求二重积分 ∫∫√4-x-y dxdy

求二重积分 ∫∫√4-x-y dxdy

嗯，幅角看错了一点，改了。 圆x+y=1 与圆x+y=2y （或x+(y-1)=1 ）的交点为 (√3/2 , 1/2) 和 (-√3/2 , 1/2) 。 这两点的极坐标分别为 (r=1, a=π/6) (r=1, a=5π/6) 而x+y=2y 则化为 r=2r sin a，即 r=2sin a。 两个交点把D分为3部分： 1. 幅角a范围为 0 到π/6， 极半径 r介于0到2sin a之间 2. 幅角a范围为 π/6 到 5π/6， 极半径 r介于0到1之间 3. 幅角a范围为 5π/6 到π， 极半径 r介于0到2sin a之间 所以用极坐标，积分化为三个积分相加 ∫∫ D √4-x-y dxdy =∫ _(0<=a<=π/6) da ∫_(0<=r<=2sina) (√4-r) rdr + ∫ _(π/6<=a<=5π/6) da ∫_(0<=r<=1) (√4-r) rdr + ∫ _(5π/6<=a<=π) da ∫_(0<=r<=2sina) (√4-r) rdr =∫ _(0<=a<=π/6) da ∫_(0<=r<=2sina) (√4-r)/2 dr +∫ _(π/6<=a<=5π/6) da ∫_(0<=r<=1) (√4-r)/2 dr +∫ _(5π/6<=a<=π) da ∫_(0<=r<=2sina) (√4-r)/2 dr =∫ _(0<=a<=π/6) da { (4-0)^(3/2) / 3 - [4-(2sina)]^(3/2) / 3 } +∫ _(π/6<=a<=5π/6) da { (4-0)^(3/2) / 3 - [4-1]^(3/2) / 3 } +∫ _(5π/6<=a<=π) da { (4-0)^(3/2) / 3 - [4-(2sina)]^(3/2) / 3 } =∫ _(0<=a<=π/6) (8/3)[1-(cosa)^3]da + (2π/3) (8/3 - √3) + ∫ _(5π/6<=a<=π) (8/3)[1+(cosa)^3]da (5π/6<=a<=π, cos a <0, 开根号取 - cos a) =(8/3)(π/6) - (8/3)∫ _(0<=a<=π/6) (cosa)^3 da + (2π/3) (8/3 - √3) + (8/3)(π/6) + (8/3)∫ _(5π/6<=a<=π) (cosa)^3 da =(8-2√3)π/3 - (8/3) { ∫ _(0<=a<=π/6) [1-(sina)^2] d(sina) - ∫ _(5π/6<=a<=π) [1-(sina)^2] d(sina) } =(8-2√3)π/3 - (8/3){sin(π/6) - [sin(π/6) ]^3 / 3 + sin(5π/6) - [sin(5π/6) ]^3 / 3 } = (8-2√3)π/3 - (8/3)*2*[1/2 - (1/2)^3/3] = (8-2√3)π/3 - 22/9

麻烦采纳，谢谢!