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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 ...
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