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Unformatted text preview: MATH 251  SECOND HOUR EXAM Version A PTOfeSSOI‘ We1be1 November 19, 2009 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 [n] 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. J acobians and Change of Variables. Suppose that under the 1—1 differentiable trans formation a: : 513(u, v) and y = y(u, u) that the region 7?, in the any—plane corresponds to
the region S in the uv—plane. Then f/f<wy)dscdy= f/ﬂxy) where the Jacobian l g—Eu’g; 306,11) 301,”) ~dud'u ﬂan _8___$ 6—3:
is det (a 8” 67’ ). For polar coordinates, dsc dy = 7" dr d6.
Bu, 81} The 3—D analogue (between coordinates u,v,w and x,y, z) amounts to the change—of variables relation dzr dy dz = ‘% du dv dw Where lW‘ is the determinant of the analogous 3 X 3 matrix of partial derivatives. For cylindrical coordinates, dac dy dz =
7" d7" d9 dz. For spherical coordinates, d3: dy dz = p2 sin(q§) dp dgb d0. Criterion for the Existence of a Potential Function. If F 2 (F1, F2) is a vector ﬁeld
in a simply connected plane region, F is a gradient ﬁeld if and only if 8F1/6y '2 BFg/Bx. If F 2 (F1, F2, F3) is a vector ﬁeld in a simply connected region in 3space, F is a gradient
vector ﬁeld if and only if the three cross partials agree: for all 71 and j, aFi/Bccj = 8F} /8m,~. Line integrals. Let F(t) parametrize a path C (to < t < t1). Then tofc f(a:, y,z)ds— —
1:230 f(r(t ))(ds/dt)dt. IfFisavector ﬁeld, then f0 F(w, y,z) ds: F(r (t)) (dr/dt)dt
IfF: VU, [CF ds = U(f"(t1)) — U(f(to)). Surface Area. If 2 = z(m, y) deﬁnes a surface S in 3—space over a region R in the (x, 3;)
plane, its surface area is //SdA= //72 ”ﬁll dmdy, Where ”ﬂ” 2 WW [7] Integrate f(a:, y)— — my over the region bounded by the line y = x and the parabola
y— — m2 (These curves meet at the origin and at (1,1).) [5]9 Sketch the domain of integration, and express the iterated integral 1n the opposite order. /_4/:ﬁ f<x y)dydx V=3 [10] A wire bracelet in the shape of the parabola y = :32 (—2 S a: S 2) has density
p(ac, y) = $132. What is its mass? 3/2 [8] Find the surface area of the surface 5' given by z = 1f a: + 2y which lies over the unit square 0 S 9:, y_ < 1 in the (x, y) plane. 9.75: SA: ‘Y‘WJXJ? . {Si”s; th‘l <1de
: %(Gl\l 0311‘; : i%(u\/T— 215)! [12] Set as = 3u+v and y = u+2v. Let R denote the region in the (sc, y)plane corresponding
to the unit square 0 < u, v < 1 in the (u, v) plane. (a) Draw the picture of the region R 1n the (m, y) plane (b) Apply the change of variables formula to ﬁnd ffRa: dcc dy. [8] Let U (:16, y) = :52 +312 and consider the vector ﬁeld 15‘ = VU. Determine f0 F‘ » ds‘ Where
C’ is the path 1"(t) = (t, 1 —t), 0 S t S 1. li(°)= (Oil)
Fm: ( W T3 x (13>; L) (Mil ’ U10,” =1~\:E>1 ‘ é [2 5] 5Consider the region bounded below by the cone m2 +y2 — 22 / 3 and above by the plane
2: \/§. Compute its volume 1n three ways: (a) Integration in Cartesian Coordinates: set this integral up, including all limits of inte
gration,_ but do not evaluate this integral. \(1’ [aﬁx (b) Integration in Cylindrical Coordinates. 1“ 1 E I
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4 : 7ﬁ<6ﬁaf~ﬂl3 [10] Find the average value of m2 + y2 inside the circle of radius R about the origin. Show your work. 9“ K
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CL [15] Consider the vector ﬁeld 1—?" = (3/2, :32, 3:2). (a) Is there a potential function U such that 1:5 = VU? Why or Why not?  g— . 3F _
N0. .322 loft ”1—1. (b) Compute the work f0}?  d§ done in moving from P = (0,0,0) to Q = (1, 2, 3) along
the path i"(t) = (t, 2t ,.3t) FURL“: <0: at St) ...
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