Vector Analysis Versus Vector Calculus (Instructor Solution Manual, Solutions)

71 Pages·2012·1.655 MB·English

by  Antonio Galbis

ANTONIO GALBIS AND MANUEL
MAESTRE
VECTOR ANALYSIS VERSUS
VECTOR CALCULUS
Springer
Solutions to exercises
SolvedexercisesofChapter1
1.3.1. Findthedivergenceofthevectorfields
(1)F(x,y)=(sin(x), ex y).
−
(2)F(x,y,z)= sin(y), cos(z), z3 .
� �
Solution:(1)F=(f ,f )where
1 2
f (x,y)=sin(x); f (x,y)=ex y.
1 2 −
Thus
∂f ∂f
1 2
DivF= +
∂x ∂y
=cosx ex y.
−
−
(2) Let us put f (x,y,z)=sin(y), f (x,y,z)=cos(z) and f (x,y,z)=z3. Then the
1 2 3
divergenceofFis
∂f ∂f ∂f
1 2 3
DivF= + +
∂x ∂y ∂z
=3z2.�
1.3.2. Findthedivergenceandthecurlatthepoint(1,1,0)forthevectorfield
F(x,y,z)=(xyz, y,z).
1
2 Solutionstoexercises
Solution:Thecomponentsofthevectorfieldare
f (x,y,z)=xyz, f (x,y,z)=y, f (x,y,z)=z.
1 2 3
Then
∂f ∂f ∂f
1 2 3
DivF= + +
∂x ∂y ∂z
=yz+2.
Hence,
DivF(1,1,0)=2.
Wenowcalculatethecurlas
e e e
1 2 3
CurlF= ∂ ∂ ∂ =(0,xy, xz).
� ∂x ∂y ∂z� −
�xyz y z �
� �
� �
� �
Inparticular, � �
(CurlF)(1,1,0)=(0,1,0).�
1.3.3. Checkthat
Div(CurlF)=0
forthevectorfield
F(x,y,z)= x2z, x, 2yz
� �
Solution:Thecomponentsofthevectorfieldare
f (x,y,z)=x2z, f (x,y,z)=x, f (x,y,z)=2yz.
1 2 3
Hence
e e e
1 2 3
CurlF= ∂ ∂ ∂ =(2z,x2,1).
� ∂x ∂y ∂z �
�x2z x 2yz�
� �
� �
That is, the components of Cur�l F = (g ,g� ,g ) are given by g (x,y,z) = 2z,
� 1 �2 3 1
g (x,y,z)=x2,g (x,y,z)=1.Consequently,
2 3
∂g ∂g ∂g
1 2 3
Div(CurlF)= + +
∂x ∂y ∂z
=0. �
1.3.4. LetF:IR3 IR3 beavectorfieldandletg:IR3 IRbeascalarfunction,
bothofclassC1on→Rn.Checkthat →
Curl(gF)=g(CurlF)+(∇∇∇g) F.
×
Solutionstoexercises 3
Solution:IfF=(f ,f ,f )then
1 2 3
e e e
1 2 3
Curl(gF)= ∂ ∂ ∂
� ∂x ∂y ∂z �
�gf gf gf �
� 1 2 3�
� �
� �
=� ∂(gf3) ∂�(gf2),∂(gf1) ∂(gf3),∂(gf2) ∂(gf1)
∂y − ∂z ∂z − ∂x ∂x − ∂y
� �
while
e e e
1 2 3
∇∇∇g F= ∂g ∂g ∂g
× �∂x ∂y ∂z �
� f f f �
� 1 2 3�
� �
� �
� ∂g �∂g ∂g ∂g ∂g ∂g
= f f ,f f ,f f .
3 2 1 3 2 1
∂y− ∂z ∂z − ∂x ∂x− ∂y
� �
Consequently,
∂f ∂f ∂f ∂f ∂f ∂f
Curl(gF) (∇∇∇g F)=g. 3 2, 1 3, 2 1
− × ∂y − ∂z ∂z − ∂x ∂x − ∂y
� �
=gCurlF.�
1.3.5. IfF,G:IR3 IR3arevectorfieldsofclassC1,provethat
→
Div(F G)= CurlF, G F, CurlG .
× � �−� �
Solution: We put F = (f ,f ,f ) and G = (g ,g ,g ). According to Definition
1 2 3 1 2 3
1.2.11,wehave
∂f ∂f ∂f ∂f ∂f ∂f
CurlF, G =g 3 2 +g 1 3 +g 2 1 .
1 2 3
� � ∂y − ∂z ∂z − ∂x ∂x − ∂y
� � � � � �
Analogously,
∂g ∂g ∂g ∂g ∂g ∂g
F, CurlG = f 3 2 +f 1 3 +f 2 1 .
1 2 3
� � ∂y − ∂z ∂z − ∂x ∂x − ∂y
� � � � � �
Moreover,fromDefinition1.1.4itfollowsthat
F G=(h ,h ,h ),
1 2 3
×
where
h = f g f g , h = f g f g , h = f g f g
1 2 3 3 2 2 3 1 1 3 3 1 2 2 1
− − −
andfromDefinition1.2.10wehave
4 Solutionstoexercises
∂h ∂h ∂h
Div(F G)= 1 + 2 + 3.
× ∂x ∂y ∂z
Finally,itisroutine(althoughatediouscalculation)tocheck
∂h ∂h ∂h
1 + 2 + 3 = CurlF, G F, CurlG .
∂x ∂y ∂z � �−� �
�
ThepreviousexerciseissolvedwithfulldetailsinExercise6.7.7usingthetheory
ofdifferentialformspresentedinChapter6.
SolvedexercisesofChapter2
2.8.1 Evaluatethelengthofthepath
π
γγγ:[0, ] IR2,γγγ(t)=(etcos(t),etsin(t)).
2 →
Solution:Wefirstevaluate
γγγ(t)=et(cost sint , sint+cost).
�
−
Hence,
1
γγγ(t) =et (cost sint)2+(sint+cost)2 2
�
� � −
� �
=et√2
andthelengthofthepathisgivenby
π
√2 2 et dt = √2 eπ2 1 .�
0 −
�
� �
2.8.2 Obtainaparameterization,counterclockwise,oftheellipse
x y
( )2+( )2=1; (a,b > 0).
a b
Solution:If(x,y)isapointoftheellipsethenthepoint
x y
( , )
a b
Solutionstoexercises 5
Y
Y
�acost,bsent� �0,1�
X
�Γ Γ
3 2
Γ
1 X
�0,0� �1,0�
(a) (b)
Fig.9.7 (a)Ellipse;(b)triangleofExercise2.8.3.
isontheunitcircle,hence
x y
, = (cost , sint)
a b
� �
forsomet [0,2π[.Consequently,aparameterizationoftheellipseisgivenby
∈
γγγ:[0,2π] IR2; γγγ(t) = (acost , bsint).�
→
2.8.3 Integratethevectorfield
FFF(x,y)=(y2, 2xy)
−
alongthetrianglewithvertex(0,0),(1,0),(0,1),orientedcounterclockwise.
Solution: The triangle is the union of three segments. A parameterization of the
segmentfrom(0,0)to(1,0)is
γγγ :[0,1] IR2; γγγ (t)=(t,0).
1 → 1
Aparameterizationofthesegmentfrom(1,0)to(0,1)is
γγγ :[0,1] IR2; γγγ (t)=(1 t)(1,0)+t(0,1)=(1 t,t).
2 → 2 − −
Now,insteadofthesegmentfrom(0,1)to(0,0)weconsidertheoppositepath(the
segmentfrom(0,0)to(0,1)),whichiseasier.Thisis
γγγ :[0,1] IR2; γγγ (t)=(0,t).
3 → 3
6 Solutionstoexercises
Then, the path along which we have to integrate the vector field FFF is the union of
thefollowingthreepaths:
γγγ=γγγ γγγ ( γγγ )
1∪ 2∪ − 3
and,consequently,
FFF = FFF+ FFF FFF.
γγγ γγγ γγγ − γγγ
� � 1 � 2 � 3
Wenowevaluate
FFF = y2dx 2xydy
γγγ γγγ −
� 1 � 1
1
= 0dt=0
0
�
andalso
F =0.
�γ3
Moreover
FFF = y2dx 2xydy
γγγ γγγ −
� 2 � 2
1
= t2( 1) 2(1 t)t dt
0 − − −
�
� �
1 1
= (t2 2t)dt= 1.
0 − 3−
�
Finally,
1
FFF = 1.�
γγγ 3−
�
2.8.4 (1) Findapathγγγwhosetrajectoryistheintersectionofthecylinderx2+y2=
1 with the plane x+y+z=1 and with the additional property that the initial
(and final) point is (0, 1,2) and the projection onto the plane XY is oriented
−
counterclockwise.
(2) Evaluate
xydx + yzdy xdz.
γγγ −
�
Solution: (1) The intersection of the cylinder with the plane is an ellipse in IR3
whoseprojectionontotheplaneXY istheunitcircle.Consequently,wefirstparam-
eterizetheunitcircleinsuchawaythattheinitialandfinalpointis(0, 1)anditis
−
orientedcounterclockwise.Thatis,weput
π 3π
x = cost ; y = sint ( t ).
−2 ≤ ≤ 2
Solutionstoexercises 7
Fig.9.8 PathofExercise2.8.4.
Now,fromtheequationoftheplanewededuce
z = 1 x y = 1 cost sint.
− − − −
Thepathwearelookingforis
π 3π
γγγ:[ , ] IR3; γγγ(t) = (cost , sint , 1 cost sint).
−2 2 → − −
(2)Wenowdenote
ωωω=xydx + yzdy xdz.
−
Wecouldevaluate
3π
2
ωωω= ωωω(γγγ(t)) γγγ(t) dt,
�
γγγ π
� �−2
� �
butamoreefficientmethodconsistsinmakingaformalsubstitutionintheintegral
x = cost ; y = sint ; z = 1 cost sint
− −
andalso
dx = ( sint)dt ; dy = cost dt ; dz = (sint cost)dt
− −
toobtain
3π
ωωω = 2 2sin2tcost sintcos2t+cos2t dt.
γγγ π − −
� �−2 � �
Moreover,
8 Solutionstoexercises
32πsin2t cost dt= sin3t 32π = 0,
π 3 π
�−2 �−2
32π sint cos2t dt= cos3��t 32π = 0
π − 3 π
�−2 �−2
and �
�
32πcos2t dt = 32π 1+cos(2t) dt = t +sin(2t) 32π = π.
π π 2 2 4 π
�−2 �−2 �−2
Finally, �
�
ωωω = π.�
γγγ
�
2.8.5 Letγγγbea simple path whose trajectoryis the intersection of the coordinate
planes with the portion of the unit sphere in the first octant, oriented according to
thesequence
(1,0,0),(0,1,0),(0,0,1),(1,0,0).
Evaluate
zdx+xdy+ydz.
γγγ
�
Fig.9.9 PathofExercise2.8.5.
Solution:Theportionoftheunitsphereinthefirstoctantis
(x,y,z) IR3; x2+y2+z2=1, x 0, y 0, z 0
{ ∈ ≥ ≥ ≥ }
anditsintersectionwithacoordinateplaneisanarcofcircumference.Inparticular,
theintersectionwiththeplanez=0isthetrajectoryofthepath
π
γγγ :[0, ] IR3; γγγ (t)=(cost,sint,0),
1 2 → 1
Solutionstoexercises 9
whiletheintersectionwiththeplanex=0isthetrajectoryof
π
γγγ :[0, ] IR3; γγγ (t)=(0,cost,sint).
2 2 → 2
Alsotheintersectionwithy=0isthetrajectoryof
π
γγγ :[0, ] IR3; γγγ (t)=(sint,0,cost).
3 2 → 3
Thepathofintegrationis
γγγ=γγγ γγγ γγγ .
1∪ 2∪ 3
Weput
ωωω=zdx+xdy+ydz
andweevaluate,bearinginmindthatz=0onγγγ ,
1
π
ωωω = xdy = 2 cos2t dt
γγγ γγγ 0
� 1 � 1 �
π2 1+cos(2t) π
= dt= .
2 4
0
�
Also,sincex=0onγγγ ,
2
ωωω = ydz = π2 cos2t dt= π.
γγγ γγγ 0 4
� 2 � 2 �
Fromthefactthaty=0alongγγγ weobtain
3
ωωω = zdx = π2 cos2t dt= π.
γγγ γγγ 0 4
� 3 � 3 �
Finally
3π
ωωω= .�
γγγ 4
�
2.8.6 Findapathwhosetrajectoryistheintersectionoftheupperhemisphereofthe
spherewithradius2a
x2+y2+z2=4a2
withthecylinder
x2+(y a)2=a2.
−
Solution:If(x,y,z)isapointonthattrajectorythen(x,y)isonthecircle
x2+(y a)2=a2.
−
Hence,