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mid2f98s

# mid2f98s - Solutions for Math 20E Midterm 2 Fall 98...

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Unformatted text preview: Solutions for Math 20E Midterm 2, Fall 98, Lindblad. 1. (a) The equation for S is x+y+ z = 1 and the unit normal is n = (i+j +k)/\/Â§ so F - n = 2(a:+y+z)/\/Â§. S is as agraph z = 1â€”mâ€”y over the regionD : {(x,y); :1:+y g 1, a: 2 0,31 2 0} in the myâ€” pâ€”lane. Then dSâ€” â€” dmdy/ cosyâ€” â€” dacdy/n kâ€” â€” x/gdxdy. Hence 1 //SF ndS= //132 ()\$+y+z dxdyz/Ol/Ol 2drdy=/2(1â€”y)dy=2yâ€”y2 0 (b) The curve 0 consists of 3 line segments that each contribute with â€”1/2 so the line integral is â€”3/2. The line segment 01 from (1,0,0) t0 (0,1,0) is given by R(t) = (1 â€” t)i +tj, 0 S t S 1 with the orientation corresponding to increasing t. We have fair-de/O F(R(1Â§))-Râ€™(t)dt=/0(i+tj+(1â€”t)k)-(â€”i+j)dt=/0 tâ€”1dt=â€”â€” 2. (a) Let S1 = {(3:,y,z); y=0, 332/4+z2/4 S 1} and S2 = {(\$,y,z); x2/4+y+z2/4= 1, y 20} The area of S1 is 47r. S2 can be viewed as a graph 3/ = g(m, z) = 1 â€” 332/4 â€” 22/4 over the disc D = {(37, z); 372/4 + 22/4 g 1} in the mzâ€”plane. With G(a:, y, z) = y â€” 9(33, 2) the unit normal is n=VG/\VG|= (â€” gwiâ€”gzkâ€”lâ€”j)/\/1+gg+gÂ§: (mi/2+zk/2+j)/\/1+332/4+22/4 Now dS= dmdz/n jâ€” _ '\/1 +( (\$2 + Z2)/ 4.da:dz Introducing polar coordinates 1n the 1133â€” â€”plane: 002 +z2)1/2 27f 4 T 3/22 4 3/2 f/S2d8= f/D( (1+ dazdzâ€” /02/0( (1+â€”) )1/2d67'drâ€”27rÂ§(1+z2) 0â€”271'Â§(2 â€”1) (b) The normal to S1 is nâ€” â€” â€”j and there F- nâ€” â€” yâ€” â€” 0 so the integral over S1 vanishes. Since F- n:( (372/2â€”lâ€”z/2â€”y) )/\/1+\$2/4+z2/4weobtain //SF ndSâ€” //D( (2/2+z/2â€”y) )dmdz://D( (\$2/2+z/2â€”(1â€”(33 +32)/4))dmdz Introducing polar coordinates 1n the xzâ€”plane; :13â€” â€” r cos 0, zâ€” â€” 7Â° sing, dazdzâ€” â€” rdrdQ; 2 27r 7,2 7' 2 27r T2 7. // (â€”(2cos26+1)+â€”sin6â€”1>d6rdr=// (â€”(c0s26+2)â€”â€”sin6â€”1)d6rd7â€˜=...=0 3. The region R corresponds to the region D = {(u, 1)); 1 S u g 2, 1 g 1) g 2}. 1 =1 0 8(u â€™0) 21:3; 332 8(x y) 8(u v) â€”1 1 Now 7 = =3 and 7 =( 7 ) =â€”s0 aw) Lil/\$2 1/33 y 3(u 7}) a(ï¬‚ay) 3y l l 2 //R:1:da:dy://Dz1:a(\$\$(ua:3;dude2/2/2â€”dud'02/12/231vâ€”dudv: Iii/01:11? 4. (a) Ru = (â€” sinu â€” vcosu)i+ (cosu â€” â€™usinu)j and R1, = â€”sinui +cosuj +k so i j k Rux R1,: â€”sinuâ€”vcosu cosuâ€”vsinu 0 =(cosuâ€”vsinu)i+(sinu+vcosu)jâ€”vk â€”sinu cosu 1 Hence dS = lRu >< va dudv = V1 + 2'02 dudv. 27r (1 2 3/21 3/2_1 (M/fszdS =/01/ vv1+2v2 dudv=27T/01v\/1+2'U2 dv=7râ€”(2+v) =73 3 3 0 ...
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