Calculus 3 Differentiation and Integration of Functions of Many - - PowerPoint PPT Presentation

Calculus 3

Differentiation and Integration of Functions of Many Variables for Physical Sciences and Engineering Hu` ynh Qang V˜ u

Faculty of Mathematics and Computer Science, University of Science, Ho Chi Minh City

05/2014

Vector functions

Definition

A vector function is a map f : D → R3 where D ⊂ R. In the standard basis of R3 we can write r(t) = x(t), y(t), z(t).

Limits

Te notion of limit can be generalized to any space with distance.

Theorem

lim

t→a r(t) = lim t→a x(t), lim t→a y(t), lim t→a z(t).

Definition

A function r is said to be continuous at t = a if limt→a r(t) = r(a).

Fact

A vector function is continuous if and only if each of its components is continuous.

Derivative

Definition

r′(t) = lim

h→0

r(t + h) − r(t) h , provided the limit exists. Te derivative represents the rate of change of r(t) with respect to t. Te derivative is the limit of secant vectors, therefore it is called a tangent vector. Physically, r′(t) is the velocity at time t.

Theorem

If r(t) = (x(t), y(t), z(t)) then r′(t) = (x′(t), y′(t), z′(t)).

Space curves

Definition

A space curve C is the set of all points r(t), t ∈ D ⊂ R. r is called the position function or a path. C is said to be parametrized by r. Tus a curve is the trace of a path.

Example

Te curve given by the following parametrization x(t) = cos t, y(t) = sin t, z(t) = t is called a helix.

Fact

A curve can be parametrized by different functions.

Smooth curves

Definition

A curve is called smooth, or regular if there is a parametrization by r(t), t ∈ D such that r′(t) is continuous and r′(t) = 0 on the interior

• f D.

With a regular parametrization, the speed is never zero, and the tangent direction is defined as the direction of the velocity vector.

Arc length

Let r : [a, b] → Rn be a smooth path. Consider a partition a = t0 < t1 < · · · < tn = b of [a, b]. On each interval [ti−1, ti], 1 ≤ i ≤ n, linearly approximate the path: r(t) ≈ r′(ti−1)(t − ti−1). Te “length” of the path r(t) from ti−1 to ti is linearly approximated by the tangent vector r′(ti−1)∆ti. Tus the “length” of the path on [a, b] is approximated by

n

• i=1

|r′(ti−1)|∆ti. Tis is exactly a Riemann sum of the function |r′(t)| on [a, b].

Definition

Te length of a smooth path r : [a, b] → Rn is defined as ˆ b

a

|r′(t)| dt. The length of a path is the integral of its speed with respect to time. It reminds the old formula: distance=speed x time.

Example

Suppose that an object is traveling on a line from time a to time b with constant speed v > 0. Ten the distance it has traveled is ´ b

a v dt = v(b − a), as we expect.

Arc length function

Theorem

Let r : [a, b] → Rn be a regular path. Te function s(t) = ˆ t

a

|r′(u)| du. is called the arc-length function of r. s(t) has an inverse function t(s), 0 ≤ s ≤ l, where l = s(b) is the length of r. Te path β(s) = r(t(s)) has the same trace as r. Te speed of β is always 1. β is said to be the re-parametrization by arc-length of r. Note that ds dt (t) = |v(t)|. Symbolically: ds = |v(t)|dt.

Curvature

Te curvature of the curve at a point p is a measure of how fast the direction of the curve turns around p. It is therefore a measure of the rate of change of the direction of the curve as one moves along the curve at constant speed. Te direction of the curve is given by the unit tangent vector T(t) = r′(t) |r′(t)|. Te constant speed is given by parametrization by arc-length.

Definition

Let r : [a, b] → Rn be a regular path and let C be its trace. Te curvature of the curve C at a point p = r(s) is defined as the number k(p) =

• dT

ds (s)

• .

Tus if the curve r(t) has constant speed then the curvature is exactly |r′′(t)|.

Usually the curve is not given in arc length form. For curve k(p) =

• dT

ds

• =
• dT

dt · dt ds

• =
• dT

dt · 1

ds dt

• =
• dT

dt · 1 |r′(t)|

• =
• T ′(t)
• |r′(t)| .

Because T(t) = r′(t)/|r′(t)|, with some calculations we get a convenient formula: k(t) = |r′(t) × r′′(t)| |r′(t)|3

In particular, for plane curve r(t) = (x(t), y(t)) k = |x′′y′ − x′y′′| (x′2 + y′2)3/2 . For a graph y = f (x): k = |f ′′(x)| (1 + f ′(x)2)3/2 .

Limit

Definition

Suppose that f (x, y) is defined on D ⊂ R2. Let (a, b) ∈ D. We say that the function f has limit L as (x, y) approaches (a, b), and write lim(x,y)→(a,b) f (x, y) = L if f (x, y) is arbitrarily close to L when (x, y) is sufficiently close to (a, b). Precisely, for any ǫ > 0 there is a number δ > 0 such that if |(x, y) − (a, b)| < δ then |f (x, y) − L| < ǫ.

Continuity

Definition

We say that f is continuous at (a, b) if lim

(x,y)→(a,b) f (x, y) = f (a, b).

Elementary functions, such as polynomials, are continuous.

Example

In this case the point (0, 0) is not in the domain of the function. lim

(x,y)→(0,0)

x2 − y2 x2 + y2 Te definition of limit needs to be extended for this case.

Example

Te main difficulty when evaluate limits is that there are infinitely many directions to approach a point on the plane, unlike the case on the line (there are only two directions). If two different directions give different limits then limit does not exist. In the above example, the direction x = 0 and the direction y = 0 give different limits −1 and 1, therefore limit does not exist.

Partial Derivatives

Suppose that (a, b) is a point in the interior of the domain of a real function f (x, y). Fix y = b then f (x, y) is a function of x only. Can take the derivative with respect to x at x = a. Call it the partial derivative of f with respect to x at (a, b):

• ∂

∂x f

• (a, b).

So ∂f

∂x (a, b) is the rate of change of f at (a, b) with respect to x.

Other notations:

• ∂

∂x f

• (a, b) = ∂f

∂x (a, b) = fx(a, b) = D1f (a, b).

Geometric meaning of partial derivatives

When y = b is fixed, we get a path (x, b, f (x, b)) with parameter x

• n the graph (x, y, f (x, y)) of z = f (x, y).

Te trace of that path is the curve which is the intersection between the plane y = b and the graph z = f (x, y). Te velocity of this path is the derivative with respect to x at x = a, which is (1, 0, fx(a, b)).

Higher partial derivatives

∂f ∂x is a function of (x, y) therefore we can talk about its partial

derivatives. ∂ ∂x ∂f ∂x

• (a, b) = ∂2f

∂x2 (a, b) = fxx(a, b) = D1,1(a, b). ∂ ∂y ∂f ∂x

• (a, b) = ∂2f

∂y∂x (a, b) = fyx(a, b) = D2,1(a, b).

Theorem

If

∂2f ∂y∂x and ∂2f ∂x∂y are continuous on an open set then they are equal

there: ∂2f ∂y∂x = ∂2f ∂x∂y .

Proof.

Idea: fxy(a, b) = lim

h→0

fy(a + h, b) − fy(a, b) h = lim

h→0 lim k→0

f (a + h, b + k) − f (a + h, b) − [f (a, b + k) − f (a, b)] hk = lim

k→0 lim h→0

f (a + h, b + k) − f (a, b + k) − [f (a + h, b) − f (a, b)] hk = fyx(a, b).

Proof.

Rigorously, let ∆(h, k) = f (a + h, b + k) − f (a + h, b) − f (a, b + k) + f (a, b). Let g(x) = f (x, b + k) − f (x, b), then by the Mean Value Teorem there is α between a and a + h and β between b and b + k such that ∆(h, k) = g(a+h)−g(a) = g′(α)h = [fx(α, b+k)−fx(α, b)]h = fyx(α, β)hk Similarly we get ∆(h, k) = fxy(α′, β′)hk. From this: lim

h→0

∆(h, h) h2 = fxy(a, b) = fyx(a, b).

Linear Approximation

Suppose that f (x, y) is differentiable in a neighborhood of (a, b). Let r(x, y) = (x, y, f (x, y)). Te trace of r(x, y) is the graph of f . Fix y = b then r(x, y) becomes a path on the graph of f . Its velocity vector is rx(a, b) = (1, 0, fx(a, b)). Tis vector is “tangent” to the graph of f at the point (a, b, f (a, b)). Similarly, fixing x = a we have another tangent vector ry(a, b) = (0, 1, fy(a, b)). Te two tangent vectors span a plane, called the tangent plane of the graph of f at the point (a, b).

Tangent plane

Tis tangent plane has a normal vector rx(a, b) × ry(a, b) = (−fx(a, b), −fy(a, b), 1). Terefore an equation for the tangent plane is: −(x − a)fx(a, b) − (y − b)fy(a, b) + (z − f (a, b)) = 0. Te main idea of linear approximation is to use the tangent plane to approximate the graph. Tus for (x, y) “near” to (a, b): f (x, y) ≈ f (a, b) + fx(a, b)(x − a) + fy(a, b)(y − b). Or ∆f (x, y) ≈ fx(a, b)∆x + fy(a, b)∆y.

The Chain Rule

Theorem

Given z = f (x, y) with x = x(t) and y = y(t). Ten dz dt = ∂z ∂x · dx dt + ∂z ∂y · dy dt .

Proof.

Idea: Linear approximations ∆z ≈ fx(x, y)∆x + fy(x, y)∆y ≈ fx(x, y)x′(t)∆t + fy(x, y)y′(t)∆t. Tus ∆z ∆t ≈ fx(x, y)x′(t) + fy(x, y)y′(t).

The Chain Rule (second version)

Corollary

Given z = f (x, y) where x = x(s, t) and y = y(s, t), then ∂z ∂t = ∂z ∂x · ∂x ∂t + ∂z ∂y · ∂y ∂t .

Differentiable functions

In Calculus 1 f ′(x) = lim

h→0

f (x + h) − f (x) h . If f ′(x) exists we can write ǫ(h) = f (x + h) − f (x) h − f (x) then f (x + h) = f (x) + f ′(x)h + ǫ(h)h, with limh→0 ǫ(h) = 0.

Similarly:

Definition

Te function f is said to be differentiable at a point x, and having derivative a vector f ′(x), if in a neighborhood of x we have f (x + h) = f (x) + f ′(x) · h + ǫ(h)|h|, with limh→0 ǫ(h) = 0.

Derivative

Theorem

If all the partial derivatives exist and are continuous then the function is differentiable and its derivative is given by the list of its partial derivatives. Precisely, the derivative of f (x, y) at (a, b) is given by the gradient vector: f ′(a, b) = ∇f (a, b) = (fx(a, b), fy(a, b)).

The Chain rule in universal form

Given f (x, y) and let g(t) = (x(t), y(t)), z = f (x, y). Ten g′(t) = (x′(t), y′(t)) f ′(x, y) = (fx(x, y), fy(x, y)). Te Chain rule df dt = ∂f ∂x dx dt + ∂z ∂y dy dt can be writen in the usual form: (f ◦ g)′(t) = f ′(g(t)) · g′(t).

Directional derivative

Let f be defined in a neighborhood of p. A direction is represented by a unit vector u.

Definition

Te directional derivative of f at p in the direction u is the rate of change of f when the variable only changes on the direction given by u: Duf (p) = lim

h→0

f (p + hu) − f (p) h Tis is the rate of change in the direction of u.

If we let g(h) = f (p + hu) then by the Chain Rule Duf (p) = g′(0) = f ′(p) d dh(p + hu)

• h=0 = ∇f (p) · u.

Theorem

Duf (p) = ∇f (p) · u

Meaning of directional derivative

Du(f )(p) = ∇f (p) · u is greatest when u is in the same direction as ∇f (p), that is, when u = ∇f (p) ||∇f (p)||. Te greatest value of Du(f )(p) is ||∇f (p)||. Tus the function increases the fastest in the direction of its gradient vector.

Gradient vectors, level curves, and level surfaces

Let f : Rn → R. Consider the level set f −1(c), where c is a constant. Let r(t) be any path that lies in f −1(c). Suppose that r(t0) = p. We have f (r(t)) = c for all t. Taking derivative of both sides with respect to t, and evaluate at t = t0: ∇f (p) · r′(t0) = 0. Tus the vector ∇f (p) is perpendicular to all tangent vectors of f −1(c) at p. Gradient vectors are perpendicular to level sets.

Example

To find an equation for the tangent plane to the sphere x2 + y2 + z2 = 3 at the point (1, 1, 1): Let f (x, y, z) = x2 + y2 + z2. Te gradient vector ∇f (1, 1, 1) is normal to the level surface f −1(3).

Extrema

Suppose that f has a local extremum at p. Ten with respect to each variable, f has a local extremum at p. Terefore each partial derivative must be 0 at p. In other words, similarly to the Fermat theorem in Calculus 1:

Theorem

If f has a local extremum at p then f ′(p) = 0. If f ′(p) = 0 then p is called a critical point of f . Extrema only happen at critical points.

Sufficient conditions for extrema

Te condition that the derivative vanishes is only a necessary condition. As in Calculus 1, we have a sufficient condition using second order derivatives.

Theorem

Suppose (a, b) is a critical point of f , that is, f ′(a, b) = ∇f (a, b) = 0. Let Hf (a, b) = fxx(a, b) fxy(a, b) fyx(a, b) fyy(a, b)

• .

Let D = det Hf (a, b) = fxx(a, b)fyy(a, b) − fxy(a, b)2.

• 1. If D > 0 and fxx(a, b) > 0 then f has a minimum at (a, b).
• 2. If D > 0 and fxx(a, b) < 0 then f has a maximum at (a, b).
• 3. If D < 0 then f has a saddle point at (a, b).

Lagrange multipliers

Just as in Calculus 1, when the domain of the function contains a nonempty boundary then we need to consider the points on the boundary separately, because the derivative only exists at the interior points.

Example

To find absolute extrema of the function f (x, y) = x + y on the domain x2 + y2 ≤ 1, we need to separately find extrema on the set x2 + y2 = 1. In general, on the boundary we need to solve: Find extrema of the function f (x, y) on the domain g(x, y) = 0. We can understand this problem geometrically: Find extrema of the function f (x, y) when the variable (x, y) is on the level curve g(x, y) = 0.

Suppose that f has an extremum at a point (x0, y0). Let r(t) = (x(t), y(t)) be a path on the level curve g(x, y) = 0, with r(t0) = (x0, y0). Ten ∇g(x0, y0) ⊥ r′(t0) (gradient is perpendicular to level set). Te function f (r(t)) has an extremum at t = t0. Terefore 0 = d dt (f (r(t0)) = ∇f (r(t0)) · r′(t0) Tus ∇f (x0, y0) ⊥ r′(t0). So ∇f (x0, y0)||∇g(x0, y0).

Theorem

An extremum of the function f (x, y) subjected to the constraint g(x, y) = 0 is a solution to the system:

• ∇f (x, y)

= λ∇g(x, y), λ ∈ R g(x, y) = 0.

Two constraints

If there are two constraints g(x, y, z) = 0 and h(x, y, z) = 0 then arguing similarly we have r′(t0) is perpendicular to ∇f (x0, y0, z0), ∇g(x0, y0, z0), ∇h(x0, y0, z0). Tus ∇f (x0, y0, z0) belongs to the plane spanned by ∇g(x0, y0, z0) and ∇h(x0, y0, z0). So we need to solve the system:      ∇f (x, y, z) = λ∇g(x, y, z) + µ∇h(x, y, z), λ, µ ∈ R g(x, y, z) = 0 h(x, y, z) = 0. Certainly what we have discussed works for functions of more variables.

Integrals on rectangles

Similar to single integral, we want a number representing a sum of values of a function over its domain.

Example

A piece of metal has the shape of a rectangular region I = [0, 1] × [0, 1]. At each point x in this region the mass density of the material is known to be ρ(x). What is the total mass of this piece

• f metal?

Te answer will be given as a real number ´

I ρ, called the integral of

f over I.

Let I be a rectangle, let f : I → R. Compute the sum of values of f

• n I.

Divide I into smaller rectangles. We hope:

◮ On each sub rectangle, the values of f would have smaller

change, thus f could be approximated by a constant function.

◮ As the divisions get finer, the approximation get beter. ◮ Te correct number is the limit of the approximations.

Geometric interpretation: If f is positive, then the volume under the graph of f could be obtained by approximation using rectangles.

Partitions of rectangles

A rectangle is a subset of Rn of the form [a1, b1] × [a2, b2] × · · · × [an, bn]. Te volume of this rectangle is the real number |I| = (b1 − a1)(b2 − a2) · · · (bn − an).

◮ When n = 1 volume=length. ◮ When n = 2 volume=area.

A partition of [a, b] is a finite subset containing a and b: a set {x0, x1, . . . , xn} such that a = x0 < x1 < x2 < · · · < xn = b. Each [xi−1, xi] is a subinterval. A partition of I = n

i=1[ai, bi] is a product of partitions of [ai, bi].

A sub rectangle of I = n

i=1[ai, bi] is a product of subintervals of

[ai, bi]. A partition gives rise to a division into subrectangles.

For each partition of a rectangle, form the Riemann sum

• R

f (xR)|R| here the sum is taken over all subrectangles R, and xR is an arbitrary point in R. Te “limit” of the Riemann sums when the partitions get finer is called the integral of f on I, denoted by ´

I f .

´

I f stands for the total value of f on I.

Te notion of limit can be made precise.

Example

If c is a constant then ´

I c = c|I|. ◮ n = 1: the usual single integral

´ b

a f (x) dx. ◮ n = 2: double integral, usually writen

˜

I f (x, y) dA or

˜

I f (x, y) dxdy . ◮ n = 3: triple integral, usually writen

˝

I f (x, y, z) dV or

˝

I f (x, y, z) dxdydz.

dx, dxdy, dxdydz, dA, dV are purely symbols to indicate the types of integrals, do not have independent meanings.

Properties

Similar to single integral:

Theorem

If f , g are integrable over I then:

◮ f + g is integrable and

´

I(f + g) =

´

I f +

´

I g. ◮ for any real number c, cf is integrable and

´

I cf = c

´

I f . ◮ If f ≤ g then

´

I f ≤

´

I g.

Theorem

Any continuous function is integrable.

Fubini Theorem: 2d

Theorem

Let f be continuous over [a, b] × [c, d] then ¨

[a,b]×[c,d]

f (x, y) dA = ˆ b

a

ˆ d

c

f (x, y) dy

• dx

= ˆ d

c

ˆ b

a

f (x, y) dx

• dy.

Te integrals on the lef hand sides are called repeated integral. Tis is the main tool for computation.

Geometric interpretation

The volume of a solid is equal to the sum of the areas of parallel cross-sections.

a b c d z = f (x, y) z y x ´ d

c f (x, y) dy

x

Proof

Suppose a = x0 < x1 < · · · < xm = b a partition of [a, b], c = y0 < y1 < · · · < yn = d a partition of [c, d]. Let x∗

i in

∆xi = [xi−1, xi], y∗

j in ∆yj = [yj−1, yj], then

ˆ b

a

ˆ d

c

f (x, y) dy

• dx

≈

m

• i=1

ˆ d

c

f (x, y) dy

• |∆xi|

≈

m

• i=1
• n
• j=1

f (x∗

i , y∗ j )|∆yj|

• |∆xi|

=

• 1≤i≤m,1≤j≤n

f (x∗

i , y∗ j )|∆xi||∆yj|

≈ ¨

[a,b]×[c,d]

f (x, y) dA.

Fubini Theorem: 3d

Tree dimensional version:

Theorem

Let g be continuous on [a, b] × [c, d] × [e, f ], then ˚

[a,b]×[c,d]×[e,f ]

g(x, y, z) dV = ˆ b

a

ˆ d

c

ˆ f

e

g(x, y, z) dz

• dy
• dx.

And similarly for the 5 remaining possibilities.

Integral on general regions

We consider bounded regions, i.e. subsets of Rn. Let D be a bounded region, and f : D → R. Since D is bounded there is a rectangle I containing D. Extend f to I as F : I → R: F(x) =

• f (x),

x ∈ D 0, x / ∈ D.

Definition

ˆ

D

f = ˆ

I

F.

Volume of general regions

Definition

Te volume of a bounded subset D of Rn is |D| = ˆ

D

1. Volume=area when n = 2 and volume=length when n = 1.

Figure : Outer and inner approximations of the volume of a region.

Roughly, continuity implies integrability.

Fubini Theorem for 2d simple regions

A subset of R2 is called

◮ a vertically simple region if any vertical line intersect this

region in a line interval.

◮ a horizontally simple region if any horizontal line intersect

this region in a line interval.

Theorem

Suppose f , g, h are continuous on a vertically simple region D = {(x, y) ∈ R2 | a ≤ x ≤ b, g(x) ≤ y ≤ h(x)}, then ¨

D

f (x, y) dA = ˆ b

a

ˆ h(x)

g(x)

f (x, y) dy

• dx.

Tis is the main tool for computation.

3d simple regions

A subset of R3 is a simple region in the direction of the z-axis if any line in the direction of the z-axis intersect this region in an interval. Similarly there are simple regions in the directions of the x-axis and the y-axis.

Theorem

Suppose f , g, h are continuous on a simple region in the direction of the z-axis E = {(x, y, z) ∈ R3 | (x, y) ∈ D, f (x, y) ≤ z ≤ g(x, y)}, then ˚

E

h(x, y, z) dV = ¨

D

ˆ g(x,y)

f (x,y)

h(x, y, z) dz

• dA.

Change of variables

If ϕ : Rn → Rm then the Jacobian of ϕ is the matrix Jϕ(x) =

• ∂ϕi

∂xj (x)

• 1≤i≤m, 1≤j≤n.

Theorem

If A is an open subset of Rn, ϕ : A → Rn is continuously differentiable and bijective, and det Jϕ(x) = 0 for all x ∈ A, and f : ϕ(A) → R is continuous, then ˆ

ϕ(A)

f = ˆ

A

(f ◦ ϕ)| det Jϕ|.

2d

When n = 2, suppose that (u, v) → (x, y) brings A to B, let ∂(x, y) ∂(u, v) = det ∂x

∂u ∂x ∂v ∂y ∂u ∂y ∂v

• ,

then ¨

B

f (x, y) dxdy = ¨

A

f (x(u, v), y(u, v))

• ∂(x, y)

∂(u, v)

• dudv.

Symbolically: dxdy =

• ∂(x, y)

∂(u, v)

• dudv.

Explaining the change of variables formula

Let A be a rectangle on (u, v)-plane, let ϕ bring A to ϕ(A) on (x, y)-plane. Divide A into subrectangles. Consider a sub-rectangle (u0, u0 + ∆u) × (v0, v0 + ∆v). ϕ brings this sub rectangle to a small curved rectangle. Approximate this curved rectangle by linear approximation.

v u x y ϕ A ϕ(A) (u0, v0) ∆u ϕ(u0, v0) ∆v

r(t) r(t + ∆t) r′(t)∆t Figure : r(t + ∆t) − r(t) ≈ r′(t)∆t.

Te curved rectangle is approximated by a parallelogram spanned by the vectors ∂ϕ

∂u (u0, v0)∆u and ∂ϕ ∂v (u0, v0)∆v. Te area of this

parallelogram is

• det

∂ϕ ∂u (u0, v0)∆u, ∂ϕ ∂v (u0, v0)∆v

• = |det Jϕ(u0, v0)|∆u∆v.

Polar coordinates

A point P = (x, y) can be described by a pair (r, θ), where r ≥ 0 is the length OP and 0 ≤ θ ≤ 2π is the angle from the vector (1, 0) to the vector − →

• OP. Tis is a change of variables

(r, θ) → (x, y) = (r cos θ, r sin θ). where ∂(x, y) ∂(r, θ) (r, θ) = r. Symbolically, dxdy = r drdθ.

Spherical coordinates

A point P = (x, y, z) in R3 could be described by the triple (ρ, φ, θ), where ρ is the distance OP, φ is the angle from the vector (0, 0, 1) to the vector − → OP, and let M = (x, y, 0) be the projection of P to Oxy then θ is the angle from (1, 0, 0) to − →

• OM. We have a change of

variables (ρ, φ, θ) → (x, y, z) = (ρ sin φ cos θ, ρ sin φ sin θ, ρ cos φ) where det Jϕ(ρ, φ, θ) = ρ2 sin φ.

Paths

◮ A path is a map from an interval [a, b] to Rn. ◮ Te set of images of the path is the trace of the path.

A path r : [a, b] → Rn is:

◮ closed if r(a) = r(b). ◮ simple if r has no self-intersection. ◮ smooth if r is a smooth function, i,e. having continuous

derivative.

◮ regular if its velocity r′(t) is never 0.

Path integrals of type 1

f : real function defined on the trace of a path r : [a, b] → Rn. Want to compute the sum of values of the function on the path. Take a partition a = t0 < t1 < · · · < tn = b.

r(ti−1) r(ti) r′(ti−1)∆ti Figure : Linear approximation: r(ti) − r(ti−1) ≈ r′(ti−1)∆ti.

On a subinterval [ti−1, ti]:

◮ approximate the path linearly:

r(t) − r(ti−1) ≈ r′(ti−1)(t − ti−1),

◮ approximate f by the constant function f (r(ti−1)).

Sum of values of f is approximated by

n

• i=1

f (r(ti−1))|r′(ti−1)|∆ti.

Definition

Path integrals of type 1: ˆ

r

f ds = ˆ b

a

f (r(t))|r′(t)| dt. If f ≡ 1 then ´

r 1 ds =

´ b

a |r′(t)| dt is the length of the path r.

Path integrals of type 2

F: a vector field defined on the trace of a path r : [a, b] → Rn. Want to compute the sum of tangent components of the field along the path. On each subinterval [ti−1, ti]:

◮ approximate the path linearly: r(t) ≈ r′(ti−1)(t − ti−1), ◮ approximate the field F by a constant field: F(r(ti−1)).

Te sum of the tangent components of F is approximated by F(r(ti−1)), r′(ti−1)∆ti

Te sum of the tangent components of F on the path is approximated by

n

• i=1

F(r(ti−1)), r′(ti−1)∆ti.

Definition

Path integrals of type 2: ˆ

r

F · dr = ˆ b

a

F(r(t)) · r′(t) dt.

Example

If an object moves on a path r under a force field F then ´

r F · dr is

the work of F.

Write ˆ

C

P(x, y) dx = ˆ b

a

P(x(t), y(t))x′(t) dt. ˆ

C

Q(x, y) dy = ˆ b

a

Q(x(t), y(t))y′(t) dt. So ˆ

C

F dr = ˆ

C

P(x, y) dx + ˆ

C

Q(x, y) dy.

Integrals over curves

Theorem

• 1. Integrals of type 1 along two simple regular paths with same

traces are equal.

• 2. Integrals of type 2 along two simple regular paths with same

traces are equal if the paths have same orientations, but are

• pposite if the paths have opposite orientations.

Now can discuss integrals over set of points (i.e. curves).

Conservative fields

Definition

A vector field F is conservative if there is a real function f , called the potential function of F, such that ∇f = F.

Example

Te gravitational field is conservative.

Theorem

Let r be a path from A to B, f be a real function on the trace of r. Ten: ˆ

r

∇f · dr = f (B) − f (A).

Corollary

For continuous conservative vector fields over smooth paths:

• 1. Path integrals only depend on initial points and terminal points of

paths.

• 2. Path integrals over closed paths equal 0.

A necessary condition for conservative fields

Theorem

If the field F = (P, Q) is smooth and conservative on an open set D ⊂ R2 then ∂P ∂y = ∂Q ∂x .

Proof.

If f is a potential function of F then ∂P ∂y = ∂2f ∂y∂x = ∂Q ∂x = ∂2f ∂x∂y .

Green theorem

Theorem

Let D be a simple region whose boundary is positively oriented. Let (P, Q) be a smooth vector field on D. Ten ˆ

∂D

P dx + Q dy = ¨

D

∂Q ∂x − ∂P ∂y

• dA.

A sufficient condition for conservative fields

A set D ⊂ R2 is called a star-shaped region if there is a point p0 ∈ D such that for all p ∈ D the segment from p0 to p is contained in D.

p0 p p + h

• i

Figure : A star-shaped region.

Example

◮ Te plane is a star-shaped region. ◮ A convex region is a star-shaped region. ◮ Te plane minus a point is not a star-shaped region.

Theorem

Let F = (P, Q) be a smooth field on an open star-shaped region D ⊂ R2. If ∂P

∂y = ∂Q ∂x on D then F is conservative.

Surfaces

A surface is a map r : D ⊂ R2 → R3. Te image r(D) is the trace of the surface. Te surface r is called:

◮ simple if r is injective. ◮ smooth if r is smooth. ◮ regular if it is smooth and the two vectors ru(u, v) = ∂r ∂u(u, v)

and rv(u, v) = ∂r

∂v(u, v) are defined and ru(u, v) × rv(u, v) is

always different from 0.

Surface areas

A sub rectangle ∆u × ∆v is mapped to a curved piece on the surface. Te surface is linearly approximated by the vectors ru(u, v)∆u and rv(u, v)∆v.

v u x z r (u, v) ∆u r(u, v) ∆v y S D

Te area of the piece is |ru(u, v) × rv(u, v)|∆u∆v.

Definition

Te area of the surface r : D → R3 is ¨

D

|ru × rv| dA.

Surface integrals of type 1

Consider a surface r : D → R3 whose trace is S = r(D). Let f be a real function on S. Want to compute the sum of values of the function on the surface. For each small curved piece on the surface:

◮ Linearly approximate the piece by a parallelogram whose area

is |ru(u, v) × rv(u, v)|∆u∆v,

◮ Approximate f by its value at r(u, v).

Definition

Surface integrals of type 1: ¨

r

f dS = ¨

D

f (r(u, v))|ru(u, v) × rv(u, v)| dA.

Example

If f ≡ 1 then ˜

r 1 dS is the area of the surface r.

Surface integrals of type 2

Consider a surface r : D → R3 whose trace is S = r(D). Let F be a vector field on S, i.e. F : S → R3. Want to compute total value of the normal components of the field on the surface.

◮ A small curved piece on the surface is linearly approximated by

a parallelogram whose area is |ru(u, v) × rv(u, v)|∆u∆v.

◮ On this piece F is approximated by its value at r(u, v).

Te unit normal vector at r(u, v) is ru(u, v) × rv(u, v) |ru(u, v) × rv(u, v)|. Te normal component of F(r(u, v)) is F(r(u, v)) · ru(u, v) × rv(u, v) |ru(u, v) × rv(u, v)|.

Te total value of the normal component of F on this piece is approximated by F(r(u, v)) · ru(u, v) × rv(u, v) |ru(u, v) × rv(u, v)||ru(u, v) × rv(u, v)|∆u∆v.

Definition

Surface integrals of type 2: ¨

r

F · d S = ¨

D

F(r(u, v)) · (ru(u, v) × rv(u, v)) dA.

Integrals on surfaces

Theorem

Consider two simple regular surfaces with same traces defined on closed bounded subsets of R2 whose boundaries are of measures zero:

• 1. Surface integrals of type 1 are same.
• 2. Surface integrals of type 2 are same if the surfaces have same
• rientations, and opposite if the surfaces have opposite
• rientations.

To compute integrals of type 1, take a any parametrization. To compute integrals of type 2, take a parametrization with same

• rientation as that of the surface.

Also: ¨

S

F · n dS = ¨

S

F · d S.

curl

Definition

Let F = (P, Q, R) be a field on R3: curl F =

• ∂R

∂y − ∂Q ∂z , ∂P ∂z − ∂R ∂x , ∂Q ∂x − ∂P ∂y

• .

Denote ∇ =

• ∂

∂x , ∂ ∂y , ∂ ∂z

• ,

then curl F = ∇ × F. curl F is also called the rotational field of F.

Theorem

If f is continuously differentiable to second order then curl ∇f = 0.

Corollary

If F is conservative then curl F = 0.

Stokes theorem

Theorem

Let F be a smooth vector field on an open set containing a surface S which is the graph of a function f : D → R where D is a plane region for which Green theorem can be applied. Suppose that the boundary of S is a curve ∂S whose orientation is induced from the orientation of ∂D via the parametrization (x, y, f (x, y)). Ten: ˆ

∂S

F · d r = ¨

S

(curl F) · n dS = ¨

S

curl F · d S.

div

Definition

Let F = (P, Q, R) be a vector field on R3: div F = ∂P ∂x + ∂Q ∂y + ∂R ∂z . Symbolically: div F = ∇ · F. div F is also called the divergence function of F.

Theorem

If F is continuously differentiable to second order then div curl F = 0.

Gauss-Ostrogradsky theorem

Theorem

Let F be smooth on an open set containing a simple region E in R3 whose boundary ∂E is oriented outward. Ten: ¨

∂E

F · n dS = ¨

∂E

F · d S = ˚

E

div F dV.