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What is the value of the line integral_C F·dr?

Resolved Question:

1- Suppose F = F(x,y, z) is a gradient field with F = Ñf , S is a level surface of f, and C is a curve on S. What is
the value of the line integral_C F·dr?

2- Consider the vector field F(x,y) = (2x2 +y2,2xy). Compute
the line integrals_C1 F· dr and Integral_C2 F· dr, where C1 is the curve

r1(t)=(t, t2) for 0t 1 andC2 is the curve r2(t)=(t, t), also
forR0 <= t <= 1.

Integral_C1 F·dr = .
Integral_C2 F·dr =

Can you decide from your answers alone whether or not F is
a conservative vector field? (Y/N)
Is F a conservative vector field? (Y/N)

3- Let F(x,y) = (-yi+xj)/(x2+y2) = P(x,y)i+Q(x,y)j and let C be the circle r (t) = (cost) i+(sint) j, 0 <= t <= 2p.
A.
dQ/dx=

Note: Your answer should be in terms of x and y; e.g. ”3xy - y”
B.
dP/dy =
Note: Your answer should be in terms of x and y; e.g. ”3xy - y”
C.
Integral_C F·dr =
Note: Your answer should be a number

3- Suppose C is any curve from (0,0,0) to (1,1,1) and
F(x,y, z) = (3z+4y) i+(2z+4x) j+(2y+3x)k. Compute the
line integral_C (F·dr).

4- Let F = (5xy,2y2) be a vector field in the plane, and C the
path y = 5x2 joining (0,0) to (1,5) in the plane.
Evaluate Integral_C (F·ds)

6- Determine whether the given set is open, connected,
and simply connected. For example, if it is open, connected, but
not simply connected, type ”YYN” standing for ”Yes, Yes, No.”
A. {(x,y) |x > 1,y < 2}
B.

(x,y) |2x2+y2 < 1

C.

(x,y) |x2-y2 < 1

D.

(x,y) |x2-y2 > 1

E.

(x,y) |1 < x2+y2 < 4

I really need it tonight my homework is due tonight. ASAP
Submitted: 9 years ago.
Category: Homework
Expert:  MrBill31 replied 9 years ago.

Hello,

Welcome to JustAnswer!

1- Suppose F = F(x,y, z) is a gradient field with F = Ñf , S is a level surface of f, and C is a curve on S. What is
the value of the line integral_C F·dr?

Zero.

If you follow a level surface, the gradient and hence F is everywhere perpendicular to the curve

F*dr = 0 so the integral must be zero

2- Consider the vector field F(x,y) = (2x2 +y2,2xy). Compute
the line integrals_C1 F· dr and Integral_C2 F· dr, where C1 is the curve

r1(t)=(t, t2) for 0t 1 andC2 is the curve r2(t)=(t, t), also
forR0 <= t <= 1.

Integral_C1 F·dr = .

Int[(i(2x2+y2)+j(2xy))*(idx + jdy)]

= Int[(2x2+y2)dx +(2xy)dy)]

Here x=t,

y=t2

dx = dt

dy = 2tdt

Integral_C1 F·dr = Int(t=0 to 1)[(2t2+(t 2)2)dt +(2tt2)2tdt)]

= Int(t=0 to 1)[(2t2+t 4 +4t4)dt]

= Int(t=0 to 1)[(2t2+5t 4dt] = ((2/3)t3 + t5)(t=1 - t=0)

= 5/3

Integral_C2 F·dr =

Int[(i(2x2+y2)+j(2xy))*(idx + jdy)]

= Int[(2x2+y2)dx +(2xy)dy)]

Here x=t,

y=t

dx = dt

dy = dt

Integral_C1 F·dr = Int(t=0 to 1)[(2t2+ t2)dt +(2tt)dt)]

= Int(t=0 to 1)[(5t2dt]

= ((5/3)t3)(t=1 - t=0)

= 5/3

Same value

Can you decide from your answers alone whether or not F is
a conservative vector field? (Y/N)

OHHHH! Nasty trick question!

The answer is NO

Yes these 2 independent paths give you the same result, but NO this is not sufficient to prove F is a conservative vector field (i.e. it is path independent).

It might be sheer coincidence that these 2 paths gave the same result.

A true proof that the path integral is completely independent for all possible paths (conservative) is to show that the gradient field can be written as the gradient of some scalar field.

Is F a conservative vector field? (Y/N)

Answer YES

Let's find out.

Can I write this field as the gradient of some function?

dF/dx = 2x2+y2

implies F = (2/3)x3 + y2x

dF/dy = 2xy

so this works

The fact that the field is a gradient of a function

(in this case F(x,y) = (2/3)x3 + y2x)

means the answer is YES

(and also proves path independence for any path integral).

3- Let F(x,y) = (-yi+xj)/(x2+y2) = P(x,y)i+Q(x,y)j and let C be the circle r (t) = (cost) i+(sint) j, 0 <= t <= 2p.

Ah hah! so this is the problem I have been having with some of your homework problems

"p" as in 0 <= t <= 2p means pi i.e 3.1415 etc,

didn't know this until now.

A.
Note: Your answer should be in terms of x and y; e.g. "3xy - y"

(-yi+xj)/(x2+y2) = P(x,y)i+Q(x,y)j

Q = x/(x2 + y2)

dQ/dx = d(x/(x2 + y2))dx = 1/(x2 + y2) - x(2x)/(x2 + y2)2

= 1/(x2 + y2) - 2x2/(x2 + y2)2

= (y2 - x2)/(x2 + y2)2

B.
dP/dy =
Note: Your answer should be in terms of x and y; e.g. "3xy - y"

(-yi+xj)/(x2+y2) = P(x,y)i+Q(x,y)j

P = -y/(x2 + y2)

dP/dy = d(-y/(x2 + y2))dy = -1/(x2 + y2) + y(2y)/(x2 + y2)2

= -1/(x2 + y2) + 2y2/(x2 + y2)2

= (y2 - x2)/(x2 + y2)2

C.
Integral_C F·dr =
Note: Your answer should be a number

Integral_C F·dr = Int[((-yi+xj)/(x2+y2))*(idx+jdy)]

= Int[((-ydx+xdy)/(x2+y2))]

r (t) = (cost) i+(sint) j

x = cos(t)

y = sin(t)

dx = -sin(t)dt

dy = cos(t)dt

Integral_C F·dr = Int[((-ydx+xdy)/(x2+y2))]

= Int(t=0 to t=2 pi)[((-sin(t)(-sin(t))dt+cos(t)cos(t))dt/(cos(t) 2+sin(t) 2))]

= Int(t=0 to t=2 pi)[1*dt/(1)]

(since sin2(t) + cos2(t) = 1)

= Int(t=0 to t=2 pi)[dt]

= t(t=2pi - t=0) = 2*pi

3- Suppose C is any curve from (0,0,0) to (1,1,1) and
F(x,y, z) = (3z+4y) i+(2z+4x) j+(2y+3x)k. Compute the
line integral_C (F·dr).

F is the gradient of what?

Well

df/dx = (3z + 4y)

implies

f = 3xz + 4xy + f1(y,z)

df/dy = (2z+4x)

implies

f = 2yz + 4xy + f2(x,z)

so we can conclude

f = 4xy + 3xz + 2yz

df/dz = 3x + 2y = (d/dz(4xy + 3xz + 2yz))

is consistant with this.

So if

f = 4xy + 3xz + 2yz is the potential function, we just plug in the begining and end points

integral_C (F·dr). = f(1,1,1) - f(0,0,0) regardless of path

= 4*1*1 + 3*1*1 + 2*1*1 - 4*0*0 + 3*0*0 + 2*0*0 = 4+3+2

= 9

Check,

let a parameterized path be

r(t,t,t)(t=0 - t=1)
integral_C (F·dr)

= Int[((3z+4y) i+(2z+4x) j+(2y+3x)k)*(idx + jdy + kdz)]

= Int[((3z+4y)dx+(2z+4x)dy+(2y+3x)dz]

x=y=z=t

dx=dy=dz=dt

integral_C (F·dr)

= (t=0 to t=1)Int[((3t+4t)dt+(2t+4t)dt+(2t+3t)dt]

= (t=0 to t=1)Int[18tdt]

= 9t2 (t=1 - t=0)

= 9

4- Let F = (5xy,2y2) be a vector field in the plane, and C the
path y = 5x2 joining (0,0) to (1,5) in the plane.
Evaluate Integral_C (F·ds)

Integral_C (F·ds) = Int[F*dr] = Int[5xydx + 2y2dy]

can parameterize the curve

y = 5x2 as

r(t,5t2) (t=0 to t=1)

x = t

y = 5t2

dx = dt

dy = 10tdt

Int[5xydx + 2y2dy] = (t=0 to t=1)Int[5(t)(5t2)dt + 2(5t2)210tdt]

= (t=0 to t=1)Int[25t3dt + 500t5dt]

= (t=0 to t=1)Int[(25t3 + 500t5)dt]

= [(25/4)t4 + (500/5)t5] (t=1 - t=0)

= 25/4 + 500/6 = 75/12 +1000/12 = 1075/12 = 89 7/12

= 89.5833333

6- Determine whether the given set is open, connected,
and simply connected. For example, if it is open, connected, but
not simply connected, type "YYN" standing for "Yes, Yes, No."
A. {(x,y) |x > 1,y < 2}

YYY

(open, one region, no holes)
B.

(x,y) |2x2+y2 < 1

NYY

(closed: a filled ellipse, one region, no holes)

C.

(x,y) |x2-y2 < 1
YYY

(open: (filled area between 2 hyperbolas, x = +- sqrt(1+Y2)), single region, no holes)

D.

(x,y) |x2-y2 > 1

YNN

(open: (2 filled hyperbolas, x = +- sqrt(1+Y2)), 2 disconnected regions)

E.

(x,y) |1 < x2+y2 < 4

NYN

(Closed: (filled annulus), single region, hole)

Customer: replied 9 years ago.
Thank you.

I have 2 more questions if you don't mind and need them now. My homework is due in 2:30 min form know. Please help. Thank you

1- Let{F} = (18 xyz + 3sin x, 9 x^2z, 9 x^2y). Find a function f so that {F} = f, and f(0,0,0) = 0.

2-

For each of the following vector fields F , decide
whether it is conservative or not. Type in a potential function f
(that is, Ñf = F). If it is not conservative, type N.
A. F(x,y) = (-12x+2y) i+(2x+2y) j

f (x,y) =

B. F(x,y) = -6yi-5xj

f (x,y) =

C. F(x,y, z) = -6xi-5yj+k

f (x,y, z) =

D. F(x,y) = (-6sin y) i+(4y-6x cos y) j

f (x,y) =

E. F(x,y, z) = -6x2i+2y2j+1z2k

f (x,y, z) =

Note: Your answers should be either expressions of x, y and z (e.g. “3xy + 2yz”), or the letter “N”
Expert:  MrBill31 replied 9 years ago.

1- Let{F} = (18 xyz + 3sin x, 9 x^2z, 9 x^2y). Find a function f so that {F} = f, and f(0,0,0) = 0.

So I guess you want the potential this is the gradient of:

df/dx = 18 xyz + 3sin x

f = 9x2yz + -3cos(x) + ?

df/dy = 9 x2z

f = 9x2yz + ?

df/dz = 9x2y

f = 9x2yz + ?

So looks like

f = 9x2yz - 3cos(x) + constant

f(0,0,0) = 0 = 0 - 3 + constant

constant = 3

f = 9x2yz - 3cos(x) + 3

2-

For each of the following vector fields F , decide
whether it is conservative or not. Type in a potential function f
(that is, Ñf = F). If it is not conservative, type N.
A. F(x,y) = (-12x+2y) i+(2x+2y) j

df/dx = -12x + 2y

f = -6x2 + 2xy + ?

df/dy = 2x + 2y

f = 2xy + y2

f(x,y) = -6x2 + 2xy + y2

B. F(x,y) = -6yi-5xj

a rotational flow, suspect not

df/dx = -6y

f = -6xy + ?

df/dy = -5x

f = -5xy + ?

These are the same term with different constants.

Integrability conditions cannot be satisfied.

There is no potential function.

f (x,y) = N (i.e. NO)

C. F(x,y, z) = -6xi-5yj+k

df/dx = -6x

f = -3x2 + ?

df/dy = -5y

f = -(5/2)y2 + ?

df/dz = 1

f = z + ?

f (x,y, z) = -3x2 + (5/2)y2 + z

D. F(x,y) = (-6sin y) i+(4y-6x cos y) j

df/dx = -6sin(y)

f = -6xsin(y) + ?

df/dy = 4y-6x cos y

f = 2y2 -6xsin(y) + ?

f (x,y) = 2y2 -6xsin(y)

E. F(x,y, z) = -6x2i+2y2j+1z2k

df/dx = -6x2

f = -2x3 + ?

df/dy = 2y2

f = (2/3)y3 + ?

df/dz = z2

f = (1/3)z3 + ?

f (x,y, z) = -2x3 + (2/3)y3 + (1/3)z3

Note: Your answers should be either expressions of x, y and z (e.g. "3xy + 2yz"), or the letter "N".

Customer: replied 9 years ago.
Could you check C in Question 2 in the second response.

C. {F} ( x, y, z) = -6 x {i} - 5 y {j} + {k}

The webwork didn't accept the answer.

Thank you very much.
Expert:  MrBill31 replied 9 years ago.
THIS ANSWER IS LOCKED!

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