Math 335 - Homework Assignment 7 - Due Oct. 16
Turn in the problems with (*)
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1. Page 501: ' 4, 7, ' 12, 13, ' 16, 17, 21, 22, ' 26, ' 28(extra points)
4. ;
Ÿ=/2,0 Ÿ0,0 cos x cos ydx Ÿ1 " sin x sin y dy,
P, Q
f x cos x cos y, fŸx, y cos x cos y, 1 " sin x sin y
; cos x cos y dx sin x cos y CŸy f y " sin x sin y C U Ÿy 1 " sin x sin y, C U Ÿy 1, CŸy ; 1dy y C,
let C 0
fŸx, y sin x cos y y
So, the line integral ;
i. ;
Ÿ=/2,0 Ÿ=/2,0 Ÿ0,0 cos x cos ydx Ÿ1 " sin x sin y dy is independent of the path.
cos x cos ydx Ÿ1 " sin x sin y dy f =2 , 0 " fŸ0, 0 sin
rŸt ¡x, 0¢, 0 t x t =
ii. C : 0 t x t = , y 0, dy 0 or 2
2
=
2
Ÿ0,0 Ÿ=/2,0 ; Ÿ0,0 cos x cos ydx Ÿ1 " sin x sin y dy =/2
;0
cosŸ0 0 " 0 1
cos xdx 0 1
Ÿx, y 2xy 3 , 3y 2 Ÿx 2 1 , 0
12. F
i. Check if Check if P y Q x :
P y 6xy 2 , Q x 6xy 2
is a gradient field. Find fŸx, y as in ii.
Since P y Q x , F
.
ii. Determine if there exists fŸx, y such that f F
fŸx, y ; 2xy 3 dx x 2 y 3 CŸy f y 3x 2 y 2 C U Ÿy 3y 2 x 2 3y 2 , C U Ÿy 3y 2 , CŸy ; 3y 2 dy y 3 C,
let C 0
fŸx, y x 2 y 3 y 3 C
is a gradient field.
F
Ÿx, y 2e 2y , xe 2y
16. F
i. Check if P y Q x :
P y 4e 2y , Q x e 2y
is not a gradient field.
Since P y p Q x , F
.
ii. Determine if there exists fŸx, y such that f F
fŸx, y ; 2e 2y dx 2e 2y x CŸy U
f y 4e 2y x C U Ÿy xe 2y , C Ÿy "3xe 2y a function of x and y.
is not a gradient field.
So F
Ÿ3,4,1 22. ;
Ÿ2x 1 dx 3y 2 dy 1z dz, P 2x 1, Q 3y 2 , R 1z
Ÿ1,2,1 fŸx, y, z ;Ÿ2x 1 dx CŸy, z x 2 x CŸy, z f y C y Ÿy, z 3y 2 , CŸy, z ; 3y 2 dy CŸz y 3 CŸz ,
fŸx, y, z x 2 x y 3 CŸz 1
f z C U Ÿz 1z , CŸz ;
1 dz C ln|z| C, let C 0.
z
fŸx, y, z x 2 x y 3 ln|z|
Ÿ3,4,1 ; Ÿ1,2,1 Ÿ2x 1 dx 3y 2 dy 1 dz fŸ3, 4, 1 " fŸ1, 2, 1 z
Ÿ3 2 3 4 3 ln 1 " Ÿ1 2 1 2 3 ln 1 66
Ÿx, y, z 2 " e z , 2y " 1, 2 " xe z ,
rŸt t, t 2 , t 3 , from Ÿ"1, 1, "1 to Ÿ2, 4, 8 26. F
dr is independent of path:
Check to see if the line integral ; F
C
fŸx, y, z ;Ÿ2 " e z dx CŸy, z 2x " e z x CŸy, z f y C y Ÿy, z 2y " 1, CŸy, z ;Ÿ2y " 1 dy CŸz y 2 " y CŸz fŸx, y, z 2x " e z x y 2 " y CŸz f z "e z x C U Ÿz 2 " xe z , C U Ÿz 2, CŸz ; 2dz 2z C,
let C 0
fŸx, y, z 2x " e z x y 2 " y 2z
dr is independent of path.
So the line integral ; F
;C
C
dr fŸ2, 4, 8 " fŸ"1, 1, "1 F
Ÿ2Ÿ2 " e 8 Ÿ2 4 2 " 4 2Ÿ8 " Ÿ2Ÿ"1 " e "1 Ÿ"1 1 2 " 1 2Ÿ"1 36 " 2e 8 " e "1 "5926. 283 854
Ÿx, y, z 8xy 3 z, 12x 2 y 2 z, 4x 2 y 3 , rŸt 2 cos t, 2 sin t, t from Ÿ2, 0, 0 to 1, 3 , =
28. F
3
W ; F dr. Check if ; F dr is independent of path:
C
C
fŸx, y, z ; 8xy 3 zdx CŸy, z 4x 2 y 3 z CŸy, z f y 12x 2 y 2 z C y Ÿy, z 12x 2 y 2 z, C y Ÿy, z 0, CŸy, z CŸz , fŸx, y, z 4x 2 y 3 z CŸz f z Ÿx, y, z 4x 2 y 3 C U Ÿz 4x 2 y 3 , C U Ÿz 0, CŸz C, let C 0.
fŸx, y, z 4x 2 y 3 z
dr is independent of path.
So the line integral ; F
C
;C F dr fŸ2, 0, 0 " f 1, 3 , =
3
3 =
2
3
Ÿ4Ÿ2 Ÿ0 Ÿ0 " 4Ÿ1 3
W
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2. Page 509: 11, 12,
2
12. ; ;
x 2 1
"1 "x 2
'
"21. 765 592 37
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18, 19, 20(evaluate the integral exactly), 35, ' 36, ' 40
fŸx, y dydx
2
4
2
2
R : "1 t x t 2, " x t y t x 1
2
-1
0
-0.5
0.5
x1
1.5
2
-2
-4
18. ; ;
x dA, y x 2 1, y 3 " x 2
y
R
x 2 1 3 " x 2 , 2x 2 2
2.5
2
x 1, x o1
2
3"x 2
1
; ;R x dA ; "1 ; x 2 1
x
y
y
;
1
"1
1.5
dydx
1
0.5
2 Ÿ3 " x 2 x " 2 Ÿx 2 1 x dx 0
-1
0
0.2 0.4 x0.6 0.8
1
R : x2 1 t y t 3 " x2, " 1
2
20. ; ; sin =x
y dA, x y , x 0, y 1, y 2
R
y 1, x 1, y 2, x 4 x,
-0.8 -0.6 -0.4 -0.2
2
x
1.8
R : 1 t y t 2, 0 t x t y 2
1.6
1.4
2
2 y
=x
; ;R sin =x
y dA ; 1 ; 0 sin y dxdy
2
2
cos y= " 1
1 "4 3=
; "y
dy
=
2
1
=3
0. 412 961 8
1.2
1
0.8
0.6
0.4
0.2
0
1
2x
3
4
1 1
35. ; ; x 2 1 y 4 dydx
0 x
R : 0 t x t 1, x t y t 1
1 1
1 y
; 0 ; x x 2 1 y 4 dydx ; 0 ; 0 x 2 1 y 4 dxdy
;
1
0
1
9
1
3
4
Ÿ1 y y
2 "
1
18
3
dy
0. 101 579 3
0.8
0.6
0.4
0.2
0
0.2
0.4
x 0.6
0.8
1
R
3
1 2
36. ; ; e "y/x dxdy
0 2y
R : 0 t y t 1, 2y t x t 2,
1 2
2 x/2
0.8
; 0 ; 2y e "y/x dxdy ; 0 ; 0 e "y/x dydx
;
2
"xe
0
" 12
x dx "2e
" 12
0.6
2
0.4
0. 786 938 7
0.2
0
0.5
1x
1.5
2
R
4 2
40. ; ;
0
y
R:
x 3 1 dxdy
y t x t 2, 0 t y t 4,
3
x 3 1 dydx 52
9
2
2 x2
;0 ;0
1
0
0.5
1x
1.5
2
R
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3. Page 514: 25, 26, ' 29, ' 34(extra points)
2 /2
1"y 2
y2
;
26. ;
dxdy
0
y
x2 y2
1
R:0tyt
y
2
2
, ytxt
1 " y2 ,
1 " y 2 , y 2 1 " y 2 , 2y 2 1, y o
0.8
2
2
R : 0 t 2 t =, 0 t r t 1
4
2
2 /2
1"y 2
=/4 1 2
y2
2 rdrd2
;0 ;y
dxdy ; ; r sin
r
0
0
2
2
x y
;
=/4 1 2
r sin 2 2drd2
0
0
;
1 = " 1 0. 0476
24
12
0.6
0.4
0.2
0
0.2
0.4
x 0.6
0.8
1
R
4
1
4"x 2
0
1"x 2
29. ; ;
2
2
4"x
x2
x2
dydx ; ;
dydx
2
2
2
1
0
x y
x y2
2
R 1 : 0 t x t 1,
1"x
2
tyt
4"x
1.8
2
1.6
R 2 : 1 t x t 2, 0 t y t 4 " x 2
R : 0 t 2 t =, 1 t r t 2
2
2
2
1
2
4"x 2
4"x
x2
; 0 ; 1"x 2 2 x 2 dydx ; 1 ; 0
dydx
x y
x2 y2
2
=/2 2 2
2 rdrd2
; ; r cos
2
0
1
r
=/2 2
; ; r cos 2 2drd2 3 = 1. 178 097
0
1
8
1.4
1.2
1
0.8
0.6
0.4
0.2
0
0.2 0.4 0.6 0.8 1x 1.2 1.4 1.6 1.8 2
- R1, - - R2
34. r 2 sin 2, r 2
2
R : 2 sin 2 t r t 2, 0 t 2 t =
2
1
=/2 2
; ;R Ÿx y dA ; 0 ; 2 sin 2 Ÿr cos 2 r sin 2 rdrd2
14
3
"
1
2
=
-2
-1
0
1
2
-1
-2
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4. Page 520: 3, 4, ' 7, 9, 10(extra points),
4. > "2y 2 dx 4xydy ; ; 8ydA
C
'
12, ' 23, ' 25,
'
30
R
the intersection of curves: y x and y "x 2 :
x "x 2, x Ÿ"x 2 2 x 2 " 4x 4
x 2 " 5x 4 0, Ÿx " 4 Ÿx " 1 0, x 1, and 4
the curve C C 1 : C 2 : C 3 :
r 1 Ÿt t, 0 , 0 t t t 1, y 0, dy 0
C1 : r 2 Ÿt t, " t 2 , t : from 2 to 1
C2 : C 3 : r 3 Ÿt t,
t , t : from 1 to 0
2
1.8
1.6
1.4
1.2
1
0.8
0.6
0.4
0.2
0
0.2 0.4 0.6 0.8
1t
1.2 1.4 1.6 1.8
2
C Ÿcounterclockwise , R
> C "2y 2 dx 4xydy ;C ;C ;C
1
0;
2
1
2
3
"2Ÿ"t 2 2 4tŸ"t 2 Ÿ"1 dt ;
0
1
"2t 4t t
1
2 t
dt 10
3
5
1
2"y
; ;R 8ydA ; 0 ; y
2
8ydxdy 10
3
7. > Ÿx 4 " 2y 3 dx Ÿ2x 3 " y 4 dy, C : x 2 y 2 4
C
2
P x 4 " 2y 3 , P y "6y 2
Q 2x 3 " y 4 , Q x 6x 2
1
> C Ÿx 4 " 2y 3 dx Ÿ2x 3 " y 4 dy
; ; Ÿ6x 2 " Ÿ"6y 2 dA
R
,
-2
0
-1
; ; 6Ÿx 2 y 2 dA
1t
2
-1
R
2= 2
; ; 6r 2 rdrd2 48=
0
-2
0
x2 y2 4
rŸt 1 2 cos t, 3 3 sin t
10. > e 2x sin 2ydx e 2x cos 2ydy, C : 9Ÿx " 1 2 4Ÿy " 3 2 36 or C
P e 2x sin 2y,
P y 2e 2x cos 2y
2x
5
2x
Q e cos 2y,
Q x 2e cos 2y
4
> C e 2x sin 2ydx e 2x cos 2ydy
3
; ; Ÿ2e 2x cos 2y " 2e 2x cos 2y dA 0
R
2
1
-1
0
1
2
3
12. > e x dx 2 tan "1 xdy, C : Ÿ0, 0 v Ÿ0, 1 v Ÿ"1, 1 v Ÿ0, 0 C
2
R : "1 t x t 0,
"x t y t 1
0.8
x2
P e , Py 0
2
1 x2
Q 2 tan "1 x, Q x 0
1
> C e dx 2 tan "1 xdy ; "1 ; "x
x2
" ln 2 1
2
0.6
0.4
2 dydx
1 x2
0.2
= 0. 877 649 1
-1
-0.8
-0.6
t
-0.4
-0.2
0
R
23. > Ÿ4x 2 " y 3 dx Ÿx 3 y 2 dy
C
6
P 4x 2 " y 3 , P y "3y 2
2
Q x 3 y 2 , Q x 3x 2
1
Since all P, Q, P y and Q x are continuous on R,
and C is smooth simple closed curve, by Green’s Thm:
> C Ÿ4x 2 " y 3 dx Ÿx 3 y 2 dy
-2
0
-1
; ; Ÿ3x 2 " Ÿ3y 2 dA
R
2= 2
2= 2
0
0
1
-2
45 = 70. 685 83
2
25. >
C
2
-1
; ; 3r 2 rdrd2 ; ; 3r 3 drd2
1
1
1 t x2 y2 t 2
"y 3 dx xy 2 dy
, C : x 2 4y 2 4
2
2 2
Ÿx y 2
Since P and Q are not defined at Ÿ0, 0 2
which is in the ellipse R : x y 2 1,
4
Green’s Thm cannot be applied to this problem.
C:
rŸt 2 cos t, sin t , 0 t t t 2=
1
-2
-1
0
1
2
-1
U
r Ÿt "2 sin t, cos t
-2
>C
"y 3 dx xy 2 dy
2
Ÿx 2 y 2 2=
;0
" sin 3 tŸ"2 sin t dt 2 cos t sin 2 t cos tdt
2
4 cos 2 t sin 2 t
2=
;0
2 sin 2 t
dt =
2
Ÿ1 3 cos 2 t "xy 2 , x 2 y
30. F
P "xy 2 , Q x 2 y
P y "2xy, Q x 2xy
R : 0 t 2 t =, 1 t r t 2
2
> C F dr ; ;R Ÿ2xy " Ÿ"2xy dA
;
;
=/2 2
1.6
1.4
1.2
1
0.8
; 4r 2 cos 2 sin 2rdrd2
0.6
; 1 2r sin 22drd2 15 7. 5
2
0.2
0
1
=/2 2
0
2
1.8
3
0.4
0
0.2 0.4 0.6 0.8
1 1.2 1.4 1.6 1.8 2
7