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Math 32B Discussion Session Week 6 Notes February 14 and 16, 2017
This week we’ll explore some special properties of gradient vector fields, and investigate their relationship with line integrals. In the Thursday section we’ll introduce surface integrals of scalar-valued functions.
Conservative Vector Fields
Recall the diagram we drew last week depicting the derivatives we’ve learned in the 32 sequence:
functions
gradient −−−−→ vector fields curl −−→ vector fields
divergence −−−−−−→ functions. (1)
Every (sufficiently nice) function has a gradient vector field, but not every vector field in the second slot above is the result of taking the gradient of some function. For reasons grounded in physics, we call those vector fields which can be written as the gradient of some function conservative. When we saw the sequence (1) last week, we observed that applying consecutive operators always resulted in zero. In particular, we noticed that
curl(grad(f )) = 0
for any function f. So we have a necessary condition for a vector field (on R^3 ) to be conservative: the vector field must have zero curl. For vector fields on R^2 , we can compute the curl as if our vector field were defined on R^3 with a z-component of 0. The condition that curl(F) = 0 then manifests itself as
0 = curlz (F) =
∂F 2
∂x
∂F 1
∂y
Now that we have a test that a vector field must pass in order to be conservative, a natural question is whether or not this test is sufficient. That is, if we have a vector field F and we find that curl(F) = 0 (or curlz (F) = 0), can we always conclude that F is conservative?
Here’s a vector field that suggests the answer is no. Consider the vector field defined on R^2 − {(0, 0)} by
F(x, y) =
−y x^2 + y^2
x x^2 + y^2
This vector field is often called a vortex vector field, as you might expect from its plot, seen in Figure 1. We can compute the cross-partials quantity for F to see that
curlz (F) =
∂x
x x^2 + y^2
∂y
−y x^2 + y^2
(x^2 + y^2 ) − (x)(2x) (x^2 + y^2 )^2
−(x^2 + y^2 ) + (y)(2y) (x^2 + y^2 )^2
Figure 1: A vortex vector field.
This suggests that the vector field is not spinning, but the plot shows clearly that F is spin- ning. Something about our spin computations doesn’t play nicely with F.
We can (non-rigorously) argue to ourselves that F can’t possibly be conservative as fol- lows. Suppose F were conservative, and that we have F = ∇f for some function f defined on R^2 − {(0, 0)}. Then F always points in the direction of greatest increase for f. If we start at the green point in Figure 1 with the goal of maximizing f , the vector field F tells us to move in a counterclockwise manner around the red curve. This might work well at first, but eventually this curve returns to our starting point, where the value of f must be the same as its original value. We cannot have been increasing the value of f all along our journey only to arrive back at the same value of f. This vector field simply can’t be a gradient vector field (except perhaps in the study of Relativity).
This informal argument that the vortex vector field is not conservative suggests that the property of being conservative — that is, of being the gradient vector field of some function — is tied to some condition on integrating around closed loops. Our objection to the claim that F was a conservative vector field had to do with the fact that F was doing positive work to move our green point all the way around the circle. Indeed, if we have a function f and a parametrization r(t), a ≤ t ≤ b of some closed curve C, then we have
∮
C
∇f · dr =
∫ (^) b
a
∇f (r(t)) · r′(t)dt =
∫ (^) b
a
d dt
(f (r(t)))dt = f (r(b)) − f (r(a)) = 0.
So we have another necessary condition for conservative vector fields: they must integrate to zero around closed curves. This condition turns out to be sufficient.
be contracted in a way that avoids this point. When we’re on a simply connected domain, the curl-free condition which is necessary of all conservative vector fields becomes sufficient.
Theorem. If F is a vector field on a simply connected domain D, then F is conservative if and only if F is curl-free (where we take curlz (F) to be the curl if F is defined on R^2 ).
We’ll conclude these notes by finding a potential function for a conservative vector field.
Example. (§17.3, Exercise 13) Determine whether or not
F = 〈z sec^2 x, z, y + tan x〉
is conservative. If F is conservative, find a potential function for F.
(Solution) First we note that
curl(F) =
i j k ∂/∂x ∂/∂y ∂/∂z z sec^2 x z y + tan x
1 − 1 , sec^2 x − sec^2 x, 0 − 0
Since the domain of F is simply connected^1 , the fact that F is curl-free tells us that F is conservative. We’ll now look for a potential function f with the property that F = ∇f. That is, we want fx = z sec^2 x, fy = z, and fz = y + tan x.
Integrating the first equality tells us that f = z tan x + g(y, z), where g is some function that does not depend on x. Then we have
z = fy = gy,
so g = zy + h(z), where h is some function that does not depend on x or y. Substituting this into f tells us that f = z tan x + zy + h(z), so
y + tan x = fz = y + tan x + h′(z).
So h′(z) = 0, meaning that h is constant. So any function of the form
f (x, y, z) = z tan x + zy + K
provides a potential function for F. ♦
(^1) Notice that the domain of F is not all of R (^3). Why is it still simply connected?
Figure 2: A patch of the unit sphere.
Surface Integrals
Recently we’ve found ourselves needing to compute integrals over less-than-ideal objects, such as funny regions or curves in Euclidean space. A technique we’ve used to compute these integrals is to find some transformation from a nice region in Euclidean space whose image is the region or curve we care about, and then to pull our integral back to the nice region using the transformation. For instance, when integrating over strange regions in the plane, we found a suitable transformation T : R^2 → R^2 which we used to rewrite our integral as an integral over a rectangle. For line integrals we want to integrate over some curve, and we use a parametrization r of this curve to write our integral as an integral over an interval in R. In both cases, the transformation gives us some stretch factor which determines the integral over the nicer-looking space.
Now that we want to integrate functions and vector fields over surfaces, we’re going to follow this familiar script. We have a parametrization T : R^2 → S of our surface S, and we’ll use this to pull integrals over S (or regions within S) back to integrals over regions in R^2. Along the way, we’ll pick up a stretch factor that looks very similar to the Jacobian factor we encountered earlier.
Let’s motivate this by way of an example. Consider the parametrization of the unit sphere in R^3 given by G(u, v) = (cos u sin v, sin u sin v, cos v),
for 0 ≤ u ≤ 2 π and 0 ≤ v ≤ π. A plot of the unit sphere can be seen in Figure 2. In this plot the sphere has a few grid lines on it, where we’ve held one variable constant and allowed the other to vary. The blue curves on the sphere correspond to fixing v and allowing u to vary, while the red curves have u fixed and allow v to vary. Just as a uv-coordinate system on R^2 subdivides the plane into small patches, these grid lines subdivide our sphere into small regions, which we consider to be our most simple regions.
Just as in the case of a transformation T : R^2 → R^2 , we need to approximate the area
So we have
dS =
∂G
∂u
×
∂G
∂v
∥ dudv^ =
cos^2 v sin^2 v + sin^4 vdudv =
sin^2 vdudv = sin vdudv.
Finally, the surface area is given by
A =
∫ (^) π
0
∫ (^2) π
0
sin vdudv = 2π
∫ (^) π
0
sin vdv = 2π [− cos v]π 0 = 2π (1 − (−1)) = 4π.
♦
As you might expect, the fudge factor
∥∂G
∂u ×^
∂G ∂v
∥ (^) finds its way into the more general
formula for surface integrals.
Theorem. Let G(u, v) be a parametrization of a surface S with parameter domain D. Assume that G is continuously differentiable, one-to-one, and regular, except possibly at the boundary of D. Then ∫ ∫
S
f (x, y, z)dS =
D
f (G(u, v))‖N(u, v)‖dudv,
where N(u, v) is the normal vector
N(u, v) =
∂G
∂u
×
∂G
∂v obtained from the parametrization.
Example. We can use surface integrals to compute the mass of a thin lamina. If our lamina is represented by the surface S and this lamina has density δ(x, y, z), then the mass of the lamina is given by integrating δ over S. Find the mass of the lamina that is the portion of the surface y^2 = 4 − z between the planes x = 0, x = 3, y = 0, and y = 3 if the density is^2 δ(x, y, z) = y.
(Solution) First we need a parametrization of our surface, which can be seen in Figure 3. Since both x and y are allowed to vary from 0 to 3 without any other condition, it makes sense to take x = u and y = v. Then we have z = 4 − y^2 = 4 − v^2 , so our parametrization is
G(u, v) = (u, v, 4 − v^2 ), 0 ≤ u ≤ 3 , 0 ≤ v ≤ 3.
We compute the relevant tangent vectors:
∂G ∂u
, ∂G
∂v
− 2 v
(^2) This example is taken from the eighth edition of Anton, Bivens, and Davis’s Calculus.
Figure 3: A thin lamina.
Then the normal vector associated to our parametrization is
N(u, v) =
∂G
∂u
×
∂G
∂v
i j k 1 0 0 0 1 − 2 v
2 v 1
Our fudge factor is the magnitude of this normal vector:
‖N(u, v)‖ =
4 v^2 + 1.
Finally, we notice that δ(G(u, v)) = δ(u, v, 4 − v^2 ) = v, so the mass is given by
M =
S
δ(x, y, z)dS =
0
0
δ(G(u, v))‖N(u, v)‖dudv
0
0
v
4 v^2 + 1dudv =
0
3 v
4 v^2 + 1dv.
Here we can make a substitution w = 4v^2 + 1, so that dw = 8vdv. As v varies from 0 to 3, w will vary from 1 to 37, so we have
M =
1
wdw =
[
w^3 /^2
] 37
1
(37^3 /^2 − 1)
for the mass of our lamina. ♦
