Table Of ContentANTONIO 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,
