Transcription of Solution to Problem Set #9 - Mathematics & Statistics
1 Solution to Problem Set # the area of the following surface.(a)(15 pts) The part of the paraboloidz= 9 x2 y2that lies abovethex yplane. 4 2024x 4 2024y 4 part of the paraboloidz= 9 x2 y2that lies abovethex yplane must satisfyz= 9 x2 y2 0. Thusx2+y2 9. Wehavez=f(x, y) = 9 x2 y2,fx= 2x,fy= 2yand 1 +f2x+f2y= 1 + ( 2x)2+ ( 2y)2= 1 + 4x2+ regionE={(x, y)|x2+y2 9}is{(r, )|0 r 3,0 2 }inpolar coordinates. Hence the area of surface is E 1 +f2x+f2ydxdy= 2 0 30 1 +r2rdrd = 2 013(1 +r2)32|30d =(13(10)32 13) 2 =2 3((10)32 1). (b)(15 pts)The part of the spherex2+y2+z2= 4that lies above theplanez= 1.
2 4 2024x 4 2024y 4 spherex2+y2+z2= 4can be written asz= 4 x2 part of the spherex2+y2+z2= 4that lies above the planez= 1must satisfy1 z= 4 x2 y2. Thusx2+y2 3. Wehavez=f(x, y) = 4 x2 y2,fx= x 4 x2 y2,fy= y 4 x2 y2andMATH 2850: page 1 of?? Solution to Problem Set #9 MATH 2850: page 2 of?? 1 +f2x+f2y= 1 + ( x 4 x2 y2)2+ ( y 4 x2 y2)2= 1 +x24 x2 y2+y24 x2 y2= 4 x2 y2+x2+y24 x2 y2 44 x2 y2= 2 14 x2 regionE={(x, y)|x2+y2 3}is{(r, )|0 r 3,0 2 }in polar coordinates. Hence the area of surface is E 1 +f2x+f2ydxdy= 2 0 302 14 r2rdrd = 2 0 302(4 r2) 12rdrd = 2 0 2 (4 r2)12| 30d = ( 2 + 4) 2 = 4.
3 The following triple integrals:(a)(15 pts) Eyzsin(x5)dVwhereEis the region{(x, y, z)|0 x 1,0 y x, x z 2x} regionEbounded by thexy,yz,xzplanes and theplane2x+y+ 2z= 4is the set{(x, y, z) R3: 0 x 2,0 y 4 2x,0 z 12(4 x 2y)}. EzdV= 20 4 2x0 12(4 x 2y)0zdz dy dx= 20 4 2x012z2 12(4 x 2y)0dy dx= 20 4 2x018(4 x 2y)2dy dx= 20 148(4 x 2y)3 4 2x0dx(by substitution u=4-x-2y)= 20 148(4 x 2(4 2x))3+148(4 x)3dx= 20 148( 4 + 3x)3+148(4 x)3dx= 148 3 4( 4+3x)4 148 4(4 x)4 20= 148 3 416 148 416 ( 148 3 4256 148 4256) = 136 112+49+43=53.
4 (b)(15 pts) EzdVwhereEis the region bounded byx= 0,y= 0,z= 0and2x+y+ 2z= 2850: page 3 of?? Solution to Problem Set # regionEbounded by thexy,yz,xzplanes and theplane2x+y+ 2z= 4is the set{(x, y, z) R3: 0 x 2,0 y 4 2x,0 z 12(4 x 2y)}. EzdV= 20 4 2x0 12(4 x 2y)0zdz dy dx= 20 4 2x012z2 12(4 x 2y)0dy dx= 20 4 2x018(4 x 2y)2dy dx= 20 148(4 x 2y)3 4 2x0dx(by substitution u=4-x-2y)= 20 148(4 x 2(4 2x))3+148(4 x)3dx= 20 148( 4 + 3x)3+148(4 x)3dx= 148 3 4( 4 + 3x)4 148 4(4 x)4 20= 148 3 416 148 416 ( 148 3 4256 148 4256)= 136 112+49+43=53.
5 3.(10 pts)A bead is made by drilling a cylindrical hole of radius1mmthrough a sphere of radius9mm Set up a triple integral in cylindricalcoordinates representing the volume of the bead. Evaluate the inte-gral. (Hint: Express the regionE={(x, y, z)|x2+y2+z2 9andx2+y2 1}(There is a typo in the original Problem .) in cylindrical coordinatesand find EdV.) cylindrical coordinates, the sphere is given by the equa-tionr2+z2= 9and the hole is given byr= 1. Hence, the bead isdescribed by the inequalities 9 r2 z 9 r2,0 2 and1 r 3.
6 We find the volume by integrating the constant densitySolution to Problem Set #9 MATH 2850: page 4 of??function1over the sphere:Volume= R1dV= 31 2 0 9 r2 9 r2r dz d dr= 31 2 0[rz] 9 r2 9 r2d dr= 312r 9 r2 2 0d dr= 2 [ 23(9 r2)3/2]31=23(8)3/22 =4 38 8 =64 3 2mm3. 4. (a)(15 pts)A spherical cloud of gas of radius3km is more dense atthe center than toward the edge. At a distance of km from thecenter, the density is ( ) = 3 . Write an integral representingthe total mass of the cloud of gas and evaluate spherical coordinates, the sphere is described by theinequalities0 3,0 2 and0.
7 Hence, the totalmass of the cloud isMass= W dV= 0 2 0 30(3 ) 2sin( )d d d =( 2 0d )( 0sin( )d )( 303 2 3d )= (2 ) [ cos( )] 0[ 3 13 4]30= (2 )(2)(27[1 34])= 27 . (b)(15 pts)A half-melon is approximated by the region between twoconcentric spheres, one a radius1and the other of a triple integral, including limits of integration, giving thevolume of the half-melon. Evaluate the spherical coordinates, the outer hemisphere is describedby the inequalities0 b,0 /2and0 2 and the innerhemisphere is described by0 a,0 /2and0 2.
8 Hence, the volume of the half-melon isVolume= W1dV= 2 0 /20 21 2sin( )d d d =( 2 0d )( /20sin( )d )( 21 2d )= (2 ) [ cos( )] /20[13 3]21=2 3(23 13) =14 3.
