Calculus Stewart ET 5e 0534393217;16. Vector Calculus; 16.3 The Fundamental Theorem for Line Integrals
1. C appears to be a smooth curve, and since Theorem 2 says that the value of
C
f is continuous, we know f is differentiable. Then
f dr is simply the difference of the values of f at the terminal
and initial points of C . From the graph, this is 50 10=40. 2. C is represented by the vector function r(t)=(t +1) i+(t +t) j , 0 t Since 3t +1 0 , we have r (t) 0 , thus C is a smooth curve. differentiable, so by Theorem 2 we have
C 2 / 2 3
1 , so r (t)=2t i+(3t +1) j .
/
2
f is continuous, and hence f is
f dr= f (r(1)) f (r(0))= f (2, 2) f (1, 0)=9 3=6 .
2
3. (6x+5y)/ y=5= (5x+4y)/ x and the domain of F is R which is open and simply connected, so by Theorem 6 F is conservative. Thus, there exists a function f such that f =F , that is, f (x, y)=6x+5y
x
and f (x, y)=5x+4y . But f (x, y)=6x+5y implies f (x,y)=3x +5xy+g(y) and differentiating both sides
y x
2
of this equation with respect to y gives f (x, y)=5x+g (y) . Thus 5x+4y=5x+g (y) so g (y)=4y and
y
/
/
/
g(y)=2y +K where K is a constant. Hence f (x, y)=3x +5xy+2y +K is a potential function for F . 4. (x +4xy)/ y=4x , (4xy y )/ x=4y. Since these are not equal, F is not conservative. 5. (xe )/ y=xe , (ye )/ x=ye . Since these are not equal, F is not conservative. 6. (e )/ y=e = (xe )/ x and the domain of F is R . Hence F is conservative so there exists a function f such that
y / y y y y 2 y y x x 3 3
2
2
2
f =F . Then f (x, y)=e implies f (x, y)=xe +g(y) and f (x, y)=xe +g (y) . But
x y
y
y
y
/
f (x, y)=xe so g (y)=0
g(y)=K . Then f (x, y)=xe +K is a potential function for F.
2 2
y
7. (2xcos y ycos x)/ y= 2xsin y cos x= ( x sin y sin x)/ x and the domain of F is R . Hence F is conservative so there exists a function f such that f =F . Then f (x, y)=2xcos y ycos x implies
x
f (x, y)=x cos y ysin x+g(y) and f (x, y)= x sin y sin x+g (y) . But f (x, y)= x sin y sin x so
y y
2
2
/
2
g (y)=0
/
g(y)=K . Then f (x, y)=x cos y ysin x+K is a potential function for F .
2
2
8. (1+2xy+ln x)/ y=2x= (x )/ x and the domain of F is { (x, y) | x>0 } which is open and simply connected. Hence F is conservative, so there exists a function f such that f =F . Then f (x, y)=1+2xy+ln x implies f (x, y)=x+x y+xln x x+g(y) and f (x, y)=x +g (y) . But f (x, y)=x so g (y)=0
x / y y 2 2 / 2 2
g(y)=K . Then f (x,y)=x y+xln x+K is a potential function for F .
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Stewart Calculus ET 5e 0534393217;16. Vector Calculus; 16.3 The Fundamental Theorem for Line Integrals
9. (ye +sin y)/ y=e +cos y= (e +xcos y)/ x and the domain of F is R . Hence F is conservative so there exists a function f such that
x / y
x
x
x
2
f =F . Then f (x, y)=ye +sin y implies f (x, y)=ye +xsin y+g(y)
x x x y
x
x
and f (x, y)=e +xcos y+g (y) . But f (x, y)=e +xcos y so g(y)=K and f (x, y)=ye +xsin y+K is a potential function for F . (xycosh xy+sinh xy) 2 (x cosh xy) 2 =x ysinh xy+xcosh xy+xcosh xy=x ysinh xy+2xcosh xy= 10. and y x the domain of F is R . Thus F is conservative, so there exists a function f such that f (x, y)=xycosh xy+sinh xy implies f (x, y)=xsinh xy+g(y)
x 2 y 2 / y 2 2
f =F . Then
f (x, y)=x cosh xy+g (y) . But
f (x, y)=x cosh xy so g(y)=K and f (x, y)=xsinh xy+K is a potential function for F . 11. (a) F has continuous first order partial derivatives and y x open and simply connected. Thus, F is conservative by Theorem 6. Then we know that the line integral of F is independent of path; in particular, the value of
C
2xy=2x=
(x ) on R , which is
2
2
F dr depends only on the endpoints
C
of C . Since all three curves have the same initial and terminal points, value for each curve. (b) We first find a potential function f , so that
x
F dr will have the same
2
f =F . We know f (x, y)=2xy and f (x, y)=x .
x y 2
Integrating f (x, y) with respect to x , we have f (x, y)=x y+g(y) . Differentiating both sides with respect to y gives f (x, y)=x +g (y) , so we must have x +g (y)=x
y 2 2 / 2 / 2
g (y)=0
/
g(y)=K , a constant.
Thus f (x, y)=x y+K . All three curves start at (1, 2) and end at (3, 2) , so by Theorem 2,
C
F dr= f (3, 2) f (1, 2)=18 2=16 for each curve.
/
12. (a) f (x, y)=y implies f (x, y)=xy+g(y) and f (x, y)=x+g (y) . But f (x, y)=x+2y so
x y y
g (y)=2y (b)
C
/
g(y)=y +K . We can take K=0 , so f (x, y)=xy+y .
2
2
F dr= f (2, 1) f (0, 1)=3 1=2 . 1 4 4 4 3 / 4 3 x y +g(y) and f (x, y)=x y +g (y) . But f (x, y)=x y so y y 4 1 4 4 g(y)=K , a constant. We can take K=0 , so f (x, y)= x y . 4
3 4
2
13. (a) f (x, y)=x y implies f (x, y)=
x
g (y)=0
/
Stewart Calculus ET 5e 0534393217;16. Vector Calculus; 16.3 The Fundamental Theorem for Line Integrals
(b) The initial point of C is r(0)=(0, 1) and the terminal point is r(1)=(1, 2) , so
C
F dr= f (1, 2) f (0, 1)=4 0=4 .
2 2 2
14. (a) f (x, y)=y /(1+x ) implies f (x, y)=y arctanx+g(y)
x
f (x, y)=2yarctanx+g (y) . But
y 2
/
f (x, y)=2yarctanx so g (y)=0
y
/
g(y)=K . We can take K=0 , so f (x, y)=y arctanx.
(b) The initial point of C is r(0)=(0, 0) and the terminal point is r(1)=(1, 2) , so
C
F d r= f (1, 2) f (0, 0)=4arctan1 0=4
4
= .
15. (a) f (x, y, z)=yz implies f (x, y, z)=xyz+g(y, z) and so f (x, y, z)=xz+g (y, z) . But f (x, y, z)=xz
x y
so g (y, z)=0
y
g(y, z)=h(z) . Thus f (x, y, z)=xyz+h(z) and f (x, y, z)=xy+h (z) . But f (x, y, z)=xy+2z
z z
y /
y
, so h (z)=2z (b)
C
/
h(z)=z +K . Hence f (x, y, z)=xyz+z (taking K=0 ).
2
2
F dr= f (4, 6, 3) f (1, 0, 2)=81 4=77 .
2 2 2
16. (a) f (x, y, z)=2xz+y implies f (x, y, z)=x z+xy +g(y, z) and so f (x, y, z)=2xy+g (y, z) . But
x y y
f (x, y, z)=2xy so g (y, z)=0
y y 2 2 / z
g(y, z)=h(z) . Thus f (x, y, z)=x z+xy +h(z) and f (x, y, z)=x +h (z) .
z 2
2
2
2
/
But f (x, y, z)=x +3z , so h (z)=3z
h(z)=z +K . Hence f (x, y, z)=x z+xy +z (taking K=0 ).
3
2
2
3
(b) t=0 corresponds to the point (0, 1, 1) and t=1 corresponds to (1, 2, 1) , so
C
F dr= f (1, 2, 1) f (0, 1, 1)=6 ( 1)=7 .
2 2
17. (a) f (x, y, z)=y cos z implies f (x, y, z)=xy cos z+g(y, z) and so f (x, y, z)=2xycos z+g (y, z) . But
x y y
f (x, y, z)=2xycos z so g (y, z)=0
y y 2 / z z
g(y, z)=h(z) . Thus f (x, y, z)=xy cos z+h(z) and
2 /
2
f (x, y, z)= xy sin z+h (z) . But f (x, y, z)= xy sin z , so h (z)=0 (taking K=0 ). (b) r(0)= 0, 0, 0 , r( )=
y 2 2
h(z)=K . Hence f (x, y, z)=xy cos z
2
, 0,
so
C y
F dr= f ( , 0, ) f (0, 0, 0)=0 0=0 .
y y
18. (a) f (x, y, z)=e implies f (x, y, z)=xe +g(y, z) and so f (x, y, z)=xe +g (y, z) . But f (x, y, z)=xe
x y y y
so g (y, z)=0
y
g(y, z)=h(z) . Thus f (x, y, z)=xe +h(z) and f (x, y, z)=0+h (z) . But f (x, y, z)=(z+1)e
z z
y
/
z
, so
3
Stewart Calculus ET 5e 0534393217;16. Vector Calculus; 16.3 The Fundamental Theorem for Line (z)=(z+1)e
/
z
h(z)=ze Integrals
h +K (using integration by parts). Hence f (x, y, z)=xe +ze (taking K=0 ).
C
z
y
z
(b) r(0)= 0, 0, 0 , r(1)= 1, 1, 1 so
2
F dr= f (1, 1, 1) f (0, 0, 0)=2e 0=2e . f =F
19. Here F(x, y)=tan yi+xsec y j . Then f (x, y)=xtan y is a potential function for F , that is, so F is conservative and thus its line integral is independent of path. Hence
C
tan ydx+xsec ydy=
2
C x
F d r= f
x
2,
4
f (1, 0)=2tan
x
4
tan 0=2 . f =F so
20. Here F(x, y)=(1 ye ) i+e j . Then f (x, y)=x+ye is a potential function for F , that is, F is conservative and thus its line integral is independent of path. Hence
C
(1 ye )dx+e dy=
3/2
x
x
C
F d r= f (1, 2) f (0, 1)=(1+2e ) 1=2/e .
C
1
21. F(x, y)=2y
i+3x y j , W =
F d r . Since (2y )/ y=3 y = (3x y )/ x , there exists a
3/2 x
3/2
function f such that
y
f =F . In fact, f (x, y)=2y
/
f (x, y)=2xy +g(y) f (x, y)=2xy
3/2
But f (x, y)=3x y so g (y)=0 or g(y)=K . We can take K=0 W=
C
y 3/2
f (x, y)=3xy +g (y) . . Thus
1/2
/
F d r= f (2, 4) f (1, 1)=2(2)(8) 2(1)=30 . y x
2 2
22. F(x, y)=
2y i j ,W= x
C
F dr . Since
2 2
y y x
2 2
=
2
2y x
2
=
x
2y x
y
, there exists a
/
function f such that
f =F . In fact, f =y /x
x 2 2
f (x, y)= y /x+g(y)
f = 2y/x+g (y)
g (y)=0 , so
/
we can take f (x, y)= y /x as a potential function for F . Thus W=
C
F dr= f (4, 2) f (1, 1)= [( 2) /4]+(1/1)=0 .
23. We know that if the vector field (call it F ) is conservative, then around any closed path C ,
C
F dr=0 . But take C to be some circle centered at the origin, oriented counterclockwise. All of the
field vectors along C oppose motion along C , so the integral around C will be negative. Therefore the field is not conservative. 24.
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Stewart Calculus ET 5e 0534393217;16. Vector Calculus; 16.3 The Fundamental Theorem for Line Integrals
From the graph, it appears that F is conservative, since around all closed paths, the number and size of the field vectors pointing in directions similar to that of the path seem to be roughly the same as the number and size of the vectors pointing in the opposite direction. To check, we calculate y x Thus F is conservative, by Theorem 6. (2xy+sin y)=2x+cos y , (x +xcos y)=2x+cos y
2
25. From the graph, it appears that F is not conservative. For example, any closed curve containing the point (2, 1) seems to have many field vectors pointing counterclockwise along it, and none pointing clockwise. So along this path the integral F dr 0 . To confirm our guess, we calculate
x 2y y 1+x +y x 2 x
2 2 2 2
=(x 2y)
y (1+x +y )
2 2 3/2
2 1+x +y + 1
2 2 2 2
=
2 2x xy (1+x +y )
2 2 3/2 2
2
,
=(x 2)
x
2 2 3/2
=
1+y +2x
2 2 3/2
.
(1+x +y ) (1+x +y ) 1+x +y 1+x +y These are not equal, so the field is not conservative, by Theorem 5. 26. f (x, y)=cos (x 2y) i 2cos (x 2y) j
C
(a) We use Theorem 2:
F dr=
1
C
f dr= f (r(b)) f (r(a)) where C starts at t=a and ends at t=b .
1
1
So because f (0, 0)=sin 0=0 and f ( , )=sin ( (0, 0) to ( , ) ; that is, r(t)= t i+ t j , 0 t (b) From (a),
C
2
2 )=0 , one possible curve C is the straight line from
1
1. 2 ,0 =1 , one possible
F dr= f (r(b)) f (r(a)) . So because f (0, 0)=sin 0=0 and f 2 ti , 0 t 1 , the straight line from (0, 0) to 2 ,0 .
curve C is r(t)=
2
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Stewart Calculus ET 5e 0534393217;16. Vector Calculus; 16.3 The Fundamental Theorem for Line Integrals
27. Since F is conservative, there exists a function f such that F=
z
f , that is, P= f , Q= f , and
x y
R= f . Since P , Q and R have continuous first order partial derivatives, Clairaut's Theorem says that P/ y= f = f = Q/ x , P/ z= f = f = R/ x , and Q/ z= f = f = R/ y .
xy yx xz zx yz zy
28. Here F(x, y, z)=yi+x j+xyz k . Then using the notation of Exercise 27, P/ z=0 while R/ x=yz . Since these aren't equal, F is not conservative. Thus by Theorem 4, the line integral of F is not independent of path. 29. D={ (x, y) | x>0, y>0 }= the first quadrant (excluding the axes). (a) D is open because around every point in D we can put a disk that lies in D . (b) D is connected because the straight line segment joining any two points in D lies in D . (c) D is simply connected because it's connected and has no holes. 30. D={ (x, y) | x 0 } consists of all points in the xy plane except for those on the y axis. (a) D is open. (b) Points on opposite sides of the y axis cannot be joined by a path that lies in D , so D is not connected. (c) D is not simply connected because it is not connected. 31. D={ (x, y) | 1<x +y <4 }= the annular region between the circles with center (0, 0) and radii 1 and 2. (a) D is open. (b) D is connected. (c) D is not simply connected. For example, x +y =(1.5) is simple and closed and lies within D but encloses points that are not in D . (Or we can say, D has a hole, so is not simply connected.) 32. D={ (x, y) | x +y
2 2 2 2 2 2 2
1 or 4 x +y
2
2 2
9 }= the points on or inside the circle x +y =1 , together with
2 2 2
2
2
the points on or between the circles x +y =4 and x +y =9 . (a) D is not open because, for instance, no disk with center (0, 2) lies entirely within D . (b) D is not connected because, for example, (0, 0) and (0, 2.5) lie in D but cannot be joined by a path that lies entirely in D . (c) D is not simply connected because, for example, x +y =9 is a simple closed curve in D but encloses points that are not in D . P Q P Q x y x y x 33. (a) P= , and Q= , . Thus = . = = 2 2 2 2 2 22 2 22 y x y x x +y x +y (x +y ) (x +y ) (b) C : x=cos t , y=sin t , 0 t , C : x=cos t , y=sin t , t=2 to t= . Then y
1 2
6
2
2
2
2
2
2
Stewart Calculus ET 5e 0534393217;16. Vector Calculus; 16.3 The Fundamental Theorem for Line Integrals
1 2 cos t+sin t Since these aren't equal, the line integral of F isn't independent of path. (Or notice that
C
F dr=
( sin t)( sin t)+(cos t)(cos t)
0 2 2
dt=
0
dt= and
C
F dr=
2
dt=
C
F dr=
3
2 0
dt=2 where C is the circle x +y =1 , and apply the contrapositive of Theorem 3). This
3 2
2
2
doesn't contradict Theorem 6, since the domain of F , which is R except the origin, isn't simply connected. 34. (a) Here F(r)=cr/|r| and r=x i+y j+z k . Then f (r)= c/|r| is a potential function for F , that is, f =F . (See the discussion of gradient fields in Section 17.1). Hence F is conservative and its line integral is independent of path. Let P =(x , y , z ) and P =(x , y , z ) .
1 1 1 1 2 2 2 2 3
W=
C
F dr= f (P ) f (P )=
2 1
c (x +y +z )
2 2 2 2 2 2 1/2
+
c (x +y +z )
1 1 1 2 2 2 1/2
=c
1 d
1
1 d
.
2
(b) In this case, c= (mMG) 1 1 W = mMG 8 8 1.52 10 1.47 10 = (5.97 1024)(1.99 1030)(6.67 10 (c) In this case, c= qQ 1 1 W = qQ 12 10 5 10
11
)( 2.2377 10
10
) 1.77 10 J
35
13
=(8.985 10 )(1)( 1.6 10
10
19
)( 10 ) 1.4 10 J.
12
4
7