Error Control for the Polar Area Formula
Suppose that we wish to derive a formula for finding the area of the region
0 ≤ r ≤ f (θ)
a≤θ≤b
Call this area A. Suppose further that
f (θ) ≤ M
|f ′ (θ)| ≤ L
for all a ≤ θ ≤ b.
Divide the interval a ≤ θ ≤ b into n equal subintervals, each of length ∆θ =
b−a
.
n
Let θi∗ be the midpoint of the ith interval. On the ith interval, θ runs from θi∗ − 12 ∆θ to
θi∗ + 12 ∆θ and the radius runs over all values of f (θ) with θi∗ − 12 ∆θ ≤ θ ≤ θi∗ + 12 ∆θ. Because
|f ′ (θ)| ≤ L all of these values of f (θ) lie between ri = f (θi∗ ) − 21 L∆θ and Ri = f (θi∗ ) + 21 L∆θ.
y
θ=b
y
Ri
θ=a
ri
x
x
So the area of the region 0 ≤ r ≤ f (θ), θi∗ − 12 ∆θ ≤ θ ≤ θi∗ + 21 ∆θ must lie between
∗
2
2
2
1
1
1
and 12 ∆θ Ri2 = 12 ∆θ f (θi∗ ) + 21 L∆θ
2 ∆θ ri = 2 ∆θ f (θi ) − 2 L∆θ
Observe that
f (θi∗ ) ± 21 L∆θ
implies that, since f (θ) ≤ M ,
Hence
2
= f (θi∗ )2 ± Lf (θi∗ )∆θ + 41 L2 ∆θ 2
2
f (θi∗ )2 − LM ∆θ + 14 L2 ∆θ 2 ≤ f (θi∗ ) ± 21 L∆θ ≤ f (θi∗ )2 + LM ∆θ + 41 L2 ∆θ 2
∗ 2
1
2 f (θi ) ∆θ
− 21 LM ∆θ 2 + 81 L2 ∆θ 3 ≤ area of sector #i ≤ 21 f (θi∗ )2 ∆θ + 12 LM ∆θ 2 + 81 L2 ∆θ 3
and the total area A obeys
n h
n h
i
i
X
X
∗ 2
2
3
∗ 2
2
3
1
1
1
1 2
1
1 2
≤
A
≤
f
(θ
)
∆θ
−
LM
∆θ
+
L
∆θ
f
(θ
)
∆θ
+
LM
∆θ
+
L
∆θ
i
i
2
2
8
2
2
8
i=1
n
X
1
f (θi∗ )2 ∆θ
2
i=1
c Joel Feldman.
− 12 nLM ∆θ 2 + 81 nL2 ∆θ 3 ≤ A ≤
2001. All rights reserved.
i=1
n
X
1
f (θi∗ )2 ∆θ
2
+ 12 nLM ∆θ 2 + 18 nL2 ∆θ 3
i=1
1
Since ∆θ =
1
2
n
X
b−a
n ,
f (θi∗ )2 ∆θ −
2
LM (b−a)
2
n
3
L2 (b−a)
8
n2
+
≤A≤
i=1
n
X
∗ 2
1
2 f (θi ) ∆θ
+
2
LM (b−a)
2
n
+
3
L2 (b−a)
8
n2
i=1
Now take the limit as n → ∞. Since
lim
n→∞
n
hX
1
f (θi∗ )2 ∆θ
2
±
2
LM (b−a)
2
n
+
3
L2 (b−a)
8
n2
i=1
=
=
Z
b
1
2
b
1
2
Z
i
2
LM (b−a)
n
n→∞ 2
f (θ)2 dθ ± lim
a
3
L2 (b−a)
2
n
n→∞ 8
+ lim
f (θ)2 dθ
a
we have that
A=
1
2
Z
b
f (θ)2 dθ
a
exactly.
c Joel Feldman.
2001. All rights reserved.
2