How do you find the area of the region shared by the circles r=2cos(θ) and r=2sin(θ)? Calculus Introduction to Integration Integration: the Area Problem 1 Answer Sasha P. Oct 24, 2015 π2−1 Explanation: Intersection: 2sinθ=2cosθ⇒sinθcosθ=1⇒tanθ=1⇒θ=π4 A=∫∫Ardrdθ A=∫π40dθ∫2sinθ0rdr+∫π2π4dθ∫2cosθ0rdr A1=12∫π40dθ⋅4sin2θ=∫π40(1−cos2θ)dθ A1=(θ−12sin2θ)∣π40=π4−12 A2=12∫π2π4dθ⋅4cos2θ=∫π2π4(1+cos2θ)dθ A2=(θ+12sin2θ)∣π2π4=π4−12 A=π4−12+π4−12=π2−1 Answer link Related questions How do you find the area of a region using integration? How do you use integration to find area under curve? Why does integration find the area under a curve? How do I evaluate ∫50|x−5|dx by interpreting it in terms of areas? How do you find the area of the parallelogram with vertices (4,5), (9, 9), (13, 10), and (18, 14)? How do you evaluate the integral of absolute value of (x - 5) from 0 to 10 by finding area? How do you find the area of the parallelogram with vertices k(1,2,3), l(1,3,6), m(3,8,6), and n(3,7,3)? How do you find the area of the parallelogram with vertices: p(0,0,0), q(-5,0,4), r(-5,1,2), s(-10,1,6)? How do you evaluate ∫5 between the interval [0,4]? What is a surface integral? See all questions in Integration: the Area Problem Impact of this question 43029 views around the world You can reuse this answer Creative Commons License