求曲面z=√(x^2+y^2)与z=8-(x^2+y^2)/4所围成的立体的的表面积和体积

Question

求曲面z=√(x^2+y^2)与z=8-(x^2+y^2)/4所围成的立体的的表面积和体积

Answer 1

∵所围成的立体在xy平面上的投影是圆S：x²+y²=4
dS1=√[1+(x/√(x²+y²))²+(y/√(x²+y²))²]dxdy=√2dxdy
dS2=√[1+(-x/2)²+(-y/2)²]dxdy=√(1+x²/4+y²/4)dxdy
∴所求表面积=∫∫dS1+∫∫dS2
=∫∫√2dxdy+∫∫√(1+x²/4+y²/4)dxdy
=√2∫dθ∫rdr+∫dθ∫r√(1+r²/4)dr (作极坐标变换)
=2√2π(2²/2-0)+4π∫√(1+r²/4)d(r²/4)
=4√2π+4π(2/3)(1+r²/4)^(3/2)│
=4√2π+(8π/3)(2√2-1)
=(28√2-8)π/3
所求体积=∫∫{[8-(x²+y²)/4]-√(x²+y²)}dxdy
=∫dθ∫(8-r²/4-r)rdr (作极坐标变换)
=2π∫(8r-r³/4-r²)dr
=2π(4r-r^4/16-r³/3)│
=2π(4*2-2^4/16-2³/3)
=2π(8-1-8/3)
=26π/3.