What is the area enclosed by r=2sin(theta+(7pi)/4) between theta in [pi/8,(pi)/4]?

1 Answer
Jan 4, 2018

= pi/8 -1/(2sqrt2)

Explanation:

Area of a polar curve is int_a^b 1/2 r^2 d theta

Accordingly, in the present case it would be int_(pi/8) ^ (pi/4) 1/2 *4sin^2 (theta + (7pi)/4) d theta

=int_(pi/8)^(pi/4) (1- cos(2 theta + (7pi)/2) d theta

[theta -1/2 sin(2theta +(7pi)/2)]_(pi/8)^(pi/4)

= (pi/4 -1/2 sin(pi/2 +(7pi)/2)-(pi/8 -1/2sin(pi/4+(7pi)/2)

(pi/4 -pi/8 +1/2 sin(4pi -pi/4))

=pi/8 +1/2 sin (- pi/4)

= pi/8 -1/(2sqrt2)