Question

Polar coordinates.

Find the area of the region that lies inside the first curve and outside the second curve.

first curve: r^2=8cos(2x)
second surve: r=2

i mainly need help finding the limits of integration to this problem.
i set 4=8cos(2x)
cos2x=1/2
x = pi/3 and -pi/3(first and fourth quadrant is where cos is positive.)
but since there is a 2 in the cos, i divide both by 2 and get pi/6 as upper limit and -pi/6 as bottom limit of the integral.

but when i integrate, i still get the wrong answer. I have even used an integral calculator to see if i was integrating properly. (i was)

Answer 1

\[.5\int\limits_{-\pi/6}^{\pi/6}8\cos(2x)-(2)^2\]

Answer 2

correct answer is 2.74

Answer 3

Thanks!!