Convert the integral I = integration from 0 to (3/(sqrt2)) integration from y to
ID: 3080681 • Letter: C
Question
Convert the integral I = integration from 0 to (3/(sqrt2)) integration from y to sqrt(9-y^2) e^(2x^2) + 2y^2 to polar coordinates, getting integration from C to D integration from A to B h(r,(theta)) drd(theta) where h(r,(theta)) A = B = C = D = and then evaluate the resulting integral to get I =.Explanation / Answer
Assuming that the ellipse has equation x^2 - xy + y^2 = 1... (The techniques are similar otherwise): -------------------------------- Letting x = uv2 - v v(2/3), and y = uv2 + v v(2/3), the ellipse transforms to (uv2 - v v(2/3))^2 - (uv2 - v v(2/3))(uv2 + v v(2/3)) + (uv2 + v v(2/3))^2 = 1 ==> (2u^2 - 4uv/v3 + (2/3) v^2) - (2u^2 - (2/3) v^2) + (2u^2 + 4uv/v3 + (2/3) v^2) = 1 ==> 2u^2 + 2v^2 = 1 ==> u^2 + v^2 = 1/2, a circle with radius 1/v2. Call the region bounded by this R. (I'll convert to polar coordinates below.) Since ?(x, y)/?(u, v) = |v2 -v(2/3)| |v2 v(2/3)| = 4/v3, we have ?? (x^2 - xy + y^2) dA = ??R (2u^2 + 2v^2) (4/v3) dv du = (8/v3) ??R (u^2 + v^2) dv du = (8/v3) ?(? = 0 to 2p) ?(r = 0 to 1/v2) r^2 (r dr d?), via polar coordinates = (8/v3) * (2p) * (r^4/4 {for r = 0 to 1/v2})