parametric equations and polar coordinates 10. 10.4 areas and lengths in polar coordinates in this...
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![Page 1: PARAMETRIC EQUATIONS AND POLAR COORDINATES 10. 10.4 Areas and Lengths in Polar Coordinates In this section, we will: Develop the formula for the area](https://reader035.vdocuments.mx/reader035/viewer/2022062318/551c5ac95503469d6a8b50c5/html5/thumbnails/1.jpg)
PARAMETRIC EQUATIONS PARAMETRIC EQUATIONS AND POLAR COORDINATESAND POLAR COORDINATES
10
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10.4Areas and Lengths
in Polar Coordinates
In this section, we will:
Develop the formula for the area of a region
whose boundary is given by a polar equation.
PARAMETRIC EQUATIONS & POLAR COORDINATES
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AREAS IN POLAR COORDINATES
We need to use the formula for the area
of a sector of a circle
A = ½r2θ
where:
r is the radius. θ is the radian measure
of the central angle.
Formula 1
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AREAS IN POLAR COORDINATES
Formula 1 follows from the fact that
the area of a sector is proportional to
its central angle:
A = (θ/2π)πr2 = ½r2θ
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AREAS IN POLAR COORDINATES
Let R be the region bounded by the polar
curve r = f(θ) and by the rays θ = a and θ = b,
where:
f is a positive continuous function.
0 < b – a ≤ 2π
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AREAS IN POLAR COORDINATES
We divide the interval [a, b] into subintervals
with endpoints θ0, θ1, θ2, …, θn, and equal
width ∆θ.
Then, the rays θ = θi divide R into smaller regions with central angle ∆θ = θi – θi–1.
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AREAS IN POLAR COORDINATES
If we choose θi* in the i th subinterval [θi–1, θi]
then the area ∆Ai of the i th region is the area
of the sector of a circle with central angle ∆θ
and radius f(θ*).
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AREAS IN POLAR COORDINATES
Thus, from Formula 1, we have:
∆Ai ≈ ½[f(θi*)]2 ∆θ
So, an approximation to the total area
A of R is: * 21
21
[ ( )]n
ii
A f
Formula 2
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AREAS IN POLAR COORDINATES
It appears that the approximation
in Formula 2 improves as n → ∞.
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AREAS IN POLAR COORDINATES
However, the sums in Formula 2 are
Riemann sums for the function g(θ) = ½[f(θ)]2.
So, * 2 21 1
2 21
lim [ ( )] [ ( )]n b
i ani
f f d
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AREAS IN POLAR COORDINATES
Therefore, it appears plausible—and can, in
fact, be proved—that the formula for the area
A of the polar region R is:
212 [ ( )]
b
aA f d
Formula 3
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AREAS IN POLAR COORDINATES
Formula 3 is often written as
with the understanding that r = f(θ).
Note the similarity between Formulas 1 and 4.
212
b
aA r d
Formula 4
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AREAS IN POLAR COORDINATES
When we apply Formula 3 or 4, it is helpful
to think of the area as being swept out by
a rotating ray through O that starts with
angle a and ends with angle b.
Formula 4
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AREAS IN POLAR COORDINATES
Find the area enclosed by one loop of
the four-leaved rose r = cos 2θ.
The curve r = cos 2θ was sketched in Example 8 in Section 10.3
Example 1
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AREAS IN POLAR COORDINATES
Notice that the region enclosed by the right
loop is swept out by a ray that rotates from
θ = –π/4 to θ = π/4.
Example 1
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AREAS IN POLAR COORDINATES
Hence, Formula 4 gives:
4 2124
4 212 4
4 2
0
4120
41 12 4 0
cos 2
cos 2
(1 cos 4 )
sin 48
A r d
d
d
d
Example 1
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AREAS IN POLAR COORDINATES
Find the area of the region that lies
inside the circle r = 3 sin θ and outside
the cardioid r = 1 + sin θ.
Example 2
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AREAS IN POLAR COORDINATES
The values of a and b in Formula 4
are determined by finding the points
of intersection of the two curves.
Example 2
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AREAS IN POLAR COORDINATES
They intersect when 3 sin θ = 1 + sin θ,
which gives sin θ = ½.
So, θ = π/6 and 5π/6.
Example 2
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AREAS IN POLAR COORDINATES
The desired area can be found by subtracting
the area inside the cardioid between θ = π/6
and θ = 5π/6 from the area inside the circle
from π/6 to 5π/6.
Example 2
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AREAS IN POLAR COORDINATES
Thus,
5 6 212 6
5 6 212 6
(3sin )
(1 sin )
A d
d
Example 2
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AREAS IN POLAR COORDINATES
As the region is symmetric about the vertical
axis θ = π/2, we can write:
2 22 21 12 26 6
2 2
6
2 2 126
2
6
2 9sin (1 2sin sin )
(8sin 1 2sin )
(3 4cos 2 2sin ) assin (1 cos 2 )
3 2sin 2 2cos
A d d
d
d
Example 2
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AREAS IN POLAR COORDINATES
Example 2 illustrates the procedure for
finding the area of the region bounded by
two polar curves.
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AREAS IN POLAR COORDINATES
In general, let R be a region that is
bounded by curves with polar equations
r = f(θ), r = g(θ), θ = a, θ = b,
where: f(θ) ≥ g(θ) ≥ 0 0 < b – a < 2π
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AREAS IN POLAR COORDINATES
The area A of R is found by subtracting
the area inside r = g(θ) from the area inside
r = f(θ).
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AREAS IN POLAR COORDINATES
So, using Formula 3, we have:
2 2
2 2
1 1[ ( )] [ ( )]2 2
1[ ( )] [ ( )]
2
b b
a a
b
a
A f d g d
f g d
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CAUTION
The fact that a single point has many
representations in polar coordinates
sometimes makes it difficult to find all
the points of intersection of two polar curves.
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CAUTION
For instance, it is obvious from this figure
that the circle and the cardioid have three
points of intersection.
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CAUTION
However, in Example 2, we solved
the equations r = 3 sin θ and r = 1 + sin θ
and found only two such points:
(3/2, π/6) and (3/2, 5π/6)
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CAUTION
The origin is also a point of intersection.
However, we can’t find it by solving
the equations of the curves.
The origin has no single representation in polar coordinates that satisfies both equations.
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CAUTION
Notice that, when represented as
(0, 0) or (0, π), the origin satisfies
r = 3 sin θ.
So, it lies on the circle.
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CAUTION
When represented as (0, 3 π/2),
it satisfies r = 1 + sin θ.
So, it lies on the cardioid.
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CAUTION
Think of two points moving along
the curves as the parameter value θ
increases from 0 to 2π.
On one curve, the origin is reached at θ = 0 and θ = π.
On the other, it is reached at θ = 3π/2.
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CAUTION
The points don’t collide at the origin since they reach the origin at different times.
However, the curves intersect there nonetheless.
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CAUTION
Thus, to find all points of intersection of two
polar curves, it is recommended that you
draw the graphs of both curves.
It is especially convenient to use a graphing calculator or computer to help with this task.
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POINTS OF INTERSECTION
Find all points of intersection of
the curves r = cos 2θ and r = ½.
If we solve the equations r = cos 2θ and r = ½, we get cos 2θ = ½.
Therefore, 2θ = π/3, 5π/3, 7π/3, 11π/3.
Example 3
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POINTS OF INTERSECTION
Thus, the values of θ between 0 and 2π that satisfy both equations are:
θ = π/6, 5π/6, 7π/6, 11π/6
Example 3
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POINTS OF INTERSECTION
We have found four points
of intersection:
(½, π/6), (½, 5π/6), (½, 7π/6), (½, 11π/6)
Example 3
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POINTS OF INTERSECTION
However, you can see that the curves have
four other points of intersection:
(½, π/3), (½, 2π/3), (½, 4π/3), (½, 5π/3)
Example 3
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POINTS OF INTERSECTION
These can be found using symmetry or
by noticing that another equation of the circle
is r = -½.
Then, we solve
r = cos 2θ and r = -½.
Example 3
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ARC LENGTH
To find the length of a polar curve r = f(θ),
a ≤ θ ≤ b, we regard θ as a parameter and
write the parametric equations of the curve
as:
x = r cos θ = f(θ)cos θ
y = r sin θ = f (θ)sin θ
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ARC LENGTH
Using the Product Rule and differentiating
with respect to θ, we obtain:
cos sin
sin cos
dx drr
d d
dy drr
d d
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ARC LENGTH
So, using cos2 θ + sin2 θ = 1,we have:2 2
22 2 2
22 2 2
22
cos 2 cos sin sin
sin 2 sin cos cos
dx dy
d d
dr drr r
d d
dr drr r
d d
drr
d
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ARC LENGTH
Assuming that f’ is continuous, we can
use Theorem 6 in Section 10.2 to write
the arc length as:
2 2b
a
dx dyL d
d d
Formula 5
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ARC LENGTH
Therefore, the length of a curve with polar
equation r = f(θ), a ≤ θ ≤ b, is:
22b
a
drL r d
d
Formula 5
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ARC LENGTH
Find the length of the cardioid
r = 1 + sin θ
We sketched it in Example 7 in Section 10.3
Example 4
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ARC LENGTH
Its full length is given by the parameter interval
0 ≤ θ ≤ 2π.
So, Formula 5 gives:2
2 2
0
2 2 2
0
2
0
(1 sin ) cos
2 2sin
drL r d
d
d
d
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ARC LENGTH
We could evaluate this integral by multiplying
and dividing the integrand by or
we could use a computer algebra system.
In any event, we find that the length of the cardioid is L = 8.
2 2sin