10.3: Examples of Arc Lengths and Areas of
Polar Curves
Instructor: Dr. Steven Rayan
Lecture Date & Time: Friday 30 March 2012, 12.10 pm – 1 pm
Venue: Sir Sandford Fleming Building, Room #2202
These notes coincide with the material covered in an extra lecture on 30
March. Students who were not present should make sure they are comfortable
with the contents of these notes.
In the regularly-scheduled half of the lecture (11.10 am – 12.00 pm), we
ended with the derivation of a couple of important formulae concerning polar
curves. Let r = f (θ) be a general polar curve. (For instance, if the polar
curve is a unit circle centered at the origin, then f (θ) = 1. If the polar curve
is the standard cardioid, then f (θ) would be 1 + cos(θ).)
• The length of the arc of the polar curve between θ0 and θ1 is
Z θ1 p
f 0 (θ)2 + f (θ)2 dθ.
L =
θ0
• The area enclosed by the polar curve and the two rays θ = θ0 and
θ = θ1 is
Z
1 θ1
A =
f (θ)2 dθ.
2 θ0
In the extra lecture that immediately followed, we calculated some examples of these quantities. Here are those examples:
E.g. Find the circumference of the cardioid r = 1 + cos(θ).
Sol’n. If we start at any point on the cardioid and then change θ, we will of
course move along the curve, to a different point on the curve. If the total
change in θ is ±2π, then we get back to the point where we started. In other
words, the circumference of the cardioid is the same as the arc length after
moving through a total angle of 2π. We may take θ0 = 0 and θ1 = 2π.
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[Remark : Just because in this example the circumference is traced out
by walking through a total angle of 2π, this does not mean that the same
is true for every example! For argument’s sake, take the circle r = 2 sin θ,
which is a unit circle centered at (0, 1). We only need to go from θ = 0 to
θ = π to get around it once — integrating from 0 to 2π would give you twice
the circumference!]
Back to the cardioid: the circumference is given by
Z
1 2π p 0 2
f (θ) + f (θ)2 dθ
L =
2 0
Z 2π p
=
(− sin(θ))2 + (1 + cos θ)2 dθ
Z0 2π p
=
2(1 + cos θ)dθ
0
Z 2π p
1 + cos θ
=
= cos2 (θ/2)
4 cos2 (θ/2)dθ ...here we used
2
0
Z 2π
= 2
cos(θ/2)dθ (∗ )
0
= 4 sin(θ/2)|2π
0
= 0
Now we need to ask ourselves “What have we done wrong?” because
the circumference of the cardioid can’t be zero! The problem lies in the line
marked (∗ ), and specifically
p in how we took the square root. In the previous
line, we were integrating 4 cos2 (θ/2), which is a nonnegative function. In
line (∗ ), we are integrating cos(θ/2), which is positive for the first half of 0
2
to 2π but negative for the second half. The end result is that one half of
the cardioid has positive length and the other has negative length — they
cancel.
Rπ
R 2π
One solution is to break the integral up into two integrals, 0 and π ,
R 2π
and to take the absolute value of the π integral so that the second half of
the arc length is positive. Another solution is to keep it all as one integral,
but to integrate over a different interval of length 2π — one on which cos(θ/2)
is completely nonnegative. Such an interval is [−π, π]. Lo and behold, if we
do this, we get
Z πp
(f 0 )2 + f 2 dθ
L =
−π
= 4 sin(θ/2)|π−π
= 4 − (−4)
= 8
E.g. Find the area inside of the cardioid r = 1 + cos(θ).
Sol’n. To get the area inside the whole cardioid, we take the area enclosed
by the cardioid and the rays θ0 and θ1 , and then let the difference between
θ0 and θ1 grow to 2π. (The two rays become the same ray.) Now the area
enclosed by the cardioid and θ0 and θ1 is the area inside the cardioid itself
(since we are sweeping all the way around). To achieve this, we can take
θ0 = 0 and θ1 = 2π for our limits of integration.
[Remark : Half of the cardioid lies above the x-axis; half, below. We
might worry that the “positive” area and the “negative” area will cancel
each other out. This is a reasonable concern, considering this can happen in
“rectangular” (x-y) integration. However, that cannot happen here. When
we derived the area formula, we summed up tiny area elements (very thin
sectors of circles). If we sweep counterclockwise, then the width of each
wedge is ∆θ > 0. Therefore, the area of the wedge is ∆A = (1/2)r2 ∆θ > 0,
and so their total sum is larger than zero. (In other words, it doesn’t matter
if part of the curve lies below the x-axis; all that matters is that we are
integrating counterclockwise the whole way through without doubling back
on ourselves.)]
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Back to the cardioid: the area inside it is
Z
1 2π
A =
f (θ)2 dθ
2 0
Z
1 2π
(1 + cos θ)2 dθ
=
2 0
Z
1 2π
(1 + 2 cos(θ) + cos2 (θ))dθ
=
2 0
Z 1
1 + cos(2θ)
1 2π 3
+ 2 cos(θ) + cos(2θ) dθ ...here we used cos2 (θ) =
=
2 0
2
2
2
2π
1 3
1
=
θ + 2 sin(θ) + sin(2θ)
2 2
4
0
3π
=
2
E.g. Find the area inside the limaçon r = 1 + 2 cos θ but outside the circle
r = 2.
Sol’n. To solve this problem, we first need to know where the two polar
curves intersect. (In the graphic below, the limaçon is the pink curve.) If
we let f (θ) = 1 + 2 cos θ and g(θ) = 2, then the limaçon and the circle meet
when
f (θ) = g(θ)
1 + 2 cos(θ) = 2
1
cos(θ) =
2
π
θ = ±
3
4
[Remark : Of course, cos(θ) is equal to 1/2 at infinitely-many values of
θ. However, the limaçon and the circle meet only twice, and so on only two
values of θ are relevant in the context of the problem: ±π/3. To see this,
draw the limaçon and the circle on the same plane, as above. In any case,
it is good practice to try and draw the limaçon yourself. It is similar to the
cardioid in many respects, but there are a couple of interesting twists and
turns!]
Now that we know the values of θ at which the curves intersect, we can
rephrase the problem like this: we want the area of the limaçon between the
rays θ0 = −π/3 and θ1 = π/3, but excluding the area of the circle between
those same rays. This is just a subtratction of one area from the other.
(From the picture it is clear that the limaçon has the larger area between
those rays, and so we subtract the circle’s area from the limaçon’s area. But
it doesn’t really matter at all about which area is the one that is subtracted:
if we subtracted the limaçon’s area from the circle’s area, then we would get
the same answer but with the wrong sign, and so we would simply take the
absolute value.)
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The desired area is
Z
1 π/3
A =
(f (θ)2 − g(θ)2 )dθ
2 −π/3
Z
1 π/3
((1 + 2 cos θ)2 − 22 )dθ
=
2 −π/3
Z
1 π/3
=
(−3 + 4 cos(θ) + 4 cos2 (θ))dθ
2 −π/3
Z
1 π/3
1 + cos(2θ)
=
(−1 + 4 cos(θ) + 2 cos(2θ))dθ ...here we used cos2 (θ) =
2 −π/3
2
1
π/3
[−θ + 4 sin(θ) + sin(2θ)]−π/3
=
2
1
=
(−π/3 + 4 sin(π/3) + sin(2π/3) − (−(−π/3)) − 4 sin(−π/3) − sin(−2π/3))
2
1
=
(−2π/3 + 8 sin(π/3) + 2 sin(2π/3)) ...because sin(x) = − sin(−x) for all x
2
= −π/3 + 4 sin(π/3) + sin(2π/3)
√
5 3 π
=
−
2
3
E.g. Find the area inside the spiral chamber
spiral r = eθ/25 .
9π
17π
≤θ≤
of the logarithmic
2
2
Sol’n. First, we need to figure out what is meant by “chamber”. Remember
that the x-coordinate of a polar curve is given by x = r cos(θ) = f (θ) cos(θ).
This means that x = 0 whenever f (θ) cos(θ) = eθ/25 cos(θ) = 0. Since eθ/25
is never zero, this means
x = 0 ⇔ cos(θ) = 0 ⇔ θ = (2k + 1)
π
2
That is, the curve crosses the y-axis every time θ is an odd multiple of π/2.
(The points where x = 0 are the y-intercepts.)
Since 9π/2 and 17π/2 are odd multiples of π/2, this means that the
portion of the curve we are interested in starts and ends on the y-axis (in
fact, the positive y-axis). Since the difference between 9π/2 and 17π/2 is
4π, this means means we have traversed two full revolutions around the
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origin before finally stopping again on the positive y-axis. Because the curve
wraps around twice, it must intersect the positive y-axis at another time in
between 9π/2 and 17π/2: this is exactly at 13π/2, which is 9π/2 + 2π. (Note
for completeness that the intercepts on the negative y-axis are 11π/2 and
15π/2.) Based on these facts, here is what the curve must look like:
As we can see, a chamber has been traced out, with endpoints on the
positive y-axis. Now that we know where the chamber is, how do we compute
its area? Suppose we were to use the area formula with limits 13π/2 to 17π/2,
i.e. ignore the inner arc from 9π/2 to 13π/2. Our thin sectors would pivot
about the origin and go around the outermost arc of the chamber, giving
us the entire area enclosed by that arc. This is far too much. We need to
remove the area inside the innermost arc. To do this, we subtract away what
the area formula would have given us if we were to have integrated from 9π/2
to 17π/2. Pictorially, we get the chamber’s area by subtracting the second
figure from the first, to the get the third:
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The final calculation is
Z
Z
1 17π/2 θ/25 2
1 13π/2 θ/25 2
A =
(e ) dθ −
(e ) dθ
2 13π/2
2 9π/2
Z
Z
1 17π/2 2θ/25
1 13π/2 2θ/25
=
e
dθ −
e
dθ
2 13π/2
2 9π/2
17π/2
13π/2
25 2θ/25 25 2θ/25 −
e
e
=
4
4
13π/2
9π/2
25 17π/25
=
e
− 2e13π/25 + e9π/25
4
∼
= 8.26
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