MATH 251 – LECTURE 23
JENS FORSGÅRD
http://www.math.tamu.edu/~jensf/
This week: 14.1–2
webAssign: 14.1–2, due 4/4 11:55 p.m.
(No webAssignments need to be handed in.)
Next week: 14.3–4
webAssign: 14.3–4, opens 4/4 12 a.m.
Help Sessions:
Sun–Thu 6–8 p.m. in BLOC 149
Office Hours:
BLOC 641C
M 12:30–2:30 p.m.
W 2–3 p.m.
or by appointment.
Vectorfields
Definition 1. A vector field on R2 is a function F which assigns to each point (x, y) a vector F (x, y).
F (x, y) = P (x, y)i + Q(x, y)j
Definition 2. A vector field on R3 is a function F which assigns to each point (x, y, z) a vector F (x, y, z).
F (x, y, z) = P (x, y, z)i + Q(x, y, z)j + R(x, y, z)k
Vectorfields
Exercise 3. Sketch the vector field F (x, y) = y2 i + x2 j.
Vectorfields
Let f (x, y) be a function of two variables. Then ∇f = hfx0 , fy0 i is the gradient vector field of f .
Let f (x, y, z) be a function of three variables. Then ∇f = hfx0 , fy0 , fz0 i is the gradient vector field of f .
Exercise 4. Find the gradient vector field of f (x, y, z) = y ln(x − z).
Vectorfields
A vector field is said to be conservative if it is the gradient vector field of some function f . That is, F is
conservative if
F = ∇f.
The function f is called a potential function for F .
Line integrals
Exercise 5. Compute the circumference of the circle with radius r.
Line integrals
-2
-2
-1
-1
0
0
1
1
2
2
4
4
2
2
0
-2
-1
1
0
0
2
-2
-2
-1
1
0
-2
-1
2
-1
0
0
1
1
2
2
4
4
2
2
0
-2
-1
0
1
0
2
-2
-1
0
1
2
Line integrals
2
1
0
-1
-2
4
3
2
1
0
-2
-1
0
1
2
So what if f (x, y) = 1?
Line integrals
Let C be a smooth parametrized curve
x = x(t),
y = y(t),
a ≤ t ≤ b.
A partition Q of [a, b] is a choice of points
a = t0 < t1 < · · · < tn = b.
The partition of the interval [a, b] determines a partition of the curve C. by points P (xi, yi) given by xi = x(ti)
and yi = y(ti). Let ∆si denote the length of the ith subarc of C.
Line integrals
The norm |Q| is the longest of the lengths ∆s1. Choose points Pi∗(x∗i , yi∗) in the ith subarc. Then we can form
the Riemann sum
n
X
f (x∗i , yi∗)∆si.
i=1
Definition 6. The line integral of f along C (w.r.t. arc length) is
Z
n
X
f (x∗i , yi∗)∆si,
f (x, y)ds = lim
|Q|→0
C
i=1
provided that the limit exists.
Since ds =
p
(x0(t))2 + (y 0(t))2 dt, we have that
Z
Z b
p
f (x, y)ds =
f (x(t), y(t)) (x0(t))2 + (y 0(t))2 dt
C
a
The value of the line integral does not depend on the choice of parametrization.
Line integrals
Exercise 7. Evaluate the line integral
R
C
xds, where C is given by x = t3, y = t, 0 ≤ t ≤ 1.