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## ENGINEERING MATH III

by: Evert Christiansen

10

0

4

# ENGINEERING MATH III MATH 251

Evert Christiansen
Texas A&M
GPA 3.92

Staff

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COURSE
PROF.
Staff
TYPE
Class Notes
PAGES
4
WORDS
KARMA
25 ?

## Popular in Mathematics (M)

This 4 page Class Notes was uploaded by Evert Christiansen on Wednesday October 21, 2015. The Class Notes belongs to MATH 251 at Texas A&M University taught by Staff in Fall. Since its upload, it has received 10 views. For similar materials see /class/226053/math-251-texas-a-m-university in Mathematics (M) at Texas A&M University.

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Date Created: 10/21/15
1 E0 00 U rh 03 MATH 251504 Practice Problems for the Final Examination Fall 2006 Use the Divergence Theorem to evaluate ffs F dS where Fy z 2zi 26y x23j 722 6y cos c k and S is the surface of the solid E that lies in the region z 2 0 and is bounded by the parabolic cylinder y 1 7 2 and the planes y 0 z 0 and z x Find curlF and div F a Fxyz ewyzi zgj cosy k b yizi2zjyizk c Fxyz 4i x 722j 4k WWW Determine whether or not F is conservative and if it is conservative nd a function f such that F Vf a Fxy 2yem2 i e2 cos yj b Fxyz ysinzixsinzj 5sycoszk C Mtg2 yzi j9cyk 2y y 1 Evaluate f0 1222 d5 where C is the line segment from 030 to 7121 Evaluate fCF dr where Fxyz lnxi yzj 3k and C is given by I39t 3itjt2kfor 72 gt 2 Use Green7s Theorem to evaluate f03x2y 12y2 dx 3 siny dy where C is the boundary of the trapezoid with vertices 00 20 02 and 24 with clockwise orientation 7 00 H E0 Use the Fundamental Theorem for Line Integrals to evaluate f0 F dr where Fxyz y4zi 4xy32 eyzj 6y 114 k and C is given by I39t costi sintj 2tj for 0 S t S 77139 Use Stokes7 Theorem to evaluate ffs curlF dS where x2y2 Z Wag72 iyz722jezk and S is the part of the sphere 2 y2 22 10 that lies below the plane 2 71 with upward orientation Solutions Using the Divergence Theorern7 1 17m2 x FdS dideV 26ydzdyd S E 0 0 0 1 142 24 1 142 1 y1m2 226g dyd 2zeydyd 2mg dx 0 0 10 0 0 0 110 1 2 1 617m 2x51 w2 7 2x dx 7 7 2 e 7 2 0 0 a curlF 7x siny 7 322 i zyemyz y sinyj 7 26M k divF yzemyz b curlF 7j 7 2k divF 71 C curlF 2 7 z i 2 7 z k divF 0 3 a Since 22y6w2 26m2 6m2 cosy and F is de ned on all of R2 F is conservative ie F Vf for some function f Since fwzy 22y6m2 we must have f2y ye2 gy for some single variable function 9 Then fyy 2 g y so that g y cos y in which case we may take 9y siny so that ay ye2 siny note that f is unique only up to addition of a constant term b Since curlF 0 and F is de ned on all of R3 F is conservative ie F Vf for some function f Since fmyz z ysinz we have fyz 22 xy sinz gyz for some two variable function 9 Then fyzyz xsinz gyz and so gyz 0 which means that gyz hz for some single variable function h Then fzyz xy cosz h z and thus h z 5 so that we may take My 52 and hence fyz zz xy sinz 52 note that f is unique only up to addition of a constant term c Since curlF lt2 0 72gt which is not zero everywhere F is not conservative r Pararnetrize C by I39t lt7t3 7 ttgt for 0 S t S 1 Then r t lt71711gt so that lr tl xZ and hence 1 3 1 1 17 3 was lt3 7 t2t2 dt xglt3t3 7 a4 25 i C 0 2 5 0 10 5 We have r t lt012tgt and hence 2 2 1 2 Fdr ltln3t33gt lt012tgtdt 9 6tdt it 3t2 0 c 72 72 4 2 6 Using Green7s Theorem 2 22 2 21 322 122 2 dx zssin d 7242 d dx 712 2 dx y y y y y y y C 0 0 0 110 2 712953 i 48952 i 4895 dz 0 2 73954 71623 7 24952 7272 0 5 Since F Vf where fyz xy4z 6 as can be found by a procedure similar to that in the solution of 3b7 we have7 by the Fundamental Theorem for Line Integrals7 CF dr fr77r 7 fr0 f710147r7 f100 147r The boundary 0 of S is a Circle which we pararnetrize by I39t lt3 cos t 3sin t 71gt for 0 S t S 27139 Then r t lt73 sin t73cos t0gt Thus7 by Stokes7 Theorern7 27r CurlFdSFdr lt797971egtlt73sint73cost70gtdt S C 0 27r 27f 27sintcos t dt 277cost sin 0 0 0 0

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