Name
ID
MATH 311
Exam 3
Section 200
Solutions
1.
Spring 2001
P. Yasskin
1
/20 3
/30
2
/10 4
/40
20 points Let V be the vector space of functions spanned by v 1 t and v 2 t 2 with the inner
product
1
f, g ™ ; fŸt gŸt dt
˜
a.
Ÿ
10 Find cos 2 where 2 is the angle between v 1 and v 2 .
1
v , v 1 ™ ; t t dt 1
3
0
˜ 1
v , v2 ™
cos 2 Pv 1 PPv 2 P
2.
1
v , v 2 ™ ; t 2 t 2 dt 1 5
0
˜ 2
1
4
˜ 1
b.
0
1
3
1
5
1
v , v 2 ™ ; t t 2 dt 1
4
0
˜ 1
15
4
10 Apply the Gram-Schmidt procedure to £v 1 , v 2 ¤ to produce an orthonormal basis £u 1 , u 2 ¤.
v1 t 3
u1 t
Pv 1 P
1
3
1
3
3
2
3 t dt 3 t t2 " 3 t
w2 v 2 " ˜v 2 , u 1 ™u 1 t 2 "
˜v 2 , u 1 ™ ; t
4
4
4
0
1
2
t 2 " 3 t dt 1
˜w 2 , w 2 ™ ;
80
4
0
w
3
2
2
80 t " t 5 Ÿ4t 2 " 3t u2 Pw 2 P
4
Ÿ
10 points In R 5 find the volume of the parallepiped with edges
a Ÿ1, 0, 2, 0, 1 b Ÿ0, 2, 1, 0, "1 c Ÿ"1, 0, 0, 2, 1 A
1
0 2 0 1
0
2 1 0 "1
AA T
"1 0 0 2 1
|AA T | 6 3 " 6 " 6 204
|AA T | 0
"1
1
0 2 0 1
0
2
0
0
2 1 0 "1
2
1
0
0
0
2
1 "1
1
"1 0 0 2 1
V
1
6
1
0
1
6
"1
0 "1
6
204
1
3.
30 points A paraboloid in R 4 with coordinates Ÿw, x, y, z , may be parametrized by
Ÿ
Ÿr, 2 Ÿr cos 2, r sin 2, r 2 , r 2 w, x, y, z R
for 0 t r t 3 and 0 t 2 t 2=.
a. Ÿ10 Find the area of the surface.
e r
e 2
A
det AA T sin 2
2r 2r
"r sin 2 r cos 2 0
e r e r e r e 2
e 2 e r e 2 e 2
AA T J
cos 2
Ÿ
0
c 2 2 s 2 2 4r 2 4r 2 "rc2s2 rs2c2
r2s22 r2c22
"rc2s2 rs2c2
1 8r 2 0
1 8r 2 Ÿr 2 r 1 8r 2
2=
3
3
3/2
Area ;; 1dS ; ; r 1 8r 2 dr d2 2= ; r 1 8r 2 dr 2= 1 Ÿ1 8r 2 24
0
0
0
= ¡125 " 1 ¢ 31=
12
3
b.
Ÿ
P;
Ÿ
2=
0
;
3
1 8r 2 cos 2 2 8r 2 sin 2 2 1 8r 2 r 1 8r 2 dr d2 2= ;
0
0
3
0
1 8r 2
2
rŸ1 8r 2 dr 2= r 2r 4
2
3
0
2= 3 18 39=
2
10 Compute I ;;Ÿxy dw dz " wz dx dy over the surface.
w r cos 2,
I;
2=
0
;
2r
3
Ÿ
0
x r sin 2,
cos 2 "r sin 2
Ÿw, z Ÿr, 2 4.
3
10 Compute P ;; 1 8w2 8x 2 dS over the surface.
1 8w2 8x 2 c.
r2
0
0
y r2,
2r 2 sin 2
z r2
Ÿx, y Ÿr, 2 sin 2 r cos 2
2r
r 3 sin 2Ÿ2r 2 sin 2 " r 3 cos 2Ÿ"2r 2 cos 2 dr d2 ;
0
2=
0
;
3
0
"2r 2 cos 2
6
2r 5 dr d2 2= r
3
3
0
18=
(40 points) The solid paraboloid V at the right
is given by
x 2 y 2 t z t 4.
It’s boundary (denoted by V) has two parts:
The paraboloid P given by z x 2 y 2 for z t 4.
The disk D given by x 2 y 2 t 4 with z 4.
Ÿxz 2 , yz 2 , zŸx 2 y 2 .
Let G
a.
G
.
5 Compute Ÿ G
Ÿxz 2 Ÿyz 2 ŸzŸx 2 y 2 2z 2 x 2 y 2
x
y
z
2
b.
Ÿ
G
dV over the solid paraboloid V.
10 Compute ;;; V
HINT: Use cylindrical coordinates.
G
dV ;
;;; V
2=
0
G
2z 2 r 2
dV r dr d2 dz
4
3
; ; 2 Ÿ2z 2 r 2 r dz dr d2 2= ; 2 z r 2 z 2 r dr
3
0 r
0
r
In cylindrical coordinates:
2
2
4
2
6
2
8
6
2 64 r 2 4 " 2 r r 4
r dr 2= 128 r r 4 " 2 r " r
3
3 2
3
6
3 8
0
64
256
32
416
2=
16 "
=
"
3
3
3
3
2= ;
c.
Ÿ
2
0
dS over the paraboloid P with normal pointing DOWN and OUT.
15 Compute ;; G
Pw
HINT: Parametrize the paraboloid with coordinates Ÿr, 2 .
Ÿr, 2 Ÿr cos 2, r sin 2, r 2 The paraboloid may be parametrized by R
e r • e 2 Ÿ"2r 2 cos 2, "2r 2 sin 2, r e 2 Ÿ"r sin 2, r cos 2, 0 e r Ÿcos 2, sin 2, 2r N
Ÿ2r 2 cos 2, 2r 2 sin 2, "r This normal points UP and IN. So reverse it: N
Ÿxz 2 , yz 2 , zŸx 2 y 2 Ÿr 5 cos 2, r 5 sin 2, r 4 G
dS ; 2= ; 2 Ÿ2r 7 " r 5 dr d2 2= r 8 " r 6
;; G
4
6
0
0
Pw
d.
N
2r 7 " r 5
G
2
2= 64 " 32
3
0
320 =
3
dS over the disk D with normal pointing UP.
5 Compute ;; G
Ÿ Du
HINT: Parametrize the disk with coordinates Ÿr, 2 .
Ÿr, 2 Ÿr cos 2, r sin 2, 4 The disk may be parametrized by R
e r • e 2 Ÿ0, 0, r which points UP.
e r Ÿcos 2, sin 2, 0 e 2 Ÿ"r sin 2, r cos 2, 0 N
Ÿxz 2 , yz 2 , zŸx 2 y 2 Ÿ16r cos 2, 16r sin 2, 4r 2 N
4r 3
G
G
dS ; 2= ; 2 Ÿ4r 3 dr d2 2=¡r 4 ¢ 2 32=
;; G
0
Du
e.
0
0
dS ;; G
dS ;; G
dS
5 Compute ;; G
Ÿ V
Pw
Du
(Note: By Gauss’ Theorem, the answers to (b) and (e) should be equal.)
dS 320 = 32= 416 =
;; G
3
3
V
3