(完整版)第一类曲面积分习题.doc

1dSz,是球面x2y2z22R被平面z h(0 h R)截出的顶部。òxydS,是圆柱面x2y21与平面z0,xz2围成的立体的全表面。F(t)f(x,y,z)dS,其中为x2y2z22t(t0),被积函数f(x,y,z)x20yzzxx22y2y2。2x12y2zdS,⑴是球面x2y2z22R;⑵是介于平面z0,z1之间的圆柱面x2y22R。z2dS,其中:x2y22z2R。 (x y z)dS, 是上半球面x2 y2 z2 2被旋转抛物面z x2 y2截出的顶部。(xy yz zx)dS, 为锥面zx22y被圆柱面x2y22ay(a 0)所截下的部分。。1dSz,是球面x2y2z22R被平面z h(0 h R)截出的顶部。解::zR2x22y,在xoy面上的投影区域 D:x2y2R22h,12zxz2y1R22x2xy2R2yx22y2R2Rx2y2R1dSzDR21x2y2R2Rx2y2dx2y2R22hDR2R2xy2dD2RRr2rdrdR20d02 2R hR2rr2dr2R(12ln(R2r2)02 2R hR(2lnR2lnh)2RòxydS, 是圆柱面x2 y2 1与平面z 0,xz 2围成的立体的全表面。解:1:1z02 3,D1:2xy21;zxzD32xydS12:z2Dxy100d1x,D2:x2y21D1xyd0x2xydSD2xy21(1)0d2D2xyd03:x2y21,33132, 31:y12x, 32:y12x,其中31、 32在xoz面上的投影区域均为 D3,且D3由xz 2,x1,x 1,z0围成。又1y2xy2z11x2x2011x2òxydS 1 2 3xydS 3xydS 31xydS 32xydSD3x1x211x2dD3x(1x2)112xd2D3xd211dx10xxdz211x(1x)dx2(23)F(t)f(x,y,z)dS,其中为x2y2z22t(t0),被积函数解:x2f(x,y,z)1 2,其中0yzzxx22y2y2。121:x2y2z22t(zx22y);2:x2y22z2t(zx22y);故f(x,y,z)dS 0dS 0;2 21:zt2x22y在xoy面的投影区域为 D:x2y22t2,则1zx22zy1t2x2x2y2y2t2tx2y21f(x,y,z)dS1(