Question

二重积分设D={(xy)x2+y2≤x+y},则(x+y)dxdy=()？

如图

Answer 1

化极坐标。
D： x^2+y^2 = x+y, 即 r = cost+sint， -π/4 ≤ t ≤ 3π/4
I = ∫<-π/4, 3π/4>dt∫<0, cost+sint> r(cost+sint) rdr
= ∫<-π/4, 3π/4>(cost+sint)dt [r^3/3]<0, cost+sint>
= (1/3)∫<-π/4, 3π/4>(cost+sint)^4dt
= (1/3)∫<π/4, 3π/4>4[sin(t+π/4)]^4dt
= (1/3)∫<π/4, 3π/4>[1-cos(2t+π/2)]^2dt
= (1/3)∫<π/4, 3π/4>[1-sin2t]^2dt
= (1/3)∫<π/4, 3π/4>[1-2sin2t+(sin2t)^2]dt
= (1/3)∫<π/4, 3π/4>[3/2-2sin2t+(1/2)cos4t]dt
= (1/3)[3t/2+cos2t+(1/8)sin4t]<π/4, 3π/4> = π

追问

谢谢点拨不过按照你的思路
得出(1/3)∫[(sint+cost)^4]dt
最后求出的结果是π/2