Scalar Fields
Def: A scalar field in R2 is just a function
f : R2 → R
Example: f (x, y) = x2 + sin(y) is a scalar field in R2 .
So: Scalar fields input a point (x, y) ∈ R2 and output a scalar f (x, y) ∈ R
that we think of as living at that point.
Def: A scalar field in R3 is just a function
f : R3 → R
Example: f (x, y, z) = x3 cos(z) + ey is a scalar field in R2 .
Physical Examples:
• Temperature
• Humidity
• Density
• Charge Density
Vector Fields
Def: A vector field in R2 is a vector function of two variables of the form:
F(x, y) = hP (x, y), Q(x, y)i
= P (x, y) i + Q(x, y) j.
A vector field in R3 is a vector function of three variables of the form:
F(x, y, z) = hP (x, y, z), Q(x, y, z), R(x, y, z)i
= P (x, y, z) i + Q(x, y, z) j + R(x, y, z) k.
Visually: Vector fields are functions that input a point (x, y, z) and output a
vector F(x, y, z) based at that point.
Physical Examples:
• Fluid velocity
• Gravitational force
• Electric field
Examples in R2 : Here are some important examples.
• F(x, y) = hx, yi
• F(x, y) = h−x, −yi
• F(x, y) = h−y, xi
Exercise: Sketch the above vector fields in R2 . (It is helpful to know what
these particular examples look like.)
Examples in R3 : Here are some important examples.
• F(x, y, z) = hx, y, zi
• F(x, y, z) = h−x, −y, −zi
• F(x, y, z) = h−y, x, 0i
• F(x, y, z) = h0, −z, yi
• F(x, y, z) = hz, 0, −xi
Exercise: Sketch the above vector fields in R3 . (Again: It is helpful to know
what these particular examples look like.)
Gradient of a Scalar Field
Def: Let f (x1 , . . . , xn ) be a scalar field in Rn . The gradient of f is the
vector field
∂f
∂f
,...,
.
∇f =
∂x1
∂xn
Example: The gradient of f (x, y, z) = x2 z + sin(y) is:
∇f = h2xz, cos(y), x2 i.
Note: The gradient operator ∇ inputs scalar fields f and outputs vector
fields ∇f :
gradient
{Scalar Field f } −−−−→ {Vector Field ∇f }
Conservative Vector Fields: Definition
Def: A vector field F is conservative (on a region R) if: F = ∇f for
some differentiable scalar field f that is defined on R.
In this case: The scalar field f is called a potential function for F.
Some vector fields are conservative, but most vector fields are non-conservative.
Example 1: The vector field F(x, y) = hx, yi is conservative. This is because
hx, yi = ∇( 21 x2 + 21 y 2 ).
The function f (x, y) = 12 x2 + 12 y 2 is a potential function for F.
Example 2: The vector field F(x, y) = h−y, xi is not conservative. In other
words, there does not exist a scalar field f (x, y) for which h−y, xi = ∇f .
(Hold on: Why does no such f exist? We’ll learn this later.)
Important Questions: We’ll answer these later in the course (Lec 17-21).
(1) Why are conservative vector fields so special?
(2) How can we tell if a vector field is conservative or not?
Line Integrals: Overview
Overview: There are three different types of line integrals:
(1) Line integrals of scalar fields f on a curve C with respect to a coordinate.
Z
Z
Z
f dx and
f dy and
f dz
C
C
C
(2) Line integrals of scalar fields f on a curve C with respect to arc
length.
Z
f ds
C
(3) Line integrals of vector fields F on a curve C.
Z
F · dr
C
In Lectures 14 and 15, we discuss (1) and (2). However, our primary interest
is (3), which we’ll discuss in Lecture 16.
Line Integrals of Scalar Fields: Definition
Let’s define line integrals of scalar fields f (x, y) in R2 . The definition for
scalar fields f (x, y, z) in R3 is exactly the same.
Given:
• Scalar field f (x, y). We assume f is continuous.
• Parametric curve C given by r(t) = hx(t), y(t)i for t ∈ [a, b].
Procedure:
• Partition [a, b] into n subintervals: a = t0 < t1 < · · · < tn = b.
• Partition C into n sub-arcs of lengths ∆si using the partition of [a, b].
• Let the projection of each sub-arc onto the x-axis and y-axis have lengths
∆xi and ∆yi , respectively.
• Choose sample points (x∗1 , y1∗ ), . . . , (x∗n , yn∗ ) on C.
We consider the Riemann sums
n
X
f (x∗i , yi∗ ) ∆xi
i=1
n
X
f (x∗i , yi∗ ) ∆yi
i=1
n
X
f (x∗i , yi∗ ) ∆sk .
i=1
Def: The line integral of f (x, y) on C with respect to x and y are:
Z
n
X
f (x, y) dx = lim
f (x∗i , yi∗ ) ∆xi
C
n→∞
Z
f (x, y) dy = lim
C
n→∞
i=1
n
X
f (x∗i , yi∗ ) ∆yi
i=1
The line integral of f (x, y) on C with respect to arc length is:
Z
n
X
f (x, y) ds = lim
f (x∗i , yi∗ ) ∆si .
C
n→∞
i=1
Physical Interpretation:
Suppose f (x, y) is the density at (x, y). Suppose
Z
C is a wire. Then
f (x, y) ds is the mass of the wire C. C