Area, Volume, and Arc Length Exercises

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:

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

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.

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:

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

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

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