Reviewing for a test tomorrow, and most of the homework is pretty simple, but I just wanted to ask if you guys would check a few that I'm unsure on.
Turn this equation from Cartesian to polar.
x + y = 9
What I did..
square both sides,
x^2 + 2xy + y^2 = 81
x^2 + y^2 is r^2, so
r^2 + (2(r)cos(theta) times (r)sin(theta)) = 81
r^2 (1+ 2costheta sintheta) = 81
r = 9/sq.root (1+2cos(theta)sin(theta))
find the area inside of the larger loop and outside of the smaller loop of the limacon r = 1/2 + cos(theta)
I solved this one through, and got pi/2 + (3root3)/4
but the answer in the back of the book has pi/4 instead of pi/2.. Could someone just show where I may have missed an extra 1/2 that would change my 1/2 into 1/4?
this was my integral
2 integral from 0 to 4pi/6 of [1/2 (1/2 + cos(theta))^2 dtheta]