Question

高数 二重积分 求∫∫(D)|x²+y²-1|dσ,D=(0<X<1,0<Y<1)

Answer 1

简单计算一下即可，答案如图所示

Answer 2

原式=∫∫(D1) (1-x^2-y^2)dxdy+∫∫(D2) (x^2+y^2-1)dxdy
其中D1：x^2+y^2<=1且x>0，y>0；D2=D-D1
原式=∫(0,π/2)dθ∫(0,1)(1-r^2)rdr+∫(0,1)dx∫(√(1-x^2),1)(x^2+y^2-1)dy
=(π/2)*(r^2/2-r^4/4)|(0,1)+∫(0,1)dx*[(x^2-1)y+y^3/3]|(√(1-x^2),1)
=π/8+∫(0,1) [x^2-2/3+(2/3)*(1-x^2)^(3/2)]dx
=π/8+∫(0,1) (x^2-2/3)dx+(2/3)*∫(0,1) (1-x^2)^(3/2)dx
令x=sint，则dx=costdt
=π/8+(x^3/3-2x/3)|(0,1)+(2/3)*∫(0,π/2) cos^4tdt
=π/8-1/3+(1/6)*∫(0,π/2) (2cos^2t)^2dt
=π/8-1/3+(1/6)*∫(0,π/2) (1+cos2t)^2dt
=π/8-1/3+(1/6)*∫(0,π/2) (1+2cos2t+cos^2(2t))dt
=π/8-1/3+(1/12)*∫(0,π/2) (3+4cos2t+cos4t)dt
=π/8-1/3+(1/12)*[3t+2sin2t+(1/4)*sin4t]|(0,π/2)
=π/8-1/3+(1/12)*(3π/2)
=π/4-1/3

Answer 3