9-4高等数学同济大学第六版本

习题习题941x2y2z2a2x2y2ax内部的那部分面积其面积是相同的za2x2y2得zxxa2x2y2yzya2x2y于是于是A2 x2y2ax1(z)2(z)2dxdyxy2x2y2a dxdya2x2y24a2d0aco01a22 d4a 2(aasin)d2a2(02)22求锥面z x2y2z22x所割下的部分的曲面的面积zzx2y2z22xzx2y22xxOy面上Dx2y22x由曲面方程x2y2得zxxx2y2yzyx2y2于是A(x1)2y21(z)2(z)2dxdyxy 2dxdy (x1)2y2133x2y2R2x2z2R2所围立体的表面积A1zR2x2相应于区域Dx2y2R2上的面积则所求表面A4A1A4D1(z)2(z)2dxd4xy1(xR2x2)202dxdyD4DRR2x2dxd4RRdxRx2R Rx21R2x2dy8RRdx16R2R44D如下求均匀薄片的质心(1)D由y 2pxxx0y0所围成解令密度为1因为区域D可表示为0xx

0,0y 2px所以0AAdxdyx0dx2pxdyx00 0 02pxdx232px30Dxx1xdxdy1x0dx2pxxdy1x0xA00A02pxdx3x5A0Dyy1ydxdy1x0dx2pxyd1x0pxd3yAA00A080D所求质心为所求质心为(3x,3y)5080(2)D是半椭圆形闭区域x,y|x2y21,ya2 b2Dyx0AD1ab(椭圆的面积)2yy1ydxdy 1 dxabaa2x2ydy1b (a2a2 2AAx)dx4ba0A2a2 a3D所求质心为所求质心为(0,4b)3(3)Drbcos(0ab)之间的闭区域解令密度为1由对称性可知y0AAdxdyb2a2(b2a2)(两圆面积的差)224Dxx1AxdxdyD 2d2Abcosrcos rdr0acosa2b2ab2(ab)所求质心是所求质心是(a2b2ab,0)2(ab)55Dyx2yx所围成它在点y)处的面密度(xy)x2y求该薄片的质心解M(x,y)dxdy1dxxx2ydy11(x4x6)dx0x202135Dxx1Mx(x,y)dxdyD1M1dxxx3ydy111(x5x7)dx350x2M 0248yy1My(x,y)dxdyD1M1dxxx2y2dy10 x2M 0311(x5x8)dx3554质心坐标为质心坐标为(35,35)48 546设有一等腰直角三角形薄片腰长为a各点处的面密度等于该点到直角顶点的距离的平方求这薄片的质心使薄片在第一象限且直角边在坐标轴上薄片上点(xy)处的函数为x2y2由对称性可知xyMM(x,y)dxdyadxax(x2y2)dy1a40 06Dxxy1Mx(x,y)dxdyD1Maxdxax(x2y2)dy2a005薄片的质心坐标为薄片的质心坐标为(2a,2a)5 57利用三重积分计算下列由曲面所围成立体的质心(设密度1)(1)z2x2y2z1z轴上xy0VVdv1(圆锥的体积)3zz1zd1d1rd1zd3VV00r4所求立体的质心为所求立体的质心为(0,0,3)4(2)(2)zA2x2y2za2x2y2(Aa0)z0zxy0Vdv2A32a32A3a3)(两个半球体体积的差333zz rsin cos drd d d sin cos d rdr13 1220 A33(A a)44VV008(A3a3)所求立体的质心为所求立体的质心为(0,0,3(A4a4))8(A3a3)(3)(3)zx2y2xyax0y0z0解Vadxaxdyx2y2dzadxax(x2y2)dy0 0 0 0 0a[x2(ax)1(ax)3]dx1a4036x11a5Vxdv 1axdxaxdyx2y2dz15V0001a642a5yyx2a5z1zdvV1adxaxdyx2y2zdzV730a2000所以立体的重心为所以立体的重心为(2a,2a,75 5 30a8设球体占有闭区域{(xz)|x2y2z22Rz}它在内部各点的密度的大小等于该点到坐标原点的距离的平方试求这球体的质心x2y2z2z轴上xy0在球面坐标下在球面坐标下可表示为02,0,0r2Rcos于是2MMdv2d2sind2Rcosr2r2dr00022325R5sincos 325dR5015zz1Mzdv1M2d2sincosd2Rcosr5dr0 0 02M20646R6sincos7d8R6332r515 45故球体的质心为故球体的质心为(0,0,5R).499设均匀薄片(面密度为常数1)所占闭区域D如下求指定的转动惯量(1)D{(x,y)|x2y2Ia2 b2D可表示为yaxa,b ax2ybaaax2于是I x2dxdyax2dxaba2x2ydy2bax2abaa2x2aaa2x2dx1a3b4D提提 aax2a2x2dxxasinta4sin22202tdt8a4(2)D(2)Dy29xx2II2解积分区域可表示为xy0x2,3 x/2y3 x/2于是I y2dxdydx23x/2xy2dy 2 2273x2dx03x/23022725DI I x2dxdyx2dx23x/26 2ydy5x2dx03x/22 0967D(3)(3)D为矩形闭区域{(xy)|0xa0yb}求I和Ixy解I y2dxdyadxby2dya1b3ab3x0033DII x2dxdyax2dxbdy1a3ba3by0033D1010已知均匀矩形板(bh板对于通过其形心且分别与一边平行的两轴的转动惯量解取形心为原点取两旋转轴为坐标轴建立坐标系I y dxdy dx y2dy1bh3b2h22xbh12D22I I x dxdy xb2 22dx dy1hb3h2y12Db2h2一均匀物体(密度为常量)占有的闭区域zx2y2z0|x|a|y|a所围成(1)求物体的体积解由对称可知V4adxadyx2y2dz0 0 04adxa(x2y2)dy4a(ax2a3)dx8a40 0 033(2)求物体的质心xy0z1Mzdv4adxadyx2y2zdzV0002adxa(x42x2y2y4)dyV002a(ax42a3x2a5)dx7a2V03515(3)z轴的转动惯量解I (x2y2)dv4adxadyx2y2(x2y2)dzz0004adxa(x42x2y2y4)dy428a6112a60 045451212ah惯量(解建立坐标系使圆柱体的底面在xOy面上z轴通过圆柱体的轴心用柱面坐标计算I (x2y2)dvr3drddzdar3drhdz1ha4z0002 1313 设面密度为常量 的匀质半圆环形薄片占有闭区域D{(x,y,0)|R1x2y2R,x0}zM0(00a)(a0)处单2FF(FFF由对称性F0而x y zyF GxDx(x2y2a2)3/dG cosdR 222(2a2)3/ d2R12G[lnR2a2R22R2R1]R2a2R11R2a22R2a21FF