midterm2_c - MATH 251 - SECOND HOUR EXAM Version c '7...

Info iconThis preview shows pages 1–5. 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
Background image of page 3

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

View Full DocumentRight Arrow Icon
Background image of page 4
Background image of page 5
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: MATH 251 - SECOND HOUR EXAM Version c '7 Professor We1bel November 19, 2009 Your Name Section Section 7 (W2 10:20AM) Section 8 (W3 12 PM) Section 9 (W4 1:40PM) There are 100 points on the exam. The notation at the start of a problem means that the problem is worth n points. Rules: No calculators. Remember—show your work to get partial credit. If you need more space, write “over” and use the back of the page. Jacobians and Change of Variables. Suppose that under the 1—1 differentiable trans— formation :1: = 3001., v) and y : y(u,v) that the region R in the any—plane corresponds to the region S in the uv—plane. Then //Rf(a:~,y)d$dy= f/Sflmflggjzi Q; a(flay) 8(uw) du do where the Jacobian ‘ a_a: is det ( g; a; > . For polar coordinates, d3: dy = r dr d6. 6_u 81) The 3—D analogue (between coordinates u,v,w and m,y,z) amounts to the change—of— variables relation dzr dy dz = l% du d'u dw where 135% the analogous 3 X 3 matrix of partial derivatives. For cylindrical coordinates, d3: dy dz = 7" dr d9 dz. For spherical coordinates, d3: dy dz = ,02 sin(gz$) dp dd) d6. is the determinant of Criterion for the Existence of a Potential Function. If E : (F1, F2) is a vector field in a simply connected plane region, E is a gradient field if and only if (9171/83; ‘2 8172 / 8:13. If E = (F 1, F2, F3) is a vector field in a simply connected region in 3—space, E is a gradient vector field if and only if the three cross partials agree: for all i and j, 8Fi/8mj = 8Fj /8:c,-. Line integrals. Let 17(75) parametrize a path C (to g t 3 t1). Then f0 f(:r,y,z)ds : t1 ,ZEO f(r’(t))(dsédt)dt. Iffi is a vector field, then [C E(w, y, z) as: $230 Fem) - (dr’/dt)dt If F = VU, f0 F - d§= U(r‘"(t1))— U(r”(t0)). Surface Area. If z : z(:I:, y) defines a surface S in 3—space over a region R in the (my) plane, its surface area is dmdy, Where [7] Integrate f (at, y) = my over the region bounded by the parabola as = y2 and the parabola y = 302. (These curves meet at the origin and at (1, [5] Sketch the domain of integration, and express the iterated integral in the opposite order: 11:14:“; “ma 3!) dydx Y [10] A wire bracelet in the shape of the parabola y = 2:32 (—2 S m g 2) has density p(m, y) = 1:2. What is its mass? 3 , M:§r¥ C 1' Ar: _-a 9‘ K ‘: S‘tl [(0451 + l C {?(t\1<£)9‘€ ’3 ._;éeé3\ _ may—51 + _L. QAHZZ,‘ «45> “ I a?» Hex: (5&7 3a “(alight lloQ‘+\ .)J\gs\~.\v\\‘°/\ Q/xcl I‘:\€3"0‘l1\7\ ‘9! 63°? '3 > [8] Find the surface area of the surface S given by z = l/E 1: + 313/2 which lies over the unit square 0 S .13, y g 1 in the (:17, 3;) plane. 7— ’9 ‘5 t “ 31,11 1 531% 3A: HéHGHA‘V“ \l V 0 3t . ( :S‘ll‘i’r‘év Al‘i‘: all” Al o U L' , 3rA \ »—~——r/ -_—_ (1+1) ‘0 - 'q l [12] Set a; : 3u—v and y : u+2v. Let R denote the region in the (as, y)-plane corresponding to the unit square 0 S u, v S 1 in the (u, v)-plane. (a) Draw the picture of the region R in the (ac, y)—plane (b) Apply the change of variables formula to find ffR 9: d1: dy. ' v Lu‘ (43 ' ’22. V T.) LL 0 \ \9 313E: 3 " : () l Dd r] 3| U [8] Let U(.z', y) = 11:23; and consider the vector field F‘ = VU. Determine f0 f‘ - ds’ where C is the path 17(75) : (t, 1 +15), 0 g t g 1. lim; (0,0 lit“: {WA} SEA; um— um c 1‘ WM— (Clam [25] Consider the region bounded below by the cone $2 +y2 : 22‘ ‘A z = 2. Compute its volume in three ways: A . and above by the plane (a) Integration in Cartesian Coordinates: set this integral up, including all limits of inte- gration, but do not evaluate this integral. [10] Find the average value of x/zcz + y2 inside the circle of radius R about the origin. Show your work. i _. . ‘ q 'R l Z -——3—‘- J‘ p {\clr clfj TR _\ x x : CL qgf ” 2*— f} R RA 9 r ’ R3 75 lo 2‘. 0 :— 3 l\ [15] Consider the vector field f‘ = (yz, x2, yz). (a) Is there a potential function U such that I?" : VU? Why or Why not? ANO‘ :1 loo—l“ 7— . (b) Compute the work f0 I? - d§ done in moving from P = (0,0,0) to Q = (3,2,1) along the path fit) 2 (3t, 2t, t). .4 F693 :<3€, '39.) .18 “em: <3 ,1) W ...
View Full Document

Page1 / 5

midterm2_c - MATH 251 - SECOND HOUR EXAM Version c '7...

This preview shows document pages 1 - 5. Sign up to view the full document.

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