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Universities of Canada in Egypt
Prince Edward Island, Cairo Campus
Faculty of Mathematics and Computational Science
Dr. Karim Hammam Seleim
Assignment 5 (Bonus)
Math 2910 Fall 2024- Multivariable and Vector Calculus
Due Date: January 2, 2025
Student Name:
Student ID:
Instructions:
- Provide your answers on these pages.
- Answer all questions clearly, showing all relevant steps.
- Submit your assignment on or before the due date.
- Ensure your work is well-organized and neatly presented.
Jasmine
Taha
Sketch the vector field F
1
2
j
- F(x, y) = 2i j
- F(x, y) = i +
1
2
yj
(
(1 ,
=
Y
&
S B
↑
& S
2 ↓
2
3
↓
s
(c B
&
↓*
↑
a
I
D S
S
&
10 .
( ,
) ( ,
(03)
( ,
- ( , =3)
·
X
a
Let F(x, y) = (3x
2
2
) i + 2xyj and let C be the curve shown.
R
C
F · dr directly.
- (b) Show that F is conservative and find a function f such that F = rf.
- (c) Evaluate
R
C
F · dr using Theorem 2.
R
C
F · dr by first replacing C by a simpler curve that has the same
initial and terminal points.
a)
y
=
Nxz
dy
=
Fdr : 3x +y
21dx + 2xydy
123xy2dx
zxy y
=
923x
(x)dx
=
J?2xdx-2xdx
=
Spdx
=
x
= 4(
= 16
b) MXF =
0
#xF
T f =
(o
(
2y)k
= 0
I 3xiy
?
2xy
o
F= Yf
fx
3x
2
y
= fy
= 2xy
f= x
3
xy f = xy2 + x
F
Yf F = X 3 + y
C so Fis conservative
ScF-dr
= f(
,
f(-
,
=
(2)
o
c)
o
c)
= 878 = IS
d) y
= 0
123x2dx
=
x
=
8 +
8 =
Evaluate the surface Integral
RR
S
xdS, S is the surface y = x
2
- 4z, 0 x 1 , 0 z 1
RR
S
xzdS, S is the boundary of the region enclosed by the cylinder y
2
2
= 9 and
the planes x = 0 and x + y = 5
/SXds ,
y
= x
, Ofely0z
Also Yx
= 2X
yz
= Y
ds = 1
ex ydA
= NdA =+ 44
16 dA =
FX
+xdzdy
S
de U = 17
"
duz SX
U
17 + 16 = 21 ,
17
=
guduu
y
= Scost z = using 0:fe
OXE5-y
= 5-3s
W
rX=
T
I
=
1-3cose-3sineh)=
(sin)
3
O>sing 3 cost
=
ghit - (ssinfsddo
= Pins-scsin
u = 5-3cosE
duz 3 sing
= 2
du
= 2
= o
next ,
we let
yercost and zersing ofres 0 set
d
In
(5-y)z
=
gett
93
sing-Frossin
drd
n
=
[sinf-cossin
.
45sing-are
costing do
=
g
U = Sing
myscose-stir
cust sing d du= coso
Escosi
=
j8u
du = 0
Rius-rus = o
=
- Use Stokes’ Theorem to evaluate
RR
S
curl F · dS.
- F(x, y, z) = x
2
sin zi + y
2
j + xyk, S is the part of the paraboloid z = 1 x
2
y
2
that lies above the xy-plane, oriented upward
- F(x, y, z) = ze
y
i + x cos yj + xz sin yk, S is the hemisphere x
2
2
2
=
16 , y 0, oriented in the direction of the positive y-axis
- A disk D, a hemisphere H, and a portion P of a paraboloid are shown. Suppose
F is a vector field on R
3
whose components have continuous partial derivatives.
Explain why this statement is true:
Z Z
D
curl F · dS =
Z Z
H
curl F · dS =
Z Z
P
curl F · dS
Explanation
:
The
integral
depends
on the boundary
of
the surfaces till it integrated
in the same
boundary
limits ,
for the corl vector.
The surface integral
of F will represent
the amount of fluid through
the surface if
the vector field represents
the flow
j
M
S
,
fodr
If ,
curlf.ds curlf =
/
curl F = Xi
X'sinz y xy
XIp . ul = pcose ,
y (p ,
Kl
psink and z(p ,
=1-p with
OIP11 and OCKE2TT
garf !
xit (xcose-ylj
· (xi
yj + zh) (apcoshi
aprsinny
ple)
=
jarf!
p3(2pcos'
sin (p2-9)+
cost sink+ 24sin3Hdpdk
g
cos Udk = 0
siricos udk = 0
,
and
Jacosksinld
--S
,
f-dr= O
r(t) =< Using, 0 ,cost
Osteat
Jucost, o
, -usint)
↓ (t)
<4 cost ,
0 ,
=
gar
cost (4cost + Of (Usintycost)odt
Icosdt
fitcos
de
=
8/t
S
= (an
= 18
Verify that the Divergence Theorem is true for the vector field F on the region
E.
- F(x, y, z) = 3 xi + xyj + 2 xzk, E is the cube bounded by the planes x = 0 , x =
1 , y = 0, y = 1 , z = 0, and z = 1
- F(x, y, z) = y
2
z
3
i+ 2 yzj+ 4 z
2
k, E is the solid enclosed by the paraboloid z = x
2
+y
2
and the plane z = 9
- F(x, y, z) = hz, y, xi, E is the solid ball x
2
2
2
=1 6
1) divf
=
(
)
+(2xz
2x = 3(X
ISS
divfdV
=
S ! S
. % :
3Hdzdydx
=
If
dudyax
= z
=
xy
yl'
=
!
X
(
x))
. 3 =
(
=
. 3 =
1
JS
.
F. ds : SS
,,
Fonds =
JS, ,
sirtstazk
: ids =
SS ,
asS = 35 .
= 3
SSsafds
= ISsefreds
=
/SxdS
=
SS
:
X dude =
[
%
=
E
Mz
= R
S
f
. dS = SSsy(3xitxyj
2xk)
fg2dS
=
%Saddy
: [x2]
=
SSsyF
(S
oitoj
ohlids = 0
JssF-dS
=
(f (3xi
Ojt2xzk)
jds = o Sjds-fS
Exityj
+Oh. uds
fS0ds
= 0
[SsF-ds
=
3 retroroto
Ifsfods
:
If
,
dirfdr
div F= Of 2z
=
10z
doward
orientation
SSSedivFd
=
JSS10zdV
=
Jans ja
18zrdz
drd
=
She
lOrzdz =
IDre] 5r(s-ri]
=
40r-Grdr
= 5
=
(idE
·
[Erd
=
2S)
= 2
:
/Ss
,
Fods
SSp(-(y2z3)(2x)
(ayzt(2y)
+ (4z4)]
dA
=
[Sp(axy
City
Pyrocity)
Play 2)2]dA
=
Cars
Costsin28-r
Pro
-grigs
(rsincos
Ssinf-4rS(drdO
="SincESE-Er]
=
gat
3294sincos +486sin28-486]d
3486(sin20)
486
= of
486(I-0)-486(2H)
= - 486
Continued
(next pages
