计算由曲面x^2+y^2=az,z=2a-√x^2+y^2(a>0)所围成的立体的全表面积

Question

计算由曲面x^2+y^2=az,z=2a-√x^2+y^2 (a>0) 所围成的立体的全表面积

Answer 1

∵解方程组x²+y²=az与z=2a-√(x²+y²),得x²+y²=a²
∴所围成的立体在xy平面上的投影是D：x²+y²=a²
令z1=(x²+y²)/a,z2=2a-√(x²+y²)
∵αz1/αx=2x/a,αz1/αy=2y/a
αz2/αx=-x/√(x²+y²),αz2/αy=-y/√(x²+y²)
∴dS1=√[1+(αz1/αx)²+(αz1/αy)²]dxdy=[√(a²+4x²+4y²)/a]dxdy
dS2=√[1+(αz2/αx)²+(αz2/αy)²]dxdy=√2dxdy
故 所求全表面积=∫∫dS1+∫∫dS2
=∫∫[√(a²+4x²+4y²)/a]dxdy+∫∫√2dxdy
=∫dθ∫[√(a²+4r²)/a]rdr+∫dθ∫√2rdr
(应用极坐标变换)
=π[(5a²)^(3/2)-(a²)^(3/2)]/(6a)+√2πa²
=(5√5+6√2-1)πa²/6.