Solutions to Midterm 1 1 Find the volume of the solid which is above the paraboloid z x2 y2 and below the half cone zpx2 y2 Solution In cylindrical coordinates E r z 0 r 1 0 2 r2 z r Therefore the volume is ZZZ dV E Z 2 Z 2 0 Z 1 Z r Z 1 r2 0 d 0 0 Z 2 Z 1 r r2 r drd r dzdrd r r2 r dr 0 0 2 1 12 6 1 2 Evaluate RRR x2 y2 z2 9 Solution In spherical coordinates E z2dV where E lies between the spheres x2 y2 z2 4 and E 2 3 0 2 0 and z2 2 cos2 Thus ZZZ z2dV E Z Z cid 18 2 0 0 Z 2 cid 19 0 Z 3 cid 18 211 2 cos2 sin d 2 Z 2 cid 19 0 3 5 Z 3 2 844 15 2 cos2 2 sin d d d cid 20 cid 21 0 cid 20 1 5 cid 21 3 2 2 0 5 d 4d cos3 3 2 3 Evaluate ZZ cid 19 cid 18 x 2y 3x y dA sin R by changing the variables The region R is the trapezoidal region with vertices 1 2 3 4 2 4 and 6 8 Solution Letting u x 2y and v 3x y we have x 1 Then 5 2v u and y 1 5 v 3u and R is the image of the trapezoidal region with vertices 5 5 5 5 10 10 and 10 10 Thus ZZ cid 19 cid 18 x 2y 3x y sin R dA Z 5 10 1 5 h v cos cid 16 u cid 17 i v v v dv x y u v 3 5 1 5 cid 12 cid 12 cid 12 cid 12 1 5 2 5 cid 17 cid 12 cid 12 cid 12 cid 12 1 Z v Z 5 cid 16 u Z 5 sin 5 v v 10 1 5 10 0 dv 0 1 5 cid 12 cid 12 cid 12 cid 12 cid 12 cid 12 cid 12 cid 12 dudv 3 4 Evaluate R C xyzds where C is given by parametric equations x 2 cos t y 2t z 2 sin t 0 t Z C Solution x 2 cos t y 2t z 2 sin t 0 t Then by Formula 9 section 16 2 xyzds 2 cos t 2t 2 sin t 8t sin t cos t s cid 18 dx cid 19 2 cid 18 dy cid 19 2 cid 18 dz cid 19 2 dt dt dt 2 sin t 2 2 2 2 cos t 2 dt dt Z Z Z 0 0 4 cos2 t sin2 t 4 dt 0 4t sin 2t 8 8 Z cid 16 2 2 0 2 t sin 2t dt 8 0 4 2 2 cid 20 1 2 cid 21 0 t cos 2t sin 2t 1 4 q q cid 17 4 5 Evaluate the line integral Z C F dr where F x y z zi yj 0k and C is the line segment from 1 0 2 to 2 4 2 Solution Since C is the line segment from 1 0 2 to 2 4 2 r t 1 t r0 tr1 h 1 t 1 2t 1 t 0 4t 1 t 2 2ti h1 t 4t 2i 0 t 1 Z C Z 1 Z 1 0 0 F dr h2 4t 0i h1 4 0i dt 2 16t dt cid 2 2t 8t2 cid 3 1 0 10 5
View Full Document
Unlocking...