sol8scan - MATH 237 Assignment 8 Winter 2007 Solutions 8 B5...

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Unformatted text preview: MATH 237 Assignment 8 Winter 2007 Solutions 8. B5 The hyperbolae $2 — y2 = 1 and —a:2 + 3y2 = 1 meet (from adding) when 2112 = 2. In the first quadrant this is at (m, y) = (fl, 1). Following the sketch 1} $11“ — x1 + 33" = l D 2 D! V 01. We find the total mass, M', of the silver is A/I=//p($,y)dA = //$ydA+//xydA D 01 D2 1 (#)1/2 «5 (1+3x2)1/2 = / / ccydydw+ mydydac 0 0 1 (z2_1)1/2 1/2 fl 1 y=< 3 ) / [am/2] dx 1 2 y=(x2—1)1/2 '11 1 2 “51 1 2 2 /—:I:—(1+x)d$+ is: 3(1+x)—(m —1) dx 0 1 $2+m‘11+/‘/§ 2 13 1 — — —a:——:I: (ix 2 4 0 1 3 3 H CR r———1 [\3It—‘ B Q to L—l cc r: lo M a + II II [I 01Dov—I 0‘le Chlb—A There is 5 / 24 grams of Silver on the plate. 9. A1. We set (11,12) : T(x, y) = (33+y) —a:+y). This mapping is linear and has Jacobian r 3—“ @ 6(u’v)=det<gfi ga>=det( i 1):2. may) a 3—3; — 1 Using the change of variables formula, we write f/(a:+y) eos(y:v— )dxdy=//ucos(v)gE ”(dado Dry Duu —1 where %E-% = [%E%] : 5 and DW is the image of the original domain of integration under the tranformtaion T. The original domain can be written as ny = {(55, y) I 0 S y S 7r,0 S .T g 7r — y}. The three vertices are mapped as: T(0,0) = (0,0), T(7r,0) = (7r, —7r) and T(0,7r) = (7r, 7r). Since this is a linear mapping, straight lines are mapped to straight lines, and so the image Dm, is as shown below: \I (TOT!) (om (0.“) ( TT, 0) We can write Dun = {(11,21) 1 0 < u < 7r —u. < v < u}. Thus the integral takes the form f/(ar—i—y) )cos(:13—y)da:dy = f/ucos(v)%dvdu O —u 0% —;—/0[usin(v) ()]v:“udu 2/511 (sin(u )— sin(—u))du 1 7r = —/ 21Lsin(u) (1'11. 2 0 7T [u cos(u)]{)r — f — cos(u) du (integration by parts) 0 ll 77 + [sin(u)]6r = 7r. 9. A3 Completing the square for the equation :32 + y2 = 2:1: gives (m — 1) + y2 = 1 which describes the circle of radius one centred at (1,0). The region of integration Dzy is as sketched below fed X‘l ; In polar coordinates, £132 +y2 = 29: becomes 7‘2 = 27" cos(6) or r = 2 cos(6), and $2 +y2 = 1 becomes 7‘ = 1. The two points of intersection occur when l = 2cos(6), i.e. 6 = 7r / 3 and 6 = —7r/ 3. Under the polarcoordinate transformation, the region of integration becomes Drg = {(r, 6) | ‘— 7r/3 g 6 S 7r/3,1S r S 2cos(6)}, as sketched below. 6_(___r ,9)__ By the change of variable formula since a—r—my — 7", we have //——— = D/fm’sfl rdrdG 1 / $2 + :92 Dry Terr/3 2cos(0) = / 7‘ c0509) d?“ d6 —7r/3 7r/3 r=2 cos(9) = f [— cos(0)] d9 —7|'/3 2 T:1 1r/3 1 ll \ [2 0053(6) — — cos(6)] d9 —1r/3 2 7r/3 [2(1 — si112(6)) 003(6) — écos(6)] d6 —7r/3 m [2(1 — 81112(0))COS(6) — écos(6)] d6 ——1r/3 || \\\ #3 [2 (108(6) — 2 sin2(6) 005(9)] d6 —7T/3 [E sin(6) — §s1n3(0)]m —7r/3 3% 23¢"; 3f 23_\/§ = ii??? {77+ 3 s ”3 9A6 In spherical coordinates, the spheres 3:2 +y2 + 22 = a,2 and 2:2 +y2 + 22 = b2 correspond to 'r = a and 7' = b. With no other restrictions, the region of integration can be described by Dr¢g = {(r, qb, 6) | b g T s (1,0 g Q5 3 7T, 0 S (9 S 2%}. Using the change of variables formula, we then have )-3/Zg_— ’2) (W ///W 31:0) Dm'yz Drrfif) fgffar-(T “Numb ¢)drd¢d6 0: >5: We mi [27TH C 3l111(r)l§ 27r(2)(ln(a ) — ln(b)) = 47T(ln(a) — ln(b)) \ \ E RM + {QM + NM T, 9:3 <1 I H 9B4 Recall that the ellipse g + bi: = 1 can be parametrized by z = a cos(6), y = 7‘ sin(theta), generalizing the standard parametrization of the unit Circle. Extending this approach to 4 spherical coordinates, we define the coordinate transformation (27,31, 2) = T(r, (b, 6) as a: = or sin(gf) cos(6) = bf" sin(rfi) sin(6) a: = a? cos(q3) Then, recalling the result for standard spherical coordinates, we find 3:2 y? 2:2 2 g+§+C—=7 . Thus in (7“, (13,6) space the given ellipsoid is the unit sphere centred at the origin: Dréé = {(7‘, (ii, 6) | O S F S 1, 0 5 qt 3 7T, 0 S 6 S 27r}. Comparing the Jacobian of the transformation T with the Jacobian of the transformation to spherical coordinates (r, o, 6), we find that the only difference is an additional factor of a, b, or c in the rows, which can be factored out to give a Jacobian of char2 sin(q'i ). Then, integrating over the given ellipsoid ny2, we find the volume as ///1de f/fabcr2 sin(qb @drd¢d6=abc(:7r ), Dzyz where the last calculation yields simply the volume of a unit sphere. 9.B5 Let D represent the mz-cross—section of the glacier as sketched below: At different points in this cross—section the ice is moving at different velocities. (Note that the motion is fastest at the origin (the “center” of the glacier), and is slower near the edges (and is in fact zero 011 the edges)). This can be understood as the effect of friction on the walls of the glacier. If we imagine dividing the cross—section up into small rectangles {IL} as shown above, then if we sample a point (33,321.) in rectangle R,- with area AAi, then the volume of ice that flows through R,- in one year is approximately 11(931, 22-) . 1 - AAi. Summing over all of the rectangles, we have the approximate total volume of ice moved: Emma, 2,)AA, In the limit of more and smaller rectangles, the volume of ice passing throught D in one year is given by [/1)(£E, 2) (M = //10—3[1 — 10%:2 + 22)] dA D D Transforming to polar coordinates we have 4/ 10-3[1 — 102(2:2 + 22)] (M II ...
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