A vector field on a domain in the plane or in space is a function that assigns a vector to each point in the domain. A field of three-dimensional vectors might have a formula like F (x, y, z) = M (x, y, z)i + N(x,y,z)j + P (x, y, z)k. This field will be continuous if the component functions M, N, and P are continuous. Also, this field will be differentiable if M, N, and P are differentiable. A field of two-dimensional vectors might have a formula like F (x, y) = M (x , y)i + N (x, y)j. Here are some examples of vector fields.

If we attach a projectile's velocity vector to each point of the projectile's trajectory in the plane of motion, then we will have a two-dimensional field defined along the trajectory.

If we attach the gradient vector of a scalar function to each point of a level surface of the function, then we will have a three-dimensional field on the surface.

If we attach the velocity vector to each point of a flowing fluid, then we will have a three-dimensional field defined on a region in space.

DEFINITION:

The gradient field of a differentiable function f (x, y, z) is the field of gradient vectors

EXAMPLE 1:

Find the gradient field of the function f (x, y, z) = xy + yz + xz.

SOLUTION:

Suppose that the vector field F = M (x, y, z)i + N (x, y, z)j + P (x, y, z)k represent a force throughout a region in space, and that r (t) = g (t) i + h (t) j + k (t) k, a £ t £ b is a smooth curve in the region. The integral of F · T, the scalar component of F in the direction of the curve's unit tangent vector, over the curve is called the work done by F over the curve from a to b.

EXAMPLE 2:

Find the work done by

from (0, 0, 0) to (1, 1, 1) over the path r (t) = t i + t ² j + t ⁴ k, 0 £ t £ 1.

SOLUTION:

EXAMPLE 3:

Find the work done by F = (y + z) i + (z + x) j + (x + y) k from (0, 0, 0) to (1, 1, 1) over the path C₃ È C₄ consisting of the line segment from (0, 0, 0) to (1, 1, 0) followed by the segment from (1, 1, 0) to (1, 1, 1).

SOLUTION:

EXAMPLE 4:

Find the work done by F = 6z i + y ² j + 12x k over the curve r = (cost)i (sin t) j + (t /6) k, 0 £ t £ 2p .

SOLUTION:

EXAMPLE 5:

Evaluate

along the curve y = x ² from (-1, 1) to (2, 4).

SOLUTION: This is a work integral of the last form, and before we go too much farther, we must determine the vector form of the curve. Let x = t, then y = t ², and r (t) = t i + t ²j, -1 £ t £ 2. Now we can proceed as we have done in the past few examples.

Now suppose that F = M i + N j + P k represents the velocity field of a fluid flowing through a region in space. Now the integral of F · T along a curve in the region gives the fluid's flow along the curve.

DEFINITION:

If r (t) = g (t) i + h (t) j + k (t) k, a £ t £ b, is a smooth curve in the domain of a continuous velocity field F = M (x, y, z) i + N (x, y, z) j + P(x, y, z) k, then the flow along the curve from t = a to t = b is the integral F · T over the curve from a to b:

The integral in this case is called the flow integral. If the curve is a closed loop, then the flow is called the circulation around the curve.

EXAMPLE 6:

F = (x - z) i + x k is the velocity field flowing through a region in space. Find the flow along the curve r = (cos t) i + (sin t) k, 0 £ t £ p .

SOLUTION:

EXAMPLE 7:

F = - y i + x j + 2 k is the velocity field of a fluid flowing through a region in space. Find the flow along the curve r = (-2cos t) i + (2sin t) j + 2t k, 0£ t £ 2p .

To find the rate at which a fluid is entering or leaving a region enclosed by a smooth curve in the xy-plane, we will calculate the line integral over C of F · n. This is the scalar component of the fluid's velocity field in the directions of the curve's outward pointing normal vector.

DEFINITION: 

If C is a smooth closed curve in the domain of a continuous vector field F = M (x, y) i + N (x, y) j in the plane, and if n is the outward-pointing unit normal vector on C, then the flux of F across C is given by the following line integral:

The difference between flux and circulation is that the flux of F across C is the line integral with respect to the arc length of F · n. The circulation of F around C is the line integral with respect to arc length of F · T.

To evaluate the flux integral, begin with the parameterization x = g (t), y = h (t), a £ t £ b that traces the curve C exactly once as t increases from a to b. Now we can find the outward unit normal vector n by crossing the curve's unit tangent vector T with the vector k. Which cross product - T ´ k or k ´ T - do we choose? It depends on which way C is traversed as t increases. If the motion is clockwise, then it is k ´ T. If it is counterclockwise, then it is T ´ k. The usual choice is n = T ´ k. Therefore the value of the arc length integral in the definition of flux does not depend on which was C is traversed.

If F = M (x, y) i + N (x, y) j, then

Therefore,

This integral is in the counterclockwise direction.

EXAMPLE 8:

Find the circulation and flux of the field F = -y i + x j around and across the closed semi-circular path that consists of the semi-circular arch r₁ = (acos t) i + (a sin t) j, 0 £ t £ p , followed by the line segment r₂ = t i, -a £ t £ a.

EXAMPLE 9:

Find the circulation and flux of the field F = -y ² i + x ² j around and across the closed semi-circular path that consists of the semi-circular arch r₁ = (acos t) i + (a sin t) j, 0 £ t £ p , followed by the line segment r₂ = t i, -a £ t £ a.