inside r2 but outside r4costhetaplease explain how and why r

inside r=2 but outside r=4costheta....please explain how and why r1 and r2 and chosen

Solution

Note: I\'ll be using x instead of theta.

To determine the area we\'ll need to find the angle for which the two curves intersect.

r = 4cosx
r = 2.

Let\'s set these equal to each other.

4cosx = 2
cosx = 1/2
x = pi/3 & 5pi/3.

In order to enclose the area that we want, we must enclose the area by moving from the smallest angle to the largest. Notice that 5pi/3 is the same as -pi/3. Therefore we\'ll be moving from -pi/3 to pi/3.

A = ? [b,a] (1/2)((r_1)^2 - (r_2)^2) dx

pi/3
..? (1/2)((4cosx)^2 - 2^2) dx
-pi/3

Let\'s neglect the bounds for now.

? (1/2)(16cos^2x - 4) dx
? 8cos^2x - 2 dx
2 ? 4cos^2x - 1 dx
2 ? 2cos(2x) + 2 - 1 dx
2 ? 2cos(2x) + 1 dx = 2sin(2x) + 2x [-pi/3 , pi/3]

2sin(2pi/3) + 2(2pi/3) - ( 2sin(-2pi/3) + 2(-2pi/3) ) =
sqrt(3) + 4pi/3 - ( -sqrt(3) - 4pi/3 ) =
2sqrt(3) + 8pi/3 = 11.84 units.