MAT 125 SAMPLE MIDTERM - MAT 215 Instructor Mike Wang NAME...

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Unformatted text preview: MAT 215 Instructor: Mike Wang NAME: « 5amp1e Mlctterm I Show all your work clearly! If you finish early. don't forget to check your work. I. Answer the following: (20 pts) (2 pt each) “(qu or False): J‘I szz sin(x -y) (b: {I}! =- J; JI x2 sin(x -y) (1y (Ix. . (True or PM: J“) L; Jx +y~' ‘1,qu =- J“) J‘; ,/x +y‘ dxdy. “ (Tfiéor False): J; J;.\“’c”dy dx =- JIxzdx J:c’dy. ‘ (True or False): JJ 1 (L4 = A(D). u ' (True or False): H k (1.4 = k/KD). l) “ (True or False): "'0 is the disk given by x" +y‘ S 4. then JJ J4.—x-’ -y‘ M = Jihr. I) " (We or False): The integral J; J; J; «I: dr 110 represents the volume enclosed by the cone : = (Ix: +y' and the plane 2 = 2. ’ (True or False): lfflx,y) 2 g(x.y) for all (x.y) in D. then ”fix.” (1/! 2 JJ g(x.y) dA. " (True or False): A C' transformation is the inverse of a Jacobian transformation. ' (True or False): lfm sflx.y) S Mfor all (x.y) in D. then mA(D) S JJ/(x,y) (M S MA(D). 2. Evaluate the iterated integral J; Jx sin(y2) dydx. (IS pts) I vol: “(11] £de = \o x%\“\‘1\\‘.t‘ ‘ \‘o “Mp 3“ V ‘3‘ {\ smut) \u ‘ iii, co<\“\\ 3. Find the area of the surface of the part of the sphere x2 + y2 + z2 = 4 that lies above the plane 2 = I. (20 pts) 2. «.4 n l _ 1 _ Jo LT} 1} £333 d! t.) 2 ‘ - 1 1(2'1) ' >424") , éLm-fi‘ at O ., r: ’ d” :7. I )2 7 \3 1 1-7) 4 - , _ 3 )\ 3 1 r {7 ' ‘ fill-1353* 0 I 'L ngrqufi “fl 0 . - - .' V , v) ' . \ - « ‘ u ru-.,_‘.~ 1. l._ .V " -',“r*:7w:.v_!n7.—.x '1‘ ‘. 'r‘m “I vi "1-; v‘ - ' ~ 'v -_._ ‘ 5. (4o pls) (i) Balm!“ ”I 8’ dV. where E is enclosed by the paraboloid z = l + x2 +y’. the cylinder x3 +y2 = 5, and the xy-planc. (20 pts) 1’v\+r1 (1:5 Li=|+r1 r—J‘S u=b du‘ 1rd: ho ‘1‘: \ 5. (continue) 2 1 2 - land (ii) Evaluate ”I 2 (11/. where E lies between the sphcrcsx +y +2 E x2 +y2 +22 - 4 in the first count. (20 pts) 92 .. I \ s p L. 2 P1 . q n 5 1 “/2, 6. Evaluate the double integral H by (1,4. D is the triangular region with vertices (0,0), D (LZ) and (0,3). (20 pts) ‘(L"‘I- 1’0 1 T=Z ‘1" b— m1 :— 31'3\ : 35;}. t \/ -. '\ \2 \‘fi ¢‘+3( vi) ' ’\ i 41": 0' D ‘3 7.7. C\' 3, 141,71 -’2~<+ b :11. 1 6}—Q.y*’~ ., 'L D 2(031": 8“ * kE'XX-L ' \1“) \ A1 1-1r-v w3 J 51 Z -y~b 1' j o {\Kq,(oy.(j'\‘nb 1"U‘a'b q“). if — .gxta " 3'1“ I 3 A b Eh -31 u 1 1 2. ’10 4’ er ’ {’30 ’1. 7. E . ‘ valuatc the double Integral H Ti}? (1,1, R z [0. l] x [0, |]_ (15 p18) R . “7 w ._ + j 3 \ i\“k\§vj) .12 \‘M ‘ ' “fly, -\\n\\mq- \m 4'1 ”\LU») du= A: J”: l lo; \If ul V I MM]. , R. :22 2 3 \“‘\{\n\|u}_‘]' ,/_ L 3 Extra Credit (5 p15) Set up the triple integral j j 1 xy dV, where E is the solid tetrahedron with venices (0.0.0), E (1,0,0), (0,2,0), and (0,0,3). YOU DON’T HAVE TO FNTEGRATE! _ '5 v r’a:L—\,:.ov YVJ—("W'V’ ‘-’\ 'L O\:V‘\*3311\‘ ,‘ O .5 ' \ («p.317 Q\i-¥n\« M‘s—‘3") ‘ aka-EA ‘ 0 y —?z + 2 t .1111 ru'?‘ ~ ) ...
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