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by: Kavon Feest

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2

# Multivariable Calculus MATH 53

Kavon Feest

GPA 3.93

D. Auroux

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COURSE
PROF.
D. Auroux
TYPE
Class Notes
PAGES
2
WORDS
KARMA
25 ?

## Popular in Mathematics (M)

This 2 page Class Notes was uploaded by Kavon Feest on Thursday October 22, 2015. The Class Notes belongs to MATH 53 at University of California - Berkeley taught by D. Auroux in Fall. Since its upload, it has received 21 views. For similar materials see /class/226592/math-53-university-of-california-berkeley in Mathematics (M) at University of California - Berkeley.

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Date Created: 10/22/15
Math 53 Practice Midterm 2 A 7 Solutions 1 1 2722 1 a The area of the triangle is 27 so y 7 ydmdy 2 0 2372 10 By symmetry i O 2 p r cos Using symmetry710 rzprdrd0 D 27r l 7r2 l 7r2 1 4 r cos ammo 4 74 cos awe 4 7cos0d0 7 0 0 0 0 0 5 5 l l 3 m z y 2 0 g m S 1 so yzgdx yzdy mzzsder m222xdm 3x5dx C 0 0 4 a Qm 6x2 byz7 Py azz 3342 Qm Py provided a 6 and b 3 10 f1 3ny y3 1 e f 23531 zyg z My Therefore 12 2363 39592 y Comparing this with Q we get 2x3 Bzyz g y 2x3 3zy2 2 so g y 2 and g 2y 1 0 So 1 2x33 1 zys z 2y constant c C starts at 170 and ends at 757207 so 1 f7equot70 7 f10 75 quot 7 1 7 2 2 5 a um 9 hy 3 y 335211 Therefore dudv 3m2y dmdy 3M dmdy 11 111 y 1 1 and hence dz dy 7 du dv 3M 4 51 41 2 b d d 7d d fl 5d 7l 5 2zy213uuv23nv3n 6 a g Mdz 7M2 dA Green s theorem 0 R b We want M such that 7My m y2 We can use eg M 7z y3 1 m3 1 e e 1 7 ad1VF2ySOjlF dS 2ydA 2ydydm mde7 c R 0 0 0 7 b For the ux through Cl ij implies F 71 yz 71 where y O The length of 01 is 17 so the total ux through 01 is f0171 ds 71 The ux through Cg is zero because i and 13 1 i c F ds F dsi F dsi F dsli7170 Ca 0102Ca C1 02 7 7 1 7r2 1 1 8 72 7 dz dr d0 0 0 0 1 9 a sphere p 2a cos 1 b plane p a sec gt 27r 7r4 2a cosq c p2s1n dpd d0 0 0 a secq 10 a 2zy 23 2x z2 2342 2zy 23 322 342 3x22 71 i 32 b 2zy237 SO 9571472 3 a a z2379z22y27s03791 Therefore gy7 z yzz hz m2 2yz 2y ail342 3x22 7 1 so F is conservative 902 9023 9y7 2 Qyz 7 and fzy z xzy x23 yzz 3m22 y2 hz y2 3m22 7 17 so hz 71 Therefore 712 11 Sis the graph ofzfmy Therefore dS ffsltv 3 HS 412 42 d1 dy since 2 1 7 m2 Shadow unit disc 2 y2 g 1 72c and fmyz m2ym23yzzizc limziyz so dS ifhify 1 dzdy lt2m72y 1 dxdy 7217239lt2m2y1o lmo ly ff52z22y22172dmdy 2 7y sWitching to polar coordinates7 we have g dS f0 f014rzrdr d0 027 r43d0 27139

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