Finding the Area Inside Two Polar Curves

[aces9113]

Homework Statement

Find the area of the region that lies inside both curves: r^2=2sin2θ, r=1

Homework Equations

A=integral 1/2 (f(θ)^2-g(θ)^2) from a to b

The Attempt at a Solution

I found the points of intersection as pi/12,5pi/12,13pi/12,17pi/12 but the fact that its inside both of them is throwing me off on setting it up. I have also sketched it but that hasn't really helped. Any help would be greatly appreciated

 

[Mark44]

Homework Statement

Find the area of the region that lies inside both curves: r^2=2sin2θ, r=1

Homework Equations

A=integral 1/2 (f(θ)^2-g(θ)^2) from a to b

The Attempt at a Solution

I found the points of intersection as pi/12,5pi/12,13pi/12,17pi/12 but the fact that its inside both of them is throwing me off on setting it up. I have also sketched it but that hasn't really helped. Any help would be greatly appreciated

One of the figures is what is called a rose with two petals. To me it looks like a propeller with two blades. The blades are in the first and third quadrants.

The region whose area you are supposed to find is everything that lies inside both curves. IOW, the parts of the propeller blades that extend beyond the circle aren't included.

 

[aces9113]

I understand that based on the sketch. But would i possibly do the area of one propeller minus (area of the outside curve-the inside curve) and then multiply that by two? I'm lost when it comes to setting this one up

 

[LCKurtz]

Adding to what Mark44 has said, remember that r goes from 0 to the inner curve as θ moves around the circle. And the inner curve changes so you have to break the integral up into appropriate pieces.

 

[aces9113]

could someone please check my answer? i got 2 +pi/3 - sqrt(3)

 

[LCKurtz]

could someone please check my answer? i got 2 +pi/3 - sqrt(3)

Good job. That's what I get too.