求曲面z=1-4x2-y2和z=0所围立体的体积．
【正确答案】：V=4∫∫D(1-4x2-y2)dxdy， 其中D={(x,y)∣0≤x≤1/2,0≤y≤√(1-4x2)} 所以V=4∫01/2dx∫0√1-4x2 (1-4x2-y2)dy =4∫01/2[(1-4x2)y-(1/3)y3]∣ 0√1-4x2dx =8/3∫01/2[√(1-4x2)3]dx =8/3∫0π/2cos3td[(1/2)sint]=4/3∫0π/2cos4tdt =1/3∫0π/2(1+cos2t)2dt=1/3∫0π/2(1+2cos2t+cos22t)dt =π/6+1/3∫0π/2cos2td2t+1/6∫0π/2(1+cos4t)dt =π/6+(1/3)sin2t∣0π/2+π/12+1/24∫0π/2cos4td4t =π/4+(1/24)sin4t∣0π/2=π/4.
