由2x+y-4=0解得x,y的取值范围为0≤x≤2,y=-2x+4∴V=∫[∫zdy]dx=∫[∫y^2dy]dx=∫[y^3/3]dx=∫[(-2x+4)^3/3]dx=-1/2∫[(-2x+4)^3/3]d(-2x+4)=-1/2[(-2x+4)^4/12]=-1/2*[0-4^4/12]=32/3
