Question

高等数学重积分的应用 求由曲面z=x²+y²,z=根号下（2-x²-y²）所围成

Answer 1

消去z,(x^2+y^2)^2=2-(x^2+y^2), (x^2+y^2)^2+(x^2+y^2)-2=0,

{(x^2+y^2)-1][(x^2+y^2)+2]=0, 后者大于零,则

x^2+y^2=1, 即为积分区域D.

S1=∫∫√[1+(z')^2+(z')^2]dxdy =∫∫√(1+4x^2+4y^2)dxdy

=∫dt∫√(1+4r^2)rdr = (π/4)∫√(1+4r^2)d(1+4r^2)

=(π/6)(5√5-1);

S2=∫∫√[1+(z')^2+(z')^2]dxdy

=∫∫√[1+x^2/(2-x^2-y^2)+y^2/(2-x^2-y^2)]dxdy

=√2∫∫[1/√(2-x^2-y^2)]dxdy

=√2∫dt∫1/√(2-r^2)rdr

= -π√2∫1/√(2-r^2)d(2-r^2)=(4-2√2)π.

所求 S = S1+S2 = (π/6)(5√5-1)+(4-2√2)π.