Question

高数曲面积分？

Answer 1

因为∑关于xoy平面对称，所以
原式=2∫∫(∑') (x+y)dS，其中∑'是∑在xoy平面上方的部分
=2∫∫(D) (x+y)√[1+x^2/(a^2-x^2-y^2)+y^2/(a^2-x^2-y^2)]dxdy，其中D是∑'在xoy平面上的投影，即x^2+(y-a/2)^2=(a/2)^2
=2a∫∫(D) (x+y)/√(a^2-x^2-y^2)dxdy
因为D关于y轴对称，所以
原式=4a∫∫(D') y/√(a^2-x^2-y^2)dxdy，其中D'是D在y轴右侧的部分
=4a∫(0,a)dy∫(0,√(ay-y^2)) y/√(a^2-x^2-y^2)dx
=4a∫(0,a)ydy*arcsin[x/√(a^2-y^2)]|(0,√(ay-y^2))
=4a∫(0,a) yarcsin{√[y/(a+y)]}dy
=2a∫(0,a) arcsin{√[y/(a+y)]}d(y^2)
=2a*arcsin{√[y/(a+y)]}*y^2|(0,a)-2a*∫(0,a) {y^2/√[1-y/(a+y)]}*(1/2)*√[(a+y)/y]*a/(a+y)^2dy
=(π/2)*a^3-a√a*∫(0,a) y^(3/2)/(a+y)dy
令y=t^2，则dy=2tdt
原式=(π/2)*a^3-2a^(3/2)*∫(0,√a) t^4/(a+t^2)dt
=(π/2)*a^3-2a^(3/2)*∫(0,√a) [t^2-a+a^2/(a+t^2)]dt
=(π/2)*a^3-2a^(3/2)*[(1/3)*t^3-at+a^(3/2)*arctan(t/√a)]|(0,√a)
=(π/2)*a^3-2a^(3/2)*[(1/3)*a^(3/2)-a^(3/2)+a^(3/2)*(π/4)]
=(π/2)*a^3-(2/3)*a^3+2a^3-(π/2)*a^3
=(4/3)*a^3