Write an integral expression for the area for the shaded region below
Answer
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
Now that that's settled we're ready to put α, β, and f(θ) into the formula to get an integral expression for the area of the shaded region:
Example 2
Write an integral expression for the area for the region below the x-axis and above the graph of the polar function r = sin θ – 1
Answer
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
Example 3
Write an integral expression for the area of one petal of the graph of the polar function r = sin(3θ)
Answer
By playing with the graphs on a calculator, we can find the values of θ that go with each petal. As a rule of thumb, though, the number in front of θ gives the number of petals in the graph.
Using our formula for integrating with polar coordinates, the area of the shaded petal is
The integrals
and
are also correct, as these give the areas of the other petals. Each petal has the same area as the other two petals. We'd be looking at a pretty lopsided flower otherwise.
Example 4
Write an integral expression for the area of one petal of the graph of the polar function r = cos(2θ)
Answer
The graph r = cos(2θ) looks like this:
We can 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:
Example 5
Write an integral expression for the area of the intersection of the regions enclosed by r = sin θ and r = cos θ
Answer
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
Example 6
Write an integral expression for the area
between the graphs r = sin θ and r = 2sin θ
Answer
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 = 2sin θ and the inner radius is r = sin θ, so the area between the graphs is
Example 7
Write an integral expression for the area of the shaded region below.
Answer
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
so
and
There we go. We have and . The outer radius is still and the inner radius is 1, so the area of the region is
and ends at the smallest angle β in the second quadrant with a cosine of 
. That would be
, and the part that's bounded by cos θ goes from 
to 
.
and the unit circle intersect.
so
and
and 
. The outer radius is still 
and the inner radius is 1, so the area of the region is