Homework Sourseuse polar coordinates to evaluate the double integral R1x2y2
use polar coordinates to evaluate the double integral
??R[(1)/(x^2+y^2)^(3/2)]dA where the region R is the triangle with the vertices A (0,0), B(3,0), and C(3,3)
??R[(1)/(x^2+y^2)^(3/2)]dA where the region R is the triangle with the vertices A (0,0), B(3,0), and C(3,3)
Solution
sol: double integral over D of (xy^2)dA = integral -1 to 1 integral 0 to sqrt(1 - y^2) of xy^2 dxdy = integral -1 to 1 of [(sqrt(1 - y^2))^2 / 2] * y^2 dy = integral -1 to 1 of [(1 - y^2)/2] * y^2 dy = integral -1 to 1 of (y^2 - y^4)/2] dy = 2 * (1/3 - 1/5)/2 = 2/15. An effective way to check an answer is to use a different method and verify that the answer is the same. In polar coordinates, this region D is represented by 0 <= r <= 1, -pi/2 <= theta <= pi/2. Also, x = rcostheta, y = rsintheta, and dA = r drdtheta. double integral over D of (xy^2)dA = integral -pi/2 to pi/2 integral 0 to 1 of r^3 costheta sin^2 theta r drdtheta = integral -pi/2 to pi/2 integral 0 to 1 of r^4 costheta sin^2 theta drdtheta = integral -pi/2 to pi/2 of (1/5) costheta sin^2 theta dtheta = integral -1 to 1 of (1/5)u^2 du, using u = sintheta and du = costheta dtheta = 2 * (1/5)(1/3) = 2/15![use polar coordinates to evaluate the double integral ??R[(1)/(x^2+y^2)^(3/2)]dA where the region R is the triangle with the vertices A (0,0), B(3,0), and C(3,3](/WebImages/37/use-polar-coordinates-to-evaluate-the-double-integral-r1x2y2-1112113-1761589925-0.webp "use polar coordinates to evaluate the double integral ??R[(1)/(x^2+y^2)^(3/2)]dA where the region R is the triangle with the vertices A (0,0), B(3,0), and C(3,3")
