Calculus: Area in Polar Coordinates?

Question

For this problem, we are asked to find the area of the region that lies inside the first curve and outside the second curve.

Note: I am using t instead of theta.

First curve: r = 2 + Sin(t)

Second curve: r = 3Sin(t)


I know pi/2 is one of the angles where these two curves intersect, but the problem is that I don’t know how to proceed from here since pi/2 is the only angle where these two curves are equal to one another for the interval between 0 and 2 pi. Please explain how to do the problem from this point forward step-by-step.

Answer 1

I'm not sure :-)
but i think we can start with this equation:
2 +sin(t) = 3 sin(t)
<=> sin(t) = 1
<=> t = pi/2 + 2k* Pi (k: integer).

Answer 2

r = 3sin(t) describes a single petal in Quadrants I & II. It retraces itself from π < t < 2π.
dA = rdrdt
In Quadrants I & II, from symmetry
A = ∫ [(2 + sin(t))^2 - 9sin^2t]dt from 0 to π/2
For Quadrants III & IV, from symmetry
A = ∫ [(2 + sin(t))^2]dt from π to 3π/2

Answer 3

The outer radius is r = 2 + sin(theta); the interior is r = 3sin(theta). It seems such as you need to use symmetry and discover the section by capability of: A = 2(indispensable)(-pi/2 to pi/2)((2 + sin(theta))^2 - (3sin(theta))^2 d(theta)