S 17-3 - Math 200, Spring 2010 Worksheet #30 V Name Key...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Math 200, Spring 2010 Worksheet #30 V Name Key Section-176 Divergence Theorem 1. Let S be the closed cylinder at2 + y2 = 4 with top and bottom 2 = 4, z = 0 , respectively. Calculate the flux of F (x, y, z) = (x, y, 2) out of S (a) directly, and (b) using the Divergence Theorem. 8.3mm 3“ - 'H'A ‘ A S 5 A??? 5‘} 2__o uni-L n--K “09‘s. 'gj<¥l,y’0>-(‘K)JS: 5504330 a. |- 2. . ISdsK X+7fihl~)2:'1-wflkR=R 81 S. “*5 a. ‘m e|‘\'2q Ion I 3 A i gqu- kasrgfims: amt) ( 1 2 I611“ lcose, QSIkeJ2)10é as (Jame, 25ers} 2).( 3 we, lsfnepMM _ I , -° Lde932fi.l-f.t+ :3211- - ‘ - - o — z — 9r) Hex; -— jgdwU-‘Hu - y(§X+$/+§%Q)JU= 5£S3du '—' 3 value (93‘3'04'9‘1') ’ '7‘?" 2. Use the Divergence Theorem to calculate the surface integral HF-dS where F(x,y,z)=x223i+2xyz3j+xz4k s and S is the surface of thebox with vertices (i1,-_t2,i3) d‘ ‘3 : A 1‘ '3 9. 3 A 7 9 3'" z “’ (t) Axlxé )t 37(1xy2) trifle“) -’ 2x2 rlxe +‘t-x2 r be ‘So/ 9, 4+6; DL'der‘aeh-ce T‘hEOWLu-n. “J J 4 IL I 1 Pas: flaws-cw: ~ I 3 555 5 v ) _‘j:_L/(_3[8xadidyclx ~ deJdngZle - L 2- " 3 c6 2%.1-1 [vi-mllLfi? -3380 #020 Q-F = 3x1+271+72+ 3-21: 3(x‘+7:+2ll ‘3 SC: k-lestETE Kl‘i-yl-pil' 2:30 a 5, (5 (“SK xz+7119)2:o ij—ln "fits-Q 5: 3 SuS, ":5 a closed Sue-kc: _ 3. Let Sbe the (non-closed) hemisphere x2 + y2 +22 =9oriented with upward pointing Dermal (not ' closed). Calculate ILF-dS for F(x,y,z) = (x3 +2er2 — yz,3x22, yzz+23)by closing S and applying the Divergence Theorem. 51 “(2.43 = 5% (Lots - F-d’s‘ : 1255‘ cixthldu - ff . (43d 5 i \ X+y+e Siliao X339,th : 3 XL ‘- '1 a s 1 21$ f *7 +2 law ff“ *3x7 4,0, o>-(-Rlds = 5f 943 a, 1. wry-re We, x-q-y sq o D Smfio 09%“de ‘9 W .5 do 4. (a) Are the points P1 and P2 sources or sinks for the vector field F shown in the figure at the right? The Ufic‘ior: +koci en; new- a are Skew-“1'9? +h°§h ‘Hne. vet-tars 'Hm‘l' ‘ flat-4f 0d: R )So'Hte net $10» is cu‘i'ward') P. is q Sew-ch. ._ \ ) The wed-or: 4%: 9nd n-Par P2. avg larger 4kg.“ *9 ream-J 4441+ ‘3th Mar 19:, So dime “91‘ 4:10» is inward; P3, 1.5 Q Stink. (b) Given that F(x, y) = (x, y2> , use the definition of divergence to verify {/2- t 'a ,- \ theanswertopart(a)- Hey) = 09% =3 amt?) 31+3y mm- _2 Tke y‘vcdde 0? P\ {.5 (3051.440?) 50 dt‘u(?) > O figs F‘ lts q gem-(4" -J— ‘ " -— . _ Ht: 1’2.) 7< 2) So &\U(F)- [+27 <0 J +50; PL ‘5 ch fink. 5. Let F = e—rz, where p = ~fo + y2 +22 and e, is the unit radial vector field.- Verify that: p (a)div(F)=0 "F": (Eva’s?) =(xf'3yffej'5 -.3_ 1 1 1 J. ‘ ‘ exf'axixwa =flx+a+2 )y*2x=fi‘ 2 fif'figfizffi ) " " "“ I - 1- 2 t LEV ixf3:f3'7qu'£ 3913.50 r: 34—1 . A 1.31 1_ 1 1 ')_ I; 3. 1 1 — 3 '1-31- amp): ige+figéx+fijgz = Waufipfito 1 R” 5% o O -r~ swag T— omato = 3r. Lumpy? : NUH) : two (0) ILF-dSEO‘fgr any closed surfaceSthat does not contain the origin. 0 - 5;“; F is" dethixed ad mew—{Jeni in treats“ to encloses! b7 swam n61 cokimkjt‘e mat... cud div (F): b on DU) In); ~H~.e Diueraeue What-em SSSF‘d§ 3 flgdivifi)du: D J Homework #30 Reread Section 17.3 (Due 05/14) Do Section 17.3 Exercises: 3, 7, 9, 13, 15, 18, 19, 22 6) '- :5 since UP: 2 cm 'Hqu Sphere 75?: ...
View Full Document

This note was uploaded on 02/24/2011 for the course MATH 200 taught by Professor Jamesdcampbell during the Spring '10 term at Santa Barbara City.

Page1 / 2

S 17-3 - Math 200, Spring 2010 Worksheet #30 V Name Key...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online