midterm 2006(3) ans

midterm 2006(3) ans - HKUST MATH 102 Third Midterm...

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Unformatted text preview: HKUST MATH 102 Third Midterm Examination Multivariable and Vector Calculus 12 April 2007 Answer all five questions Time allowed – 120 minutes Directions – This is a closed book examination. No talking or whispering are allowed. Work must be shown to receive points. An answer alone is not enough. Please write neatly. Answers which are illegible for the grader cannot be given credit. Note that you can work on both sides of the paper and do not detach pages from this exam packet or unstaple the packet. Student Name: Student Number: Tutorial Session: Question No. Marks 1 /20 2 /20 3 /20 4 /20 5 /20 Total /100 – 1 – Problem 1 (20 points) Your Score: The intersection of the two surfaces (elliptic cylinders) x 2 + y 2 2 = 1 and z 2 + y 2 2 = 1 consists of two curves. (a) Parameterize each curve in the form r ( t ) = ( x ( t ) , y ( t ) , z ( t )). (b) What is the arc length of one of the curves? (c) Find the volume bounded by the two elliptic cylinders above. (d) Find the surface area of the part of the elliptic cylinder z 2 + y 2 2 = 1 that lies within the elliptic cylinder x 2 + y 2 2 = 1. Solution: (a) Fix first x ( t ), y ( t ) to satisfy x 2 + y 2 2 = 1. i.e. x = cos t y = √ 2 sin t z = ± cos t ∴ r ( t ) = (cos t, √ 2 sin t, ± cos t ) , where 6 t < 2 π or r ( t ) = ( t, ± p 2(1- t 2 ) , ± t ) , where- 1 6 t < 1. (b) r ( t ) = (- sin t, √ 2 cos t, ± sin t ) s = Z 2 π k r ( t ) k dt = Z 2 π p sin 2 t + 2 cos 2 t + sin 2 t dt = √ 2 Z 2 π dt = 2 π √ 2 . (c) V = 8 Z √ 2 Z √ 1- y 2 / 2 Z √ 1- y 2 / 2 dz dx dy = 8 Z √ 2 Z √ 1- y 2 / 2 r 1- y 2 2 dx dy = 8 Z √ 2 1- y 2 2 dy – 2 – Problem 1 (20 points) Your Score: = 8 y- y 3 6 √ 2 = 16 √...
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midterm 2006(3) ans - HKUST MATH 102 Third Midterm...

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