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23. if=4xy , -2,3 xyz >
X. - y , ×=z , x ?_ z n=< I , -1,0 >
Curle . |¥dx¥y¥z| . - < 3×2-+1 , - (3×7+0) , ok >Txf my .
Z 3×yZ = < 3×2-+1 ,-3×7,7×3
zz:X}.×
parameterise |¥×( 3×7+1 , -3×2 , -7X >.< 1 , -1,0> =
6×2-+1%(6×2+1) dzox°
x'ffE÷33 . ( Ep , if ' ±s=fCisyinete.Cont . diff . on open set
containing S• C be simple . smooth closed
. S be oriented and smooth
C : lity ' : 1 Curve in
×2+y2±othe plane zoo
vector field f isnt defined on Zaxis . But anyS of which ( is a
boundary Must include at
least one pt . on z
-
25.ffcurlfnds Sistheicap
"
of thes Unit sphere below the
Xyplaneandinsidexty"YgWith the outward normal Cylinder
)
Flay ,⇒=-yz2i+xz2j+5×YE-
.×
f¥dFLscwficitzcosoftrz ) : t.no.gs#jtsinotcos0t2rD
of , < stcosoitssino ,o >Edt : - atcososf - tasiiosfto
27"¥ ''
ftp.oo.tl?Ih#
0
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Practice Questions for Exam 4 Math 231
1. Find Fcurl and Fdiv if ( , , ) sin sin sinx y zF x y z e y e
z e xi j k 2. Evaluate the line integral 3
C
x zds where C is the curve
2sin 2cosr t t t ti j k for 0 /2t . 3. Evaluate 2 2( )
S
x z y z dSwhere S is part of the plane 4z x y that lies
inside the rectangle 0 1, 0 1x y .
StokesNo opposite
divF-r.F-o0yltxsny1thfyleiYsinHtfzle7sinxj.EYmy-i9sinz-e3sinxCur1F-8xE-mYxYoyYozCO-eiYcoszitetosx-o1o-eTosyitInye-YsnzeiB.nx-5eTosza-eTosq-eEosyyXyzdS-HrttHrtDh2cost.l.dsint7dsiFEtEFn2tf.x3z-8sin3t.2cost.fE-o5.16sin3tcostfx3zos-jIY2sin3tcostrsottkfsin4trshMi4IFLxiytcx.y.4txtysrI-a.o.isrTico.l.isioff@4txtyIty44txtyDr3.dydxIxFy.4.t
,l >
d.si#.xryH=BdA
-
4. Use Green’s Theorem to evaluate 2 2x ydx xy dy where C is the
circle
2 2 4x y with counterclockwise orientation. 5. Use Stokes’
Theorem to evaluate ( )
S
curl F ndS where
2 2 3( , , ) , , xyF x y z x yz yz z e , S is part of the sphere
2 2 2 5x y z that lies above the plane 1z and S is oriented
upward.
6. Evaluate the line integral
C
F dr where
3 2 3 4 2 2 3(4 2 ) (2 3 4 )F x y xy x y x y yi j and C: ( ) (
sin ) (2 cos )r t t t t ti j , 0 1t .
247×2/Fdx+Edy=HCH÷.tt#A:ffty2-x2wydx-2 - fixf , :X2y272fz : .
xyz iff .r2rdrd÷
7 Flo ) :< 2650,25Mt , I >-
to9$isa Circle a circle ofXHy?2' radius 20h< S j the
Olanlztlzsinozcoso,:¥¥t-÷Z" ( jcojgzsinoasino ,e > .
asinopcosoo >Skog.ge#i6coszsirio+4sin0ws0t0)d0vsepower -× <
cusgsino > reducingCCW and half . angleCw < Sino
,wso>→< cososinto ) > id .
TKOSO , - Sino >
where fis the potential
if Fis cons . ( curif .. e)
Of F
then ,f( B)- fl A) - §Yf .dr : ftFdr
Yx±f.FI#oguy.sxykogxg.+yyYooz/=Eogojbx5*5HxHisB
-
where fis the potential
if Fis cons . ( curif .. e)
Of F
then f( B)- fl A) = [ Dpfdr : FF. or
just'Ft¥Yk×y.si#ogxg.+yyYooz/=Eogojbx5*5Hxt6Bf(4x3yt2xy3)dx=x4ytx2y3+cf(2x4y.3x2y2t4y3)dy=x4ytx2y3+y4+cfcxip=x4yhi5ty4tCr(o).-
$0,13f(B) - f( A) = r( 1) : ( l , I )fan ) - FCO . ) :C
iitiiti's
.
.Co - on )1=0
-
7. Evaluate the line integral lnC
xy x dy where C is the arc of the parabola
2y x from (1, 1) to (3, 9). 8. Use Green’s Theorem to evaluate
the line integral
22 cosxC
y e dx x y dy along the positively oriented curve C where C
is
the boundary of the region enclosed by the parabolas 2y x and 2x
y . 9. Show that the vector field 2 2 2( , , ) 2 2 3F x y z xz y xy
x zi j k is
conservative and evaluate C
F dr where C: 2, 1, 2 1x t y t z t for
0 1t .
E- < xy.tn >
fFidr-fxydxtIndydr.cox.dy7FCxkcxsx3ltXE3zg.d.2x7dxI-cx.x.1nxz1.32xox-oy.fy3t2xlnxrxfffEoxoED.ffl2ihoydx-fStdydxFi2xtC0sy2cc0x2dFzif-x-2b.ff1.dxdyooEnWExtuioy.x1var.Bcurvescur1f-oaf7fdriFlBtftHsu.faces-2WirH-42.Ltyli2zttl7F-7fCur1E-oifConservatrvelsowefor.7xF-ffxHoyFzf-c0il2x-a.zy-zyyoV2xztuizxyx2t3Ef@Xzty3dXf2xydyfxtt3z3dZfE.or-4YsYdnn.n
,zz+×y2+c|. .xy2.ie/=x2ztz3tcrios:A-Coil : '
>÷.f=×2zt×y2+ZtC
.
. f( 1,2 ,1 ) . FLQI .
-11=(1,21+1.4+13)
- ⇐-0 ' . . . ) 'T
-
10. Evaluate the surface integral 2 2S
x y dS where S is the surface z xy
inside 2 2 4x y for 0, 0x y . 11. Evaluate
S
curl F ndS where 2, ,F x y z x y xyzi j k and S is part of
the
paraboloid 2 24z x y with 0z . Use the upward unit normal
vector. 12. Find the surface area of the cap cut from the
paraboloid 2 2 3y z x by the
plane 1x .
2 fix
FK ,y)=< x ,y,xy >
§kityIs=ffHty[xFy±ydI0 0
typifies;Psri*=Aix , ' > '¥- ffrtrroroods.tt#r.HFEtxI$E,
°°biretta
text . =P tzlshitrou
Using Stokes 'thm→ the boundary Curve is z=o=4-Xfy2⇒ 4=×4y2→
Flo)=< Icosoirsina , 0 > ⇐ CCWF=< IWSO , 4sin2O , O >
dI=< - 25inch 200,07
Eidra -
Hsinocosotssinocoso
So ffdr = f*( . 4s , nocosotssirioasodo= ¥6520102
" tesinolo't
=o
Surface area - ftds'
rty,⇒=CHy#HP> tyttzztx
Try =L 254 , 1,0 >i=×
Isictzz -0,1 > So 1=55+52-2
rjxrI.cl , . zgy ,- zgz >
3=5+2'
B Tyr2h53dstlryxr .IM#ty4aF:fffEtaIdzdy:ffftEr2rdrooA
- # 0 0
-
s :) arc lengthds=HF' H "dtftp.ddgysur.ruuam
FiFd5=Fidr Fd5=Fjf¥µ ,, .lk 'Hdt=dr±dsvs'±dFy#Fds = ' Hdt '
ds = HFUXTHDA⇐
adS=Fa±in ,jHFu×FvlldA=FuXrvdA
fidsflux throughftnds s Surfaces ffdeifeids fifinds Fwxinc C s
direction of
normal
