9-4高等數(shù)學(xué)同濟(jì)大學(xué)第六版本

習(xí)題習(xí)題941x2y2z2a2x2y2ax內(nèi)部的那部分面積其面積是相同的za2x2y2得zxxa2x2y2yzya2x2y于是于是A2 x2y2ax1(z)2(z)2dxdyxy2x2y2a dxdya2x2y24a2d0aco01a22 d4a 2(aasin)d2a2(02)22求錐面z x2y2z22x所割下的部分的曲面的面積zzx2y2z22xzx2y22xxOy面上Dx2y22x由曲面方程x2y2得zxxx2y2yzyx2y2于是A(x1)2y21(z)2(z)2dxdyxy 2dxdy (x1)2y2133x2y2R2x2z2R2所圍立體的表面積A1zR2x2相應(yīng)于區(qū)域Dx2y2R2上的面積則所求表面A4A1A4D1(z)2(z)2dxd4xy1(xR2x2)202dxdyD4DRR2x2dxd4RRdxRx2R Rx21R2x2dy8RRdx16R2R44D如下求均勻薄片的質(zhì)心(1)D由y 2pxxx0y0所圍成解令密度為1因?yàn)閰^(qū)域D可表示為0xx

0,0y 2px所以0AAdxdyx0dx2pxdyx00 0 02pxdx232px30Dxx1xdxdy1x0dx2pxxdy1x0xA00A02pxdx3x5A0Dyy1ydxdy1x0dx2pxyd1x0pxd3yAA00A080D所求質(zhì)心為所求質(zhì)心為(3x,3y)5080(2)D是半橢圓形閉區(qū)域x,y|x2y21,ya2 b2Dyx0AD1ab(橢圓的面積)2yy1ydxdy 1 dxabaa2x2ydy1b (a2a2 2AAx)dx4ba0A2a2 a3D所求質(zhì)心為所求質(zhì)心為(0,4b)3(3)Drbcos(0ab)之間的閉區(qū)域解令密度為1由對(duì)稱性可知y0AAdxdyb2a2(b2a2)(兩圓面積的差)224Dxx1AxdxdyD 2d2Abcosrcos rdr0acosa2b2ab2(ab)所求質(zhì)心是所求質(zhì)心是(a2b2ab,0)2(ab)55Dyx2yx所圍成它在點(diǎn)y)處的面密度(xy)x2y求該薄片的質(zhì)心解M(x,y)dxdy1dxxx2ydy11(x4x6)dx0x202135Dxx1Mx(x,y)dxdyD1M1dxxx3ydy111(x5x7)dx350x2M 0248yy1My(x,y)dxdyD1M1dxxx2y2dy10 x2M 0311(x5x8)dx3554質(zhì)心坐標(biāo)為質(zhì)心坐標(biāo)為(35,35)48 546設(shè)有一等腰直角三角形薄片腰長為a各點(diǎn)處的面密度等于該點(diǎn)到直角頂點(diǎn)的距離的平方求這薄片的質(zhì)心使薄片在第一象限且直角邊在坐標(biāo)軸上薄片上點(diǎn)(xy)處的函數(shù)為x2y2由對(duì)稱性可知xyMM(x,y)dxdyadxax(x2y2)dy1a40 06Dxxy1Mx(x,y)dxdyD1Maxdxax(x2y2)dy2a005薄片的質(zhì)心坐標(biāo)為薄片的質(zhì)心坐標(biāo)為(2a,2a)5 57利用三重積分計(jì)算下列由曲面所圍成立體的質(zhì)心(設(shè)密度1)(1)z2x2y2z1z軸上xy0VVdv1(圓錐的體積)3zz1zd1d1rd1zd3VV00r4所求立體的質(zhì)心為所求立體的質(zhì)心為(0,0,3)4(2)(2)zA2x2y2za2x2y2(Aa0)z0zxy0Vdv2A32a32A3a3)(兩個(gè)半球體體積的差333zz rsin cos drd d d sin cos d rdr13 1220 A33(A a)44VV008(A3a3)所求立體的質(zhì)心為所求立體的質(zhì)心為(0,0,3(A4a4))8(A3a3)(3)(3)zx2y2xyax0y0z0解Vadxaxdyx2y2dzadxax(x2y2)dy0 0 0 0 0a[x2(ax)1(ax)3]dx1a4036x11a5Vxdv 1axdxaxdyx2y2dz15V0001a642a5yyx2a5z1zdvV1adxaxdyx2y2zdzV730a2000所以立體的重心為所以立體的重心為(2a,2a,75 5 30a8設(shè)球體占有閉區(qū)域{(xz)|x2y2z22Rz}它在內(nèi)部各點(diǎn)的密度的大小等于該點(diǎn)到坐標(biāo)原點(diǎn)的距離的平方試求這球體的質(zhì)心x2y2z2z軸上xy0在球面坐標(biāo)下在球面坐標(biāo)下可表示為02,0,0r2Rcos于是2MMdv2d2sind2Rcosr2r2dr00022325R5sincos 325dR5015zz1Mzdv1M2d2sincosd2Rcosr5dr0 0 02M20646R6sincos7d8R6332r515 45故球體的質(zhì)心為故球體的質(zhì)心為(0,0,5R).499設(shè)均勻薄片(面密度為常數(shù)1)所占閉區(qū)域D如下求指定的轉(zhuǎn)動(dòng)慣量(1)D{(x,y)|x2y2Ia2 b2D可表示為yaxa,b ax2ybaaax2于是I x2dxdyax2dxaba2x2ydy2bax2abaa2x2aaa2x2dx1a3b4D提提 aax2a2x2dxxasinta4sin22202tdt8a4(2)D(2)Dy29xx2II2解積分區(qū)域可表示為xy0x2,3 x/2y3 x/2于是I y2dxdydx23x/2xy2dy 2 2273x2dx03x/23022725DI I x2dxdyx2dx23x/26 2ydy5x2dx03x/22 0967D(3)(3)D為矩形閉區(qū)域{(xy)|0xa0yb}求I和Ixy解I y2dxdyadxby2dya1b3ab3x0033DII x2dxdyax2dxbdy1a3ba3by0033D1010已知均勻矩形板(bh板對(duì)于通過其形心且分別與一邊平行的兩軸的轉(zhuǎn)動(dòng)慣量解取形心為原點(diǎn)取兩旋轉(zhuǎn)軸為坐標(biāo)軸建立坐標(biāo)系I y dxdy dx y2dy1bh3b2h22xbh12D22I I x dxdy xb2 22dx dy1hb3h2y12Db2h2一均勻物體(密度為常量)占有的閉區(qū)域zx2y2z0|x|a|y|a所圍成(1)求物體的體積解由對(duì)稱可知V4adxadyx2y2dz0 0 04adxa(x2y2)dy4a(ax2a3)dx8a40 0 033(2)求物體的質(zhì)心xy0z1Mzdv4adxadyx2y2zdzV0002adxa(x42x2y2y4)dyV002a(ax42a3x2a5)dx7a2V03515(3)z軸的轉(zhuǎn)動(dòng)慣量解I (x2y2)dv4adxadyx2y2(x2y2)dzz0004adxa(x42x2y2y4)dy428a6112a60 045451212ah慣量(解建立坐標(biāo)系使圓柱體的底面在xOy面上z軸通過圓柱體的軸心用柱面坐標(biāo)計(jì)算I (x2y2)dvr3drddzdar3drhdz1ha4z0002 1313 設(shè)面密度為常量 的勻質(zhì)半圓環(huán)形薄片占有閉區(qū)域D{(x,y,0)|R1x2y2R,x0}zM0(00a)(a0)處單2FF(FFF由對(duì)稱性F0而x y zyF GxDx(x2y2a2)3/dG cosdR 222(2a2)3/ d2R12G[lnR2a2R22R2R1]R2a2R11R2a22R2a21FF