Question

Find the area lying outside r=6cos(theta) and inside r=3+3cos(theta)

Answer 1

So in polar we'll use:\[\Large A \quad=\quad \frac{1}{2} \int\limits r^2\;d \theta\]

So ummm, we need an r.
Have you tried graphing these functions? :)
https://www.desmos.com/calculator/jgur6caz6i

So we're interesting in the area outside of that red circle, but inside of the blue, yes?

Answer 2

So that radial length is given by the blue function minus the red.
\[\Large r=3+3\cos \theta-6\cos \theta\]

Answer 3

@zepdrix, for the area between two curves we need an outer \(r\) and an inner \(r\). In this case, \(r=6\cos\theta\) is the inner and \(r=3+3\cos\theta\) is the outer.

Then, the area would be
\[\frac{1}{2}\int_\alpha^\beta(r_\text{outer}^2-r_\text{inner}^2)~d\theta\]

Answer 4

i like your demo @zepdrix thats super cool

Answer 5

r=3-cos(theta)?

Answer 6

Oh did I make a boo boo there? :(
lol sorry, just woke up.

the graph megan? :O ikr? cool little tool

Answer 7

just a slight one : )

Answer 8

(1/2) integral of (6cos(theta))^2 - (3+3cos(theta))^2 ????

Answer 9

Hmm I think you might have those backwards.
The outer function is r=3+3cos theta.
We would want to subtract the smaller from that.

Answer 10

the outer function is 6cos(theta)

Answer 11

It is? D:
See how the blue is OUTSIDE of the red?