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# ENGINEERING MATH III MATH 251

Texas A&M
GPA 3.6

David Kerr

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COURSE
PROF.
David Kerr
TYPE
Study Guide
PAGES
4
WORDS
KARMA
50 ?

## Popular in Mathematics (M)

This 4 page Study Guide was uploaded by Vivien Bradtke V on Wednesday October 21, 2015. The Study Guide belongs to MATH 251 at Texas A&M University taught by David Kerr in Fall. Since its upload, it has received 29 views. For similar materials see /class/226006/math-251-texas-a-m-university in Mathematics (M) at Texas A&M University.

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Date Created: 10/21/15
1 2 00 r U 103 00 MATH 251 Practice Problems for Examination 2 Spring 2007 Find the absolute minimum and maximum values of the function ay xy 7 6x2 on the closed region bounded by the parabola y 9x2 7 1 and the x axis Calculate the double integral a ffR 3xem2 dA7 where R 717 0 gtlt 01 b ffR 5 cosy2 dA7 where R is the region bounded by the lines y 7x z 07 and y 71 c ffR em2y2 tan 1 dA7 where R is the region described in polar coordinates by r00 0 70 r 0 d ffR5x2y2zy5 dA7 where R is the region between the two parabolas y 2 10 and y 2x2 6 Find the volume of the solid which lies between the paraboloid z 6x2 y2 and the xy plane and above the region 0 S y S 17 y2 S x S Find the center of mass of a lamina occupying the region 0 S x S 57 0 S y 3 2 with density function py 2 Determine whether the following is true or false 72 S ffosinemy2 x5 dA S 27 where R 01 gtlt 72 0 Compute f 6W2 xy dy dz Find the volume of the solid bounded by the parabolic cylinder 2 y2 and the planes 24y7z1and2 Let E be the volume in R3 lying between the parabolic cylinders z y2 and z y2 2 and above the region in the xy plane bounded by the lines z 07 y 07 and y 17 Compute xy dV Solutions 1 We have fwzy y 7 12x and fy x and both of these are zero at the point my 07 07 which is on the boundary of the region For the part of the boundary of the region that lies on the s axis we have the function f7 0 76z2 for 7 S x S i which has maximum value 0 at 0 and minimum value 7 at and 7 On the other part of the boundary we have the function g f7 9x2 7 1 9x3 7 6x2 7 z for 7 S x S Then g z 27H 7 12x 7 17 which is zero at 0 HEW Since g 0 lt 0 the value 9z0 003754 is maximum for 9 while 97 7 is the minimum Thus for f the absolute minimum value is 7 while the absolute maximum value is approximately 003754 2 a 10 2d d 1 3 2m0 d 3 3xew zy 76m y717e 0 1 0 2 171 2 0 7y 0 F721 0 5 cosy2 dx dy 5x cosy2 dy 75y cosy2 dy 7 0 71 F0 71 5 0 5 7 siny2i1L 5 sin1 I 9 I 7 I l 9 l 4 re drdt 79572 d6 76592 7 1 d0 0 0 0 2 70 0 2 I 2 2692 lazy 1 1 4 0 4 6 4 d 2 2 1 10 2 5 1 ym210 5x211 2zy5 dy dz 7z2y2 7zy6 dz 2 2m26 2 2 3 y2126 ggzamz 10 7 2x2 62gt dz 106 7 2x2 66gt dz Expand to evaluate the rst integral7 and for the second use substitution 3 The volume is given by 1 y 1 y 1 V 6z2y2dzdy 2z3y2z dy 3y372y67y4dy 0 y2 0 F112 0 7 3 4 2 7 1 5 37 7 4y 7y 5y 0 714039 4 The mass and moments are 5 12 5 1 m xzdyd 4dx7z5 547 0 0 0 5 0 5 12 5 2 5 7 1 4 1 1 5 5 Mm xzydydz 7952112 dz 76d7z7 7 0 0 0 2 110 0 2 14 0 14 5 m2 5 1 My xsdyd 5dx7x6 6 0 0 0 A 25 125 and so the center of mass is F j 5 The statement is true On R the values of the function are bounded between 71 and 17 and the area of R is 2 Therefore 72 71 gtlt areaR S ffosinewy2 x5 dA S 1 gtlt areaR 2 6 Using Fubini7s theorem7 0 1 2 1 0 2 1 2 1 F0 e y zydydz 67y zydzdy ze y 7z2y dy 71 ix 0 7y 0 2 m y 1 2 1 3 7y 7 d 7 7y 7 0 ye 29 y 25 say 7 The volume is given by 2 4 2 2 2 13y 4 i ddz 2 77 dzi 10y y y 1 y 3y 210 3 1 171 y22 1 171 zy22 95de 9596125195195 dydz E 0 0 22 0 0 2212 1 17m 1 y1im dy dz w dz 0 0 0 210 1 22 3 dz H o H D

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