The problem statement, all variables and given/known data Find area of region bounded by [tex]x^2 + y^2 = 4[/tex]above y=1 The attempt at a solution OK, so i drew the graph. http://s1.ipicture.ru/uploads/20111226/oR99IdxJ.jpg The red part of the graph is the area that i need to find. Put y = 1 in the equation of circle; x^2 = 3 x = -√3 and +√3 Now, from what i understood from struggling with this problem, i must always describe the region area in the following way: For x fixed, x varies between -√3 to √3 Then, y varies between 1 and √(4 - x^2) Since this problem is found at the end of the section, "Double Integrals - Polar Coordinates", i'm going to transform to polar coordinates, instead of solving it directly: x^2 +y^2 = 4 r^2 = 4 r = -2 or +2 Considering r= +2, since the graph is symmetric about the y-axis, i will evaluate the area of the positive half found in the first quadrant and then later when i get the answer, i will multiply the latter by 2 to get the final answer. θ varies between 0 and ∏/2 So, rewriting the double integrals: [tex]\int\int dydx = \int\int rdrd\theta[/tex] And using the following limits: For r: y = 1 rsinθ=1 So,r = 1/sinθ (1/sinθ)≤r≤2 and 0≤θ≤(∏/2) For the inner integral, i get the answer: 2 - 1/[2sin^2(θ)] which i integrate again w.r.t.θ 2θ + (1/2)cotθ Applying the limits, i get: ∏ which is incorrect. I don't know where i made the mistake/s.