Which of the following represents the area inside r 12 and

Which of the following represents the area inside r = 12, and between the lines theta = pi/2 and r = 6 sec(theta)? 2(integral_0^pi/2 1/2 (6 sec(theta))^2 d theta + integral_0^pi/3 1/2 (12)^2 d theta) integral_0^pi/2 1/2 ((6 sec(theta))^2 - (12)^2) d theta integral_0^pi/2 1/2 (6 sec (theta))^2 d theta + integral_0^pi/3 1/2 (12)^2 d theta integral_0^pi/3 1/2 (6 sec(theta))^2 d theta + integral_pi/3^pi/2 1/2 (12)^2 d theta 2 (integral_0^pi/3 1/2 (6 sec(theta))^2 d theta + integral_pi/3^pi/2 1/2 (12)^2 d theta)

Solution

Answer is d)

Area = integration of (1/2)r^2 dtheta (from a to b)

12 = 6 sec(theta)

Sec (theta) = 2

Theta = pi/3

Hence region of integration is from 0 to pi/3 and pi/3 to pi/2

Hence d)