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

1个回答

  • ∵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.