Polar Equations

Question

Find the area of one small petal and one large petal of r = 2+4sin3θ

Answer 1

The big petals occur with positive r and the small ones with negative r, as we have the leading coefficient of 2. Rather than going from 0 to 2 pi, we will go from - pi/18 to 35 pi/18

For the large petal, 2 + 4 sin 3 theta >= 0 when 4 sin 3 theta >= -2, or sin 3 theta >= -1/2

sin theta = - 1/2 at theta = -pi/6 and 7pi/6

Thus, if 3 theta = -pi/6, 5 pi/6, theta = -pi/18 and 7 pi/18

Other big petals are, adding 2 pi/3 and 4 pi/3, from 11 pi/18 to 19 pi/18 and 23 pi/18 and 31 pi/18

Thus, for the area, using polar coordinates, r dr dtheta = dxdy, and

I[-pi/18,7 pi/18]I[0,2 + 4 sin 3 theta] r dr d theta =

I[-pi/18,7 pi/18] 1/2r^2 E[0,2 + 4 sin 3 theta] dr =

I[-pi/18,7 pi/18] 2 + 8 sin 3 theta + 8 sin^2 3 theta d theta = as sin^2 3 theta = 1/2 - 1/2 cos 6 theta,

I[-pi/18,7 pi/18] 2 + 8 sin 3 theta + 8 (1/2 - 1/2 cos 6 theta) d theta =

I[-pi/18,7 pi/18] 6 + 8 sin 3 theta - 4 cos 6 theta) d theta =

6 theta - 8/3 cos 3 theta - 2/3 sin 6 theta E[-pi/18,7 pi/18] =

6(7 pi/18 - -pi/18) - 8/3 cos 7 pi/6 + 8/3 cos -pi/6 - 2/3 sin 7 pi/3 + 2/3 sin -pi/3 =

8 pi/3 + 2 sqrt(3)

The first small petal is from 7 pi/18 to 11 pi/18

We are integrating the same function, just changing where we evaluate.

6 theta - 8/3 cos 3 theta - 2/3 sin 6 theta E[7pi/18, 11 pi/18] =

6(11 pi/18 - 7 pi/18) - 8/3 cos 11 pi/6 + 8/3 cos 7 pi/6 - 2/3 sin11 pi/3 + 2/3 sin 7pi/3 =

4 pi/3 - 2 sqrt(3)