Question

find the area inside the oval limacon r=6+2cos(theta)

Answer 1

polar integral eh

Answer 2

yes

Answer 3

trying to recall:\[\int\int r(t)dtdr\]or is that going to far?

Answer 4

the formula is \[\int\limits_{a}^{b}1/2 (r(\theta))^2 d \theta\]

Answer 5

\[\int\limits_{a}^{b}\frac12 (r(\theta))^2 d \theta\] by chance?

Answer 6

yes

Answer 7

and t is from 0 to 2pi; and they give you r(t)

\[2*\int\limits_{0}^{pi}\frac12 (6+2cos(t)))^2 dt\]
\[2*\int\limits_{0}^{pi}\frac12 4cos^2(t)+24cos(t)+36 dt\]
\[2*\int\limits_{0}^{pi} 2cos^2(t)+12cos(t)+18 dt\]

might have to reduce power on the cos^2

Answer 8

\[cos(2t)=cos^2(t)-sin^2(t)\]
\[cos(2t)=cos^2(t)-(1-cos^2(t))\]
\[cos(2t)=cos^2(t)-1+cos^2(t))\]
\[cos(2t)=2cos^2(t)-1\]
\[\frac{1+cos(2t)}{2}=cos^2(t)\]

Answer 9

im curious what the 1/2 signifies in there

Answer 10

i think my doubling it was premature

Answer 11

would you agree?

Answer 12

but then, im only integrating the top half, so i would need to double it ...

Answer 13

\[2*\int\limits_{0}^{pi} 2\frac{1+cos(2t)}{2}+12cos(t)+18 dt\]
\[2*\int\limits_{0}^{pi} cos(2t)+12cos(t)+19 dt\]
\[2*(\frac{sin(2t)}{2}+12sin(t)+19t)\]
\[sin(2t)+24sin(t)+38t;\ [0,pi]\]

Answer 14

at 0 it all equals 0; so at pi it equals 38pi

Answer 15

if i did it right; its been awhile :)