请问这道题怎么做呀
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先计算区域σ的四个顶点
2x=(4-x)^2,2x=16-8x+x^2,x^2-10x+16=0,x=2或8,y=2或-4
2x=(12-x)^2,2x=144-24x+x^2,x^2-26x+144=0,x=8或18,y=4或-6
所以四个顶点为(2,2),(8,-4),(8,4),(18,-6)
原式=∫(2,8)dx∫(4-x,√(2x))(x+y)dy+∫(8,18)dx∫(-√(2x),12-x)(x+y)dy
=∫(2,8)dx*[xy+(1/2)*y^2]|(4-x,√(2x))+∫(8,18)dx*[xy+(1/2)*y^2]|(-√(2x),12-x)
=∫(2,8)[√2*x^(3/2)+x-4x+x^2-8+4x-(1/2)*x^2]dx+∫(8,18)[12x-x^2+72-12x+(1/2)*x^2+√2*x^(3/2)-x]dx
=∫(2,8)[√2*x^(3/2)+(1/2)*x^2+x-8]dx+∫(8,18)[√2*x^(3/2)-(1/2)*x^2-x+72]dx
=[(2√2)/5*x^(5/2)+(1/6)*x^3+(1/2)*x^2-8x]|(2,8)+[(2√2)/5*x^(5/2)-(1/6)*x^3-(1/2)*x^2+72x]|(8,18)
=826/5+5678/15
=8156/15
2x=(4-x)^2,2x=16-8x+x^2,x^2-10x+16=0,x=2或8,y=2或-4
2x=(12-x)^2,2x=144-24x+x^2,x^2-26x+144=0,x=8或18,y=4或-6
所以四个顶点为(2,2),(8,-4),(8,4),(18,-6)
原式=∫(2,8)dx∫(4-x,√(2x))(x+y)dy+∫(8,18)dx∫(-√(2x),12-x)(x+y)dy
=∫(2,8)dx*[xy+(1/2)*y^2]|(4-x,√(2x))+∫(8,18)dx*[xy+(1/2)*y^2]|(-√(2x),12-x)
=∫(2,8)[√2*x^(3/2)+x-4x+x^2-8+4x-(1/2)*x^2]dx+∫(8,18)[12x-x^2+72-12x+(1/2)*x^2+√2*x^(3/2)-x]dx
=∫(2,8)[√2*x^(3/2)+(1/2)*x^2+x-8]dx+∫(8,18)[√2*x^(3/2)-(1/2)*x^2-x+72]dx
=[(2√2)/5*x^(5/2)+(1/6)*x^3+(1/2)*x^2-8x]|(2,8)+[(2√2)/5*x^(5/2)-(1/6)*x^3-(1/2)*x^2+72x]|(8,18)
=826/5+5678/15
=8156/15
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