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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 mzcross—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/ 103[1 — 102(2:2 + 22)] (M II ...
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