MATH 253 Honors
EXAM 3
Fall 1999
Sections 201-202
Solutions
P. Yasskin
Multiple Choice: (8 points each)
B
ds
; F
1. Compute the line integral
Ÿyz, "xz, z F
of the vector field
A
H
the helix H parametrized by
and
B Ÿ"3, 0, 4= .
a. 20=
b. 26=
c. 26= 2
d. "20=
e. "10= 2
correctchoice
rŸt Ÿ3 cos t, 3 sin t, 4t rŸt Ÿ3 cos t, 3 sin t, 4t vŸt Ÿ"3 sin t, 3 cos t, 4 rŸa Ÿ3 cos a, 3 sin a, 4a A Ÿ3, 0, 0 rŸt F
A Ÿ3, 0, 0 Ÿ12t sin t, "12t cos t, 4t a0
®
rŸb Ÿ3 cos b, 3 sin b, 4b B Ÿ"3, 0, 4= between
along
b=
®
B
ds ; = F
v dt ; = "36t sin 2 t " 36t cos 2 t 16t dt ; = Ÿ"36t 16t dt
; F
A
0
0
H
0
=
=
0
0
; Ÿ"20t dt "10t 2
"10= 2
rŸt Ÿ3 cos t, 3 sin t, 4t between
2. Find the total mass of the helix H parametrized by
2
and
if the linear mass density is
A Ÿ3, 0, 0 B Ÿ"3, 0, 4= >z .
3
a. 16=
b.
c.
d.
e.
3
40= 3
3
80= 3
3
16= 3
80= 3
correctchoice
rŸt Ÿ3 cos t, 3 sin t, 4t vŸt Ÿ"3 sin t, 3 cos t, 4 |v| 9 sin 2 t 9 cos 2 t 16 5
> z 2 16t 2
=
=
=
3
M ; > ds ; > |v|dt ; 16t 2 5 dt 80 t
3
0
0
0
=
0
3
80=
3
1
>Ÿ"xy 2 dx x 2 y dy the triangle whose vertices are
Theorem.
a. 4
b. 8
c. 16
correctchoice
d. 32
e. 64
3. Compute
P "xy 2
Q x2y
counterclockwise around the complete boundary of
Ÿ0, 0 , Ÿ2, 4 and Ÿ0, 4 .
HINT: Use Green’s
x Q " y P 2xy " "2xy 4xy
>Ÿ"xy 2 dx x 2 y dy >ŸPdx Q dy ;; x Q " y Pdx dy
2
4
2
4
0
2x
0
y2x
; ; 4xy dy dx ; 2xy 2
2
dx ; Ÿ2x16 " 2x4x 2 dx
4
3
2
0
1
2
2
0
0
; Ÿ32x " 8x 3 dx 16x 2 " 2x 4
64 " 32 32
0
along the
;Ÿyz dx xz dy xy dz rŸt e sin 4t , cos 5t, ln 1 =t
curve
between t 0 and t =.
scalar potential for F Ÿyz, xz, xy .
correctchoice
a. " ln 2
b. 1 " ln 2
c. "1 " ln 2
d. 1 ln 2
e. "1 ln 2
1
2
4. Compute
HINT: Find a
A rŸ0 Ÿe sin 0 , cos 0, lnŸ1 Ÿ1, 1, 0 B rŸ= e sin 4= , cos 5=, ln 1 =
=
Ÿ1, "1, ln 2 f Ÿyz, xz, xy F
The scalar potential is f xyz since By the Fundamental Theorem of Calculus for Curves:
B
ds
;Ÿyz dx xz dy xy dz ; F
A
f ds fŸB " fŸA Ÿ1 Ÿ"1 Ÿln 2 " Ÿ1 Ÿ1 Ÿ0 " ln 2
; B
A
2
5. (25 points) Stokes’ Theorem states that if S is a surface in 3-space and S is its
boundary curve traversed counterclockwise as seen from the tip of the normal to S
then
•F
dS > F
ds
;; S
S
Verify Stokes’ Theorem if
F Ÿy, "x, x 2 y 2 and S is the cone z x 2 y 2 for z t 2
with normal pointing up and in.
The cone may be parametrized by:
Ÿr, 2 Ÿr cos 2, r sin 2, r R
;;
5a. (16 points) Compute
•F
dS
using the following steps:
S
i
•F
j
k
x y
z
iŸ2y " 0 " jŸ2x " 0 kŸ"1 " 1 Ÿ2y, "2x, "2 y "x x 2 y 2
r Ÿcos 2, sin 2, 1 R
2 Ÿ"r sin 2, r cos 2, 0 R
r • R
2 iŸ"r cos 2 " jŸr sin 2 k r cos 2 2 r sin 2 2
R
N
Ÿ"r cos 2, "r sin 2, r This is oriented correctly as up and in.
•F
;;
S
Ÿr, 2 R
Ÿ2r sin 2, "2r cos 2, "2 •F
•F
dS ;; N
dr d2 ; ;Ÿ"2r 2 sin 2 cos 2 2r 2 sin 2 cos 2 " 2r dr d2
2=
2
2
0
0
0
; ; Ÿ"2r dr d2 2= "r 2
"8=
and S is the
5b (9 points) Recall
F Ÿy, "x, x 2 y 2 2
2
cone
z x y
with normal pointing up and in.
Compute
using the following steps:
> F ds
S
(Remember to check the orientation of the curve.)
rŸ2 Ÿ2 cos 2, 2 sin 2, 2 vŸ2 Ÿ"2 sin 2, 2 cos 2, 0 rŸ2 Ÿ2 sin 2, "2 cos 2, 4 F
ds ; 2= F
v d2 ; 2= "4 sin 2 2 " 4 cos 2 2 d2 ; 2= Ÿ"4 d2 "8=
> F
0
0
0
S
3
6. (25 points) Gauss’ Theorem states that if V is a solid region and V is its boundary
surface with outward normal then
F
dS
dV ;; F
;;; V
V
Verify Gauss’ Theorem if
F Ÿxz, yz, x 2 y 2 and V is the solid hemisphere
0tzt
4 " x2 " y2 .
Notice that V consists of two parts:
z
the hemisphere H:
4 " x2 " y2
x 2 y 2 t 4 with z 0
and a disk D:
F
dV.
;;; 6a. (5 pts) Compute
V
F
2z 2> cos I
2
F
dV ; =/2 ; 2= ; 2 2> cos I > 2 sin I d> d2 dI Ÿ2= sin I
;;; 2
0
0
0
V
6b. (8 pts) Compute
dS.
;; F
=/2
0
2> 4
4
2
8=
0
(HINT: You parametrize the disk.)
D
Ÿr, 2 Ÿr cos 2, r sin 2, 0 R
r Ÿcos 2, sin 2, 0 R
2 Ÿ"r sin 2, r cos 2, 0 R
Ÿ0, 0, r N
R
Ÿr, 2 F
This is upward. Reverse it:
Ÿ0, 0, "r N
Ÿxz, yz, x 2 y 2 Ÿ0, 0, r 2 dS ;; F
N
dr d2 ; 2= ; 2 Ÿ"r 3 dr d2 "2= r 4
;; F
4
0
0
D
D
6c. (9 pts) Compute
dS
;; F
2
"8=
0
over the hemisphere parametrized by
H
ŸI, 2 Ÿ2 sin I cos 2, 2 sin I sin 2, 2 cos I R
I Ÿ2 cos I cos 2, 2 cos I sin 2, "2 sin I R
2 Ÿ"2 sin I sin 2, 2 sin I cos 2, 0 R
i 4 sin 2 I cos 2 " j "4 sin 2 I sin 2 k 4 sin I cos I cos 2 2 sin 2 2
N
4 sin 2 I cos 2, 4 sin 2 I sin 2, 4 sin I cos I
This is outward.
R
ŸI, 2 F
Ÿxz, yz, x 2 y 2 4 sin I cos I cos 2, 4 sin I cos I sin 2, 4 sin 2 I
N
16 sin 3 I cos I cos 2 2 16 sin 3 I cos I sin 2 2 16 sin 3 I cos I 32 sin 3 I cos I
F
dS ;
;; F
H
2=
0
;
=/2
0
32 sin 3 I cos I dI d2 2= 32
sin 4 I
4
=/2
16=
0
4
6d. (3 pts) Combine
dS
;; F
and
H
dS.
;; F
to obtain
dS.
;; F
V
D
Be sure to discuss the orientations of the surfaces (here or above) and give a
formula before you plug in numbers.
D so it points
H points up. We reversed N
The normals must point outward. N
downward.
dS ;; F
dS ;; F
dS 16= " 8= 8=
;; F
V
7.
H
which agrees with (6a).
D
(20 points) The paraboloid at the right is
the graph of the equation
z 4x 2 4y 2 .
It may be parametrized as
Ÿr, 2 Ÿr cos 2, r sin 2, 4r 2 .
R
Find the area of the paraboloid for z t 16.
r Ÿcos 2, sin 2, 8r R
2 Ÿ"r sin 2, r cos 2, 0 R
iŸ"8r 2 cos 2 " jŸ8r 2 sin 2 k r cos 2 2 r sin 2 2
N
N
64r 4 cos 2 2 64r 4 sin 2 2 r 2 Ÿ"8r 2 cos 2, "8r 2 sin 2, r 64r 4 r 2 r 64r 2 1
3/2
2
dr d2 ; 2= ; 2 r 64r 2 1 dr d2 2= 2Ÿ64r 1 A ;; N
3 128
0
0
2
0
= Ÿ257 3/2 " 1 96
5