1. ## Area of limacon

Find area of limacon.
$
r = 4 + 2 cos\Theta
$

Well I cant seem to get the latex to work for the integral that I set up so here it is rougly.

Upper limit Pi
Lower limit 0

1/2 int (4+2cos theta)^2 dtheta

by symmetry I multiply by 2 which gets rid of the 1/2 outside

Then

foiling

int (16) + int (16 cos theta) + int (4 cos^2theta) dtheta

16 theta - 16 sin theta + 4 int 1 + cos(2theta)/2 dtheta

Am I going the right way with this integral? If not where am I going wrong?
Thanks

2. Notice you complete the outer loop between 0 and 2pi/3 (where r=0)

By integrating from 0 to pi you get the area inside the limacon plus the area of the innerloop

3. Originally Posted by Calculus26
Notice you complete the outer loop between 0 and 2pi/3 (where r=0)

By integrating from 0 to pi you get the area inside the limacon plus the area of the innerloop

Well the question was find the area inside of the limacon. So isn't that what we want to be doing 0 to Pi? When I graph this limacon I do not see any loops since its an oval limacon.

4. I was reading 2+4cos(theta) my apologies

The integral looks good

5. Originally Posted by Calculus26
I was reading 2+4cos(theta) my apologies

The integral looks good
ok for some reason I am not getting the correct answer which is

18 Pi

so again..from above

16 theta - 16 sin theta + 8 int(1+cos(2theta) )

16 theta - 16 sin theta + 8 theta - 4 sin (2 theta)

I have integrated incorrectly somewhere but I am not sure where..

6. 4 int ((1 + cos(2theta))/2 dtheta = 2theta + sin(theta) yields 2pi

you multiplied by 2 instead of dividing by 2

In your original integral you had 4 int (1 + cos(2theta)/2 dtheta

the 1 was divided by 2 also

7. Originally Posted by Calculus26
4 int ((1 + cos(2theta))/2 dtheta = 2theta + sin(theta) yields 2pi

you multiplied by 2 instead of dividing by 2