(1)先对X积分,再对Y积分,(D)双积分:(X^2+Y^2-X)dσ=(D)双积分:(x^2+y^2-x)dxdy=积分:(0,2)dy积分(y/2,y)(x^2+y^2-x)dx=积分:(0,2)dy[x^3/3+xy^2-x^2/2]|(y/2,y)=积分:(0,2)[(y^3/3+y^3-y^2/2)-(y^3/24+y^3/2-y^2/8)]dy=积分:(0,2)[19y^3/24-3y^2/8]dy=[(19/24)(y^4/4)-(3/8)(y^3/3)]|(0,2)=(19/24)(16/4)-(3/8)(8/3)=19/6-1=13/6.(2).∫∫{√(1-x^2-y^2)}dσ ... |