Find the area inside one curve & outside another

[MarkSA]

Hello,

Find the area that lies inside curve r^2 = 8cos(2theta) but outside curve r = 2.

From my calculator this seems to be a circle and a horizontal figure eight.

I'm really unsure if i'm doing these multiple curve problems correctly. What i've tried to do on this one is to say that there are two lines of symmetry at y =0 and x = 0 that break the graph up into 4 pieces. The intersection point in Quadrant 1 seems to be pi/6, so for limits i'm using 0 to pi/6. I'm thinking the area for this is then:
A = Integral from 0 to pi/6 of: 1/2 * [8cos(2theta) - 4] dtheta
And then multiplying the integral by 4 to account for the other pieces due to symmetry...

Does this look like a legitimate way to set this problem up, or am I missing some detail? (BTW: does anyone know if a TI-89 can integrate polar curves? I would love to be able to check my answers on these)

 

[tkhunny]

The intersection point in Quadrant 1 seems to be pi/6

This is the only part that bothered me. What does "seems to be" mean? Prove it!

 

[MarkSA]

I wish I could..

Finding which exact limits to use for integrating is what i'm having trouble with (see my other posts)

Right now i'm just eyeballing it from the graphing calculator...

 

[galactus]

Then solve \(\displaystyle 2=\sqrt{8cos(2{\theta})}\)

\(\displaystyle 4=8cos(2{\theta})\)

\(\displaystyle \frac{1}{2}=cos(2{\theta})\)

\(\displaystyle \frac{cos^{-1}(\frac{1}{2})}{2}={\theta}=\frac{\pi}{6}\)

 

[MarkSA]

Ok thanks, I see what you did.