Transcription of ANSWERS TO EVEN ASSIGNED PROBLEMS CHAPTER 12 …
1 ANSWERS TO even ASSIGNED PROBLEMS CHAPTER 12 review (pages 723-724) the projection on theyzplane we can evaluate the integralas follows: EzdV= 10 1 y20 2 y0z dx dz dy= 10 1 y20(2 y)z dz dy= 1012(2 y)(1 y2)dy= 1012(2 y 2y2+y3)dy= spherical coordinates we have that 0 1(since the solid is bounded by the sphere of radius 1),0 2since the solid lies above thexyplane) and 0 le2 . Thus Hz3 x2+y2+z2dV= 2 0 /20 10( 3cos3 ) ( 2sin )d d d = 2 0d /20cos3 sin d 10 6d = 2 [ 14cos4 ] /20(17)= of the solid is the disk of radius 2 centered at theorigin. In cylindrical coordinates the planey+z= 3 has equationz= 3 rcos.
2 Thus the volume is given byV= EdV= 2 0 20 3 rsin 0r dz dr d = 2 0 20(3r r2sin )dr d = 2 0[6 83sin ]d = 6 ]2 0+ 0 = 12 paraboloid and the half-cone intersect whenx2+y2= x2+y2, that is whenx2+y2= 1. Thus the projection of thesolid on thexyplane is a circle of radius 1. In cylindrical coordi-nates, the equation of the paraboloid isz=r2and the equation ofthe cone isz=rand the volume is given byV= EdV= 2 0 10 rr2r dz dr d = 2 0 10(r2 r3)dr d = 2 0(13 14)d =112(2 ) = 642. (a)The surface is a vertical plane at an angle of 4with the positivex-axis. In cartesian coordinates,the plane has equationy=x.
3 (b)The surface is a cone at an angle of 4radians with the positivez-axis. In cartesian coordinates thecone has equationz= x2+
