Name
ID
Section
MATH 253
EXAM 3
Fall 1998
Sections 501-503
Solutions
P. Yasskin
Multiple Choice: (7 points each)
1. If
a.
b.
c.
d.
F
"yz sin x, sin x, sin x
e.
F
2. If
then
F yz cos x, y sin x, z sin x
"yz sin x
correctchoice
Ÿ2 " yz sin x
"yz sin x, " sin x, sin x
0, Ÿz " y cos x, Ÿy " z cos x
Ÿyz cos x
x
F
Ÿy sin x
y
Ÿz sin x
z
then
yz cos x, y sin x, z sin x
0, Ÿy " z cos x, Ÿy " z cos x
b. 2Ÿy " z sin x
c. "yz sin x, " sin x, sin x
d. 0, Ÿz " y cos x, Ÿy " z cos x
"yz sin x sin x sin x
•F
Ÿ2
" yz sin x
correctchoice
a.
e. 0
•F
î
F
k
x
y
z
îŸ0
" 0 " F Ÿz cos x " y cos x
k Ÿy cos x " z cos x
yz cos x y sin x z sin x
0, Ÿy " z cos x, Ÿy " z cos x
3. Find a scalar potential for
F
yz cos x, y sin x, z sin x .
a. "yz sin x
b. yz sin x
y2
sin x,
2
y2
d. yz sin x sin x 2
e. Does Not Exist
c.
yz sin x,
z 2 sin x
2
z 2 sin x
2
correctchoice
Since • F p 0, there is no potential.
1
B
ds
; F
4. Compute the line integral
F
of the vector field
A
Ÿy, "x, z
along the
H
helix H parametrized by
rŸt Ÿ3 cos 2, 3 sin 2, 2
B Ÿ"3, 0, 3= .
a. 2= 2 " 18=
b. 2=
c. 3=
2
d. 9= " 27=
correctchoice
2
2
e. 9=
2
between
A
Ÿy, "x, z Ÿ3 sin 2, "3 cos 2, 2
F
v Ÿ"3 sin 2, 3 cos 2, 1
B
3=
3=
3=
;
ds
F
v d2 ;
"9 sin 2 2 " 9 cos 2 2 2 d2 ; Ÿ"9 ; F
A
0
0
0
H
3=
9= 2
22
"92 "27= 2 0
2
r Ÿt
5. Find the total mass of the helix H parametrized by
between
and
A Ÿ3, 0, 0
B Ÿ"3, 0, 3=
> 3 2z.
a. 9Ÿ= = 2
and
Ÿ3, 0, 0
2 d2
Ÿ3 cos 2, 3 sin 2, 2
if the linear mass density is
b. 27Ÿ= = 2
c. 9 10 Ÿ= = 2
correctchoice
10 Ÿ6= 4= 2
e. 6= 4= 2
d.
r and
>
hence v are the same as in #4. So:
3 2z
3 22
M
;
3=
0
Ÿ3 22
>Ÿ4x " 3y dx Ÿ3x " 2y dy
triangle with vertices
Ÿ0, 0 ,
Ÿ0, 3
(HINT: Use Green’s Theorem.)
a. 3
b. 6
c. 9
d. 12
e. 18
correctchoice
;;
triangle
6
1
2
23
Q
" P
y
x
dx dy
10 d2
9 sin 2 2 9 cos 2 2 1
10 32 2 2
3=
0
10
10 Ÿ9= 9= 2
counterclockwise around the edge of the
6. Compute
> Pdx Q dy
|v|
and
Ÿ2, 0
.
;; Ÿ3 " "3 dx dy
6 area of triangle
triangle
18
2
• F dS
;; 7. Compute
F
for the vector field
Ÿ"y, x, z
over the paraboloid
P
for
with normal pointing in and up.
z x2 y2
zt9
(HINT: Use Stokes’ Theorem.)
a. 2=
b. 3=
c. 4=
d. 9=
e. 18=
correctchoice
• F dS
;; By Stokes’,
ds
>F
So we parametrize the boundary circle:
P
P
r Ÿ2
Ÿ3 cos 2, 3 sin 2, 9
2=
• F dS ;
v
F
;; 0
P
d2
F
v Ÿ"3 sin 2, 3 cos 2, 0
2=
;
9 d2 18=
0
Ÿ"y, x, z
Ÿ"3 sin 2, 3 cos 2, 9
8. (25 points) Green’s Theorem states that if R is a nice region in the plane and R is its
boundary curve traversed counterclockwise then
Q
" P dx dy > Pdx Q dy
;;
y
x
R
R
and
Verify Green’s Theorem if
and R is the region inside
Q xy 2
P "x y
2
2
the circle
x y 4.
Q
a. (5 pts) Compute
" P .
(HINT: Use rectangular coordinates.)
y
x
2
Q
" P
y
x
Ÿxy 2 " Ÿ"x 2 y
y
x
Q
" P
y
x
;;
b. (10 pts) Compute
R
y2
x2
dx dy.
(HINT: Switch to polar coordinates and don’t forget the Jacobian.)
;;
R
Q
" P
y
x
dx dy
;;Ÿx 2 y 2 dx dy
R
> Pdx Q dy .
c. (10 pts) Compute
2=
2
0
0
; ; Ÿr 2 r dr d2
4
2= r
4
2
0
8=
(HINT: Parametrize the boundary circle.)
R
r Ÿ2
F
Ÿ2 cos 2, 2 sin 2
Ÿ" x
2
y, xy
2
vŸ2
Ÿ"2 sin 2, 2 cos 2
2
"8 cos 2 sin 2, 8 cos 2 sin 2
2
16 cos 2 sin 2 16 cos 2 2 sin 2 2 32 cos 2 2 sin 2 2 8Ÿ2 sin 2 cos 2
1 " cosŸ42
2
8 sin Ÿ22 8
4 " 4 cosŸ42
2
2=
2=
> Pdx Q dy ; 4 " 4 cosŸ42 d2 42 " sinŸ42 0 8=
F v
R
2
2
2
0
Note: The answers to (b) and (c) are the same.
3
9.
(15 points) The spiral ramp at the right
may be parametrized as
R
Ÿr, 2
Ÿr cos 2, r sin 2, 2
0trt2
for
0 t 2 t 3=.
and
;; x 2 y 2 dS
Compute
over
this spiral ramp.
F
k
sin 2,
0
r cos 2,
1
î
R
r
Ÿcos 2,
R
2
Ÿ"r sin 2,
N
îŸsin 2 " F Ÿcos 2
N
sin 2 2 cos 2 2 r 2
3=
2
0
0
3= 1 2 Ÿ1 r 2
2 3
3/2
;; x 2 y 2 dS
k Ÿr
1 r2
; ; r 1 r 2 dr d2
2
0
=Ÿ5 3/2 " 1
10. (25 points) Gauss’ Theorem states that if V is a solid region and V is its boundary
surface with outward normal then
F dV
;;; dS
;; F
V
V
2
2 3
and V is the solid region above the
Verify Gauss’ Theorem if
Ÿxz , yz , z
F
2
2
paraboloid
below the plane
Notice that V consists of two
z x y
z 4.
parts: a piece of the paraboloid P and a disk D.
a. (7 pts) Compute
F dV.
;;; F
(HINTS: Compute
in rectangular
V
coordinates. Integrate in cylindrical coordinates.)
F
z2
F dV
;;; V
z2
3z 2
2=
2
0
0
5z 2
4
; ; ; 2 5z 2 r dz dr d2
10= 32r 2 " r 8
8
3
r
2
0
2= ;
10= ¡128 " 32 ¢
3
2
0
3
5z r
3
4
zr 2
10= ¡96 ¢
3
2
2= 5 ; ¡64r " r 7 ¢dr
3 0
dr
320=
4
dS.
;; F
b. (8 pts) Compute
(HINT: Here is the parametrization of the
P
paraboloid.)
Ÿr, 2 Ÿr cos 2, r sin 2, r 2
R
î
F
k
R
r
Ÿcos 2,
sin 2,
2r
R
2
Ÿ"r sin 2,
r cos 2,
0
N
k
Ÿr
îŸ"2r 2 cos 2 " F Ÿ" "2r 2 sin 2
F
RŸr, 2
dS
;; F
P
Ÿxz
2
2=
2
0
0
, yz 2 , z 3
Ÿr
N dr d2
; ; F
2=
2
0
0
; ; "r 7 dr d2
5
2
cos 2, "2r 2 sin 2, r
cos 2, r 5 sin 2, r 6
2=
2
0
0
; ;
"2r 7 cos 2 2 " 2r 7 sin 2 2 r 7
2
8
2= " r
8
dS.
;; F
c. (7 pts) Compute
Ÿ"2r
0
"2=Ÿ32
dr d2
"64=
(HINT: You parametrize the disk.)
D
Ÿr, 2
R
Ÿr cos 2, r sin 2, 4
k
F
î
R
r
Ÿcos 2,
R
2
Ÿ"r sin 2,
r cos 2, 0
N
Ÿ0,
0,
F
RŸr, 2
dS
;; F
D
sin 2,
Ÿxz
2
2=
2
0
0
0
r
Ÿ16r cos 2, 16r sin 2, 64
, yz 2 , z 3
N dr d2
; ; F
d. (3 pts) Combine
2=
2
0
0
; ; 64r dr d2
dS
;; F
and
P
2=¡32r 2 ¢ 20
dS.
;; F
to obtain
265=
dS.
;; F
V
D
Be sure to discuss the orientations of the surfaces and to give a formula before
you plug in numbers.
The normal to V must point outward. The normal to P points in and up but should
point down and out. So an extra minus is needed. The normal to D points up
which is correct. So:
dS
;; F
V
" ;; F
dS ;; F
dS
P
" "64= 256=
320=
D
Note: The answers to (a) and (d) are the same.
5