TABLE OF CONTENTS
Write an integral expression for the area for ...
the shaded region
The function we need for the integrand is
f(θ) = sin θ.
We need to find the limits that make sense for θ in this region. The region starts at and ends at the smallest angle β in the second quadrant with a cosine of . That would be
We put α, β, and f(θ) into the formula to get an integral expression for hte area of the shaded region:
the region below the x-axis and above the graph of the polar function r = sin θ – 1
If we graph r = sin θ – 1 for 0 ≤ θ ≤ π, we get the boundary of the region we're looking for, traced from left to right:
This means we want α = 0 and β = π. The area of the region is
one petal of the graph of the polar function r = sin(3θ)
By playing with the graphs on a calculator, we find the values of θ that go with each petal. The area of the shaded petal is
are also correct, as these give the areas of the other petals. Each petal has the same area as the other two petals.
one petal of the graph of the polar function r = cos(2θ)
The graph r = cos(2θ) looks like this:
We play with the graph and find the values of θ that correspond to each petal. Using this information to determine the limits of integration, any of the following integrals would be correct:
\item the intersection of the regions enclosed by r = sin θ and r = cos θ
This is the region whose area we're looking for:
If we look closely, we see that this region consists of 2 parts. One part is bounded just by the graph of cos θ, and the other part is bounded just by the graph of sin θ.
The part that's bounded by sin θ goes from θ = 0 to , and the part that's bounded by cos θ goes from to .
The area of the part bounded by sin θ is
and the area of the part bounded by cos θ is
Adding these, we find that the area of the region is
Write an integral expression for the area
between the graphs r = sin θ and r = 2sin θ
We want the area of this region:
The angle θ goes from 0 to π as we go around this region once. The outer radius is r = 2sinθ and the inner radius is r = sin θ, so the area between the graphs is
of the shaded region
This region goes from θ = α to θ = β where these are the angles at which and the unit circle intersect.
Since r = 1 on the unit circle, we need
Solving, we find
We have and . The outer radius is and the inner radius is 1,
so the area of the region is