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二重积分问题 (1)计算∫∫根号下(y^2-xy) dxdy,区域D={y=x,x=0,y=1} (2)区域D={(X,Y)| X^2+Y^2
人气:462 ℃ 时间:2019-08-21 20:47:10
解答
∫∫根号下(y^2-xy) dxdy=∫(0,1)[∫(0,y)根号下(y^2-xy) dx]dy
=∫(0,1)[∫(0,y)(-y)*y根号下(1-x/y) d(1-x/y]dy
=∫(0,1)[∫(0,y)(-y)*y根号下(1-x/y) d(1-x/y]dy
=∫(0,1)[(-y^2*2(1-x/y)^1.5/3|(0,y)dy
==∫(0,1)[-2y^2/3]dy=-2y^3/9|(0,1)=2/9
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