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Unformatted text preview: 1. Let D be the region deﬁned by
D={(z,y):231+2ys4.0§$7y53). (b) (8 points) Find a rectangular region D' = [(2,2)] x [c, d] and a linear
transformation T such than D = T(D‘)r v u:¥*zj a u+lv=3X
3 vzx—‘j X: “1,3,, 3 62 [2,11% Y0151 (c) (8 points) Compute ffD (Ii—ghdzdy by making a change of varir ables. .‘ .7: ‘ _’I 30‘) = )—
“(iii— 3 =» 1m 9
} '7 I4
‘1 2 v _ or“: _ L2: 3
5130 V ‘50“, § 1; (ZuL)D&V
_! ‘1: 3
= %§:ﬁ1d“:%(u)r g Pngc 2 2. Let T : 1W — 1R2 he the transformation T(u,v) = (u+2v,2u~uz). (a) (10 points) Draw and label the image of the rectangle D‘ = [0, 2] X
[0,1] under T. V ("9 L.)
\
@@ @t L“ (9. (+,o),oe~€£—z ranILHIOeteL @ ( (Mu)
: 2,4. , ostH  1 
) —) (2am t 3,021“ «“(a‘m7wﬂ:z>
@3 (in) ,obléL —; (++Z,2£—I)Iol;téL ‘5 “POLE‘3
(ha)
 u
®= (o,t),osket » {ﬁfth/0261:: z (“Ml W” ) 3:41,? (1)) (10 points) Is T one~to~one on D"? Explain. sane» TOM)": :— 1(uL,Vpl 9° ‘MZV: “Aviva SHE—w. Lulum 2u\'V.L =Zu2—VL" . 7 3206 “Ms “4‘7 VF": 
' a. So v\=v1_ laud ul— Y{$. "PK {\MLl'u» +fﬂl: ’X‘q
vc surku hank f \s uol Vb" 0" R I bul l = ,
we: unf. ulaz//‘ Hf (S ‘_%V‘ m loin ' — ML LA“
3° Page 5 3. Let 51(t) and am be the paths
CW) = (cos(t).sin(t)).03tsvr,
52“) = (cos(t),7s'm(t)),0 51,5 7r.
(:1) (5 points) Draw graphs of (710) and 52(t). Be sure to label Lhe (b) (5 points) Show than the vccmr ﬁeld F(w,y) = (33%;, ﬂ) is a
gradient vector ﬁeld  # may 9x )3 ’ (guy)L Q'uqz)‘
f 1: XI' 1 L _ X 1" j l '— o
“my” , 04 *3le ("‘*7‘)1
“fl ‘ 3:: E 0 4 moldd Vaclav (c) (5 points) Compute #13115 and F— I ds‘ directly (by using the line integral formula). _
"(T/t) = (Sln+,(0'7“‘) a7“  (’SW‘I'C‘S‘t) .. _. 1r
S‘Z' E ml; = §W($M§, dost) ’ (943, (east )Jf ga’F '0‘ 5 = 306947455" )' (‘5‘;‘3
‘ a 05.45 I wk)  (—Suk’ #951 = $5)“ :Eﬂ 5:0 Wu»: ' EVA (d) (5 points) Why doesn't your answer to part (c) contradict the
damental Thevrem of Line '.7 Tu wax luau. lam = w' ()5) :3 ml A»);ch
’QD/ 3’30, whbk ‘5 a} “La Meolu") 06¢“me ,
§D 4H1. MW dos «cl aRI7 . {I} \.\.plwl, “cum/k) 4. Let S be the part of the cone 2 = 3 I2 + if that lies below the
plane 2 : 6. (a) (7 points) Find a parametrization of S. (b) (7 points) Draw a picture of S, showing the grid lines for your
paramctr ization. i=9'7
,Rx r ad. data} (M “44 W
i‘)‘ 9 %' ink: —?.=§r \vl an”) 6 points) Sigposc S is oriented so that the “inside” of the cone
is the part that would hold water. Is you parametrization orientation:
preserving or orientation reversing? Explain. i, ’Ymmb u‘, 7T; 70:49 r’g‘iftwloolmhvsk 90 L7 “all” M (Via. [2&7} KT“; VD“?
‘ib'umls Wukdz aha)“  g OWiR’ins P014193 O
.. .. g
14(40ka7, a»qu TrxTD : Q (twelﬁq'e/g) x (m.tiér(osa,a)
.: (3m591‘5r5m9’ f) Lot/WA (mm) 1'» (M U?) 5. Let S be the surface of the unit cube [0,1] x [0,1] >< [0,1]. Let
F(r,y.z) = (I, ZyVO)‘ (a) (4 points) Explain Why the surface integrals of F over the bottom
(2 = (J) and top (2 : 1) of the cube are both zero. :‘vl, L‘A F TLLL “M3,. 4m\ vulvrs m “ l
\«us We v~ coapm  90 (5.: :o m bah £0549; WA SfER‘JSW)
Slagz C;‘ (5“ LbHovk, g7, "s (b) (11 points) Compute the surface integrals of I? over the left side
(y = 0) and right side (y = l) of the cube (2+ 9} \,4 M bﬁ— doll{QM 9‘4 4“
. _ % OEXe', 092’
9; . @ (X/e) — (X/o/ )1 (DIU0) v0!” ‘ 1 ﬁle = u,o,o)xzom = P $83.41 = S“ (x,o,o"(°/'I°)""‘°‘* 1E
S; o o ruﬁhfSVb  L L
el/o,%/\ : ,2 : Cxl'I‘l’) ’0éx
g“ ) LAM :l—yXFa = (0"llol 'POMK ’
SS” Edi; ;'S‘0‘(o\(¥,2,o)(0,I/o)clxol%
5% : _§\§0\(;1)AYA£ :E]
D my.» a (#5 continued) (0) (4 points) Compute the surface imegrals of 15 over the back (I = 0) and fron ($21) of he Cube. _
m %; ML mic/M Sb l“ l“ l“ ££b\ $51 Q—(yptl : (0,7,%),Déj£\, 0 “A N g  (o V o) x(a D \) : (“0,0) (30ml) muo — I, / / A l \ : S‘S‘E.Dl§=—Sa§b(0,27,0)(/O,0)A7e 101 s; 09 5‘, obZél
Saga/7,2): (we), 3 J ,n = 0,00) 179'
T” \e ’ .(l/D’o)d7JQ:§D I \ J gm
SC §;§;(\,2’Io) go"? —2 (e) (/1 points) Suppose F is the velomty ﬁeld of a liquid, What does
the sign of your answer imply about the flow of the liquid relative to
ihe cube? Page 7 ...
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 Fall '07
 Enright
 Vector Calculus, hank, lhe, damental Thevrem

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