10.4 Areas and Lengths in
Polar Coordinates
Areas and Lengths in Polar Coordinates
In this section we develop the formula for the area of a region whose boundary is given by a polar equation. We need to use the formula for the area of a sector of a circle:
A = r2θ
where, as in Figure 1, r is the radius and θ is the radian measure of the central angle.
Areas and Lengths in Polar Coordinates
Formula 1 follows from the fact that the area of a sector is proportional to its central angle:
A = (θ/2π)π r2 = r2θ.
Areas and Lengths in Polar Coordinates
Let be the region, illustrated in Figure 2, bounded by the polar curve r = f(θ) and by the rays θ = a and θ = b, where f is a positive continuous function and where 0 < b – a ≤ 2π.
We divide the interval [a, b] into subintervals with endpoints θ0, θ1, θ2, . . . , θn and equal width ∆θ.
Figure 2
Areas and Lengths in Polar Coordinates
The rays θ = θi then divide into n smaller regions with central angle ∆θ = θi – θi –1. If we choose in the ith
subinterval [θi – 1, θi], then the area ∆Ai of the ith region is approximated by the area of the sector of a circle with central angle ∆θ and radius f( ). (See Figure 3.)
Figure 3
Areas and Lengths in Polar Coordinates
Thus from Formula 1 we have
∆Ai ≈ [f( )]2 ∆θ
and so an approximation to the total area A of is
It appears from Figure 3 that the approximation in (2) improves as n → .
Areas and Lengths in Polar Coordinates
But the sums in (2) are Riemann sums for the function g(θ) = [f(θ)]2, so
It therefore appears plausible that the formula for the area A of the polar region is
Areas and Lengths in Polar Coordinates
Formula 3 is often written as
with the understanding that r = f(θ). Note the similarity between Formulas 1 and 4.
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.
Example 1
Find the area enclosed by one loop of the four-leaved rose r = cos 2θ.
Solution:
Notice from Figure 4 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 – Solution
Therefore Formula 4 gives
cont’d
Example 1 – Solution
cont’dArc Length
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
Using the Product Rule and differentiating with respect to θ, we obtain
Arc Length
So, using cos2θ + sin2θ = 1, we have
Arc Length
Assuming that f′ is continuous, we can write the arc length as
Therefore the length of a curve with polar equation r = f(θ), a ≤ θ ≤ b, is
Example 4
Find the length of the cardioid r = 1 + sin θ. Solution:
The cardioid is shown in Figure 8.
r = 1 + sin θ
Example 4 – Solution
Its full length is given by the parameter interval 0 ≤ θ ≤ 2π, so Formula 5 gives
cont’d
Example 4 – Solution
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.
cont’d
