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Unformatted text preview: MATH 251  SECOND HOUR EXAM Version B Professor Weibel 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 77, 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 :c = x(u, v) and y = y(u, 11) that the region R in the Jay—plane corresponds to
the region S in the uv—plane. Then //[email protected]=//fwww where the J acobian ‘g—g’—Z; y) 8W: 1)) du dv 6—1:
is det (37 23 6_y 8—”). For polar coordinates, d2: dy— — 1' d?" d0 611, 6—11 The 3—D analogue (between coordinates u,v,'w and say, z) amounts to the changeof— a($,y,z) 8(x’y’z)
6(u,’u,'w) 8(u,’u,w) the analogous 3 x 3 matrix of partial derivatives. For cylindrical coordinates, dw dy dz =
7" dr d6 dz. For spherical coordinates, dm dy dz = p2 sin(¢) dp d¢ d0. variables relation d3: dy dz = } is the determinant of du dv dw where 1 Criterion for the Existence of a Potential Function. If E 2 (F1, F2) is a vector ﬁeld
in a simply connected plane region, F is a gradient ﬁeld if and only if 8F1 /6‘y ': 8F2/8x. —; If E 2 (F1, F2, F3) is a vector ﬁeld in a simply connected region in 3—space, F is a gradient
vector ﬁeld if and only if the three cross partials agree: for all 2' and j, 8F, / 81:, = BFj/ami. Line integrals. Let f"(t) parametrize a path 0 (to S t 3 t1). Then f0 f(3:, y,z)ds =
t1 H0 f(f~(t))(ds/dt)dt. Iff‘ is a vector ﬁeld, then [C E(m, y, z)d§= tie, F(r (t)) (dr/dt)dt
If F = VU, f0 F . d§ = U(f’(t1))— 0mm». Surface Area. If z : z(a:, y) deﬁnes a surface S in 3space over a region R in the (m, 3;)
plane, its surface area is //SdA=//Rllﬁlldwdy, Whereﬁ=WIWj [7] Integrate f (m, y) = my over the region bounded by the line y = a: and the parabola
y = 1:2 ~ 2:. (These curves meet at the origin and at 0417.) [5] Sketch the domain of integration, and express the iterated integral in the opposite order:
5 25
/ f (iv, y) dy div
III [10] A Wire bracelet in the shape of the curve y = sin(a:) (0 _<_ x g 7r/2) has density
p(a:, y) = y. What is its mass? (Be careful With minus signs.) 3/2 [8] Find the surface area of the surface 8' given by z = 23: + By which lies over the unit square 0 3 33,3; _<_ 1 in the (ac 3/) plane. 953: 32”“ SA JJW‘JW
25:3 :XS‘WAXA —jWAx O “‘ zik‘t ”(Ch H03 :Di‘l O
2 gammdiomJ [12] Set as = 311. + 2'0 and y = —u + '0. Let R denote the region in the (x,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 (7:, y)plane (b) Apply the change of variables formula to ﬁnd ffR a: dx dy. ‘l (M) ,< E \
21:) O \ .l 0 [8] Let U (x, y) 2 ﬁg‘ and consider the vector ﬁeld I? = VU. Determine f0 F  d§ Where
C is the path f’(t) = (1 — t,t), 0 S t S 1. lite] : (l ' O)
5—3 L“ ‘3‘ (OI‘) $EA§3r U(O)\\——U(J‘O)
G”: :14 :EJ [25] Consider the region between the (m, y)—p1ane and the cone m2 + 3/2 = 2'2 over the disk
0 S m2 + y2 S 4. Compute its volume in three ways: (a) Integration in Cartesian Coordinates: set this integral up, including all limits of inte—
gration, but do not evaluate this integral. (b) Integration in Cylindrical Coordinates. a.“ '1 r ’«k P
V: j szch—dg = aﬁj‘ (ale (if 1 3m jargdr ‘— QT: (31:: “33;: 0 (0) Integration in Spherical Coordinates. Hint: 4'—=—pees(9§—_§2 F‘ r. f S'IALQ E: 1
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’ 351C Lso — 3 ban 115‘ 3 [10] Consider the average value of 1 / :3 inside the circle 7‘ = 2 cos(6) of radius 1 about the point (1,0). Show your work. ”Wk noose
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a [15] Consider the vector ﬁeld F = (yz, mz, a: + z). (a) Is there a potential function U such that F = VU? Explain your answer. No. 957 last EA. 21' 9* (b) Compute the work f0 F  ds’ done in moving from P = (0,0,0) to Q = (3, 1, 5) along
the path f(t) 2 (3t, 75, 5t). ELFLLW ’‘ <56) 58‘} $E>
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