第1个回答 2012-04-17
x^2+y^2=9
x=-√(9-y^2)
dx/dy=-y/√(9-y^2)
ds=√(dx)^2+(dy)^2=√[(dx/dy)^2+1] dy
=√[y^2/(9-y^2)+1]dy
=[3/√(9-y^2)] dy
∫y^2ds=∫[-3,3] y^2*3dy/√(9-y^2)
=3∫[-3,3]y^2dy/√(9-y^2)
=3*[(-1/2)y√(9-y^2)+(1/2)arcsin(y/3)]|[-3,3]
=3*[(1/2)*(π/2)-(-1/2)(π/2)]
=3π/2
∫y^2dy/√(9-y^2)=-∫√(9-y^2)dy+∫dy/√(9-y^2)
=-y√(9-y^2)-∫y^2dy/√(9-y^2)+∫dy/√(9-y^2)
2∫y^2dy/√(9-y^2)=-y√(9-y^2)+∫dy/√(9-y^2)
∫y^2dy/√(9-y^2)=(-1/2)y√(9-y^2)+(1/2)∫d(y/3)/√(1-(y^2/9))
=(-1/2)y√(9-y^2)+(1/2)arcsin(y/3)