
2
By the AMGM inequality
p
2 · area(D) · 2π ≤
2area(D) + 2π
,
2
and so
p
Equality is achieved for circles.
4π · area(D) ≤ length(∂D).