From Math 2220 Class 36 Integrals Other Problems Dr. Allen Back - - PowerPoint PPT Presentation

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From Math 2220 Class 36 Integrals Other Problems Dr. Allen Back - - PowerPoint PPT Presentation

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From Math 2220 Class 36 V2c Surface Integrals Why Surface From Math 2220 Class 36 Integrals Other Problems Dr. Allen Back Why Greens Greens Problems Flow Lines Nov. 19, 2014 Div and Curl Surface of Revolution Case Graph Case

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From Math 2220 Class 36 V2c Surface Integrals Why Surface Integrals Other Problems Why Green’s Greens Problems Flow Lines Div and Curl Surface of Revolution Case Graph Case Surface Parametriza- tion

From Math 2220 Class 36

  • Dr. Allen Back
  • Nov. 19, 2014
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From Math 2220 Class 36 V2c Surface Integrals Why Surface Integrals Other Problems Why Green’s Greens Problems Flow Lines Div and Curl Surface of Revolution Case Graph Case Surface Parametriza- tion

Surface Integrals

Picture of Tu, Tv for a Lat/Long Param. of the Sphere.

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From Math 2220 Class 36 V2c Surface Integrals Why Surface Integrals Other Problems Why Green’s Greens Problems Flow Lines Div and Curl Surface of Revolution Case Graph Case Surface Parametriza- tion

Surface Integrals

Basic Parametrization Picture

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From Math 2220 Class 36 V2c Surface Integrals Why Surface Integrals Other Problems Why Green’s Greens Problems Flow Lines Div and Curl Surface of Revolution Case Graph Case Surface Parametriza- tion

Surface Integrals

Parametrization Φ(u, v) = (x(u, v), y(u, v), z(u, v)) Tangents Tu = (xu, yu, zu) Tv = (xv, yv, zv) Area Element dS = Tu × Tv du dv Normal N = Tu × Tv Unit normal ˆ n = ±

  • Tu ×

Tv|

  • Tu ×

Tv (Choosing the ± sign corresponds to an orientation of the surface.)

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From Math 2220 Class 36 V2c Surface Integrals Why Surface Integrals Other Problems Why Green’s Greens Problems Flow Lines Div and Curl Surface of Revolution Case Graph Case Surface Parametriza- tion

Surface Integrals

Two Kinds of Surface Integrals Surface Integral of a scalar function f (x, y, z) :

  • S

f (x, y, z) dS Surface Integral of a vector field F(x, y, z) :

  • S
  • F(x, y, z) · ˆ

n dS.

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From Math 2220 Class 36 V2c Surface Integrals Why Surface Integrals Other Problems Why Green’s Greens Problems Flow Lines Div and Curl Surface of Revolution Case Graph Case Surface Parametriza- tion

Surface Integrals

Surface Integral of a scalar function f (x, y, z) calculated by

  • S

f (x, y, z) dS =

  • D

f (Φ(u, v)) Tu × Tv du dv where D is the domain of the parametrization Φ. Surface Integral of a vector field F(x, y, z) calculated by

  • S
  • F(x, y, z) · ˆ

n dS = ±

  • D
  • F(Φ(u, v)) ·

Tu × Tv|

  • Tu ×

Tv

  • Tu ×

Tv du dv where D is the domain of the parametrization Φ.

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From Math 2220 Class 36 V2c Surface Integrals Why Surface Integrals Other Problems Why Green’s Greens Problems Flow Lines Div and Curl Surface of Revolution Case Graph Case Surface Parametriza- tion

Surface Integrals

3d Flux Picture

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From Math 2220 Class 36 V2c Surface Integrals Why Surface Integrals Other Problems Why Green’s Greens Problems Flow Lines Div and Curl Surface of Revolution Case Graph Case Surface Parametriza- tion

Surface Integrals

The preceding picture can be used to argue that if F(x, y, z) is the velocity vector field, e.g. of a fluid of density ρ(x, y, z), then the surface integral

S

ρ F · ˆ n dS (with associated Riemann Sum

  • ρ(x∗

i , y∗ j , z∗ k)

F(x∗

i , y∗ j , z∗ k) · ˆ

n(x∗

i , y∗ j , z∗ k) ∆Sijk)

represents the rate at which material (e.g. grams per second) crosses the surface.

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From Math 2220 Class 36 V2c Surface Integrals Why Surface Integrals Other Problems Why Green’s Greens Problems Flow Lines Div and Curl Surface of Revolution Case Graph Case Surface Parametriza- tion

Surface Integrals

From this point of view the orientation of a surface simple tells us which side is accumulatiing mass, in the case where the value of the integral is positive.

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From Math 2220 Class 36 V2c Surface Integrals Why Surface Integrals Other Problems Why Green’s Greens Problems Flow Lines Div and Curl Surface of Revolution Case Graph Case Surface Parametriza- tion

Surface Integrals

2d Flux Picture There’s an analagous 2d Riemann sum and interp of

  • C
  • F · ˆ

n ds.

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From Math 2220 Class 36 V2c Surface Integrals Why Surface Integrals Other Problems Why Green’s Greens Problems Flow Lines Div and Curl Surface of Revolution Case Graph Case Surface Parametriza- tion

Surface Integrals

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From Math 2220 Class 36 V2c Surface Integrals Why Surface Integrals Other Problems Why Green’s Greens Problems Flow Lines Div and Curl Surface of Revolution Case Graph Case Surface Parametriza- tion

Surface Integrals

Problem: Calculate

  • S
  • F(x, y, z) · ˆ

n dS for the vector field F(x, y, z) = (x, y, z) and S the part of the paraboloid z = 1 − x2 − y2 above the xy-plane. Choose the positive orientation of the paraboloid to be the one with normal pointing downward.

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From Math 2220 Class 36 V2c Surface Integrals Why Surface Integrals Other Problems Why Green’s Greens Problems Flow Lines Div and Curl Surface of Revolution Case Graph Case Surface Parametriza- tion

Surface Integrals

Problem: Calculate the surface area of the above paraboloid.

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From Math 2220 Class 36 V2c Surface Integrals Why Surface Integrals Other Problems Why Green’s Greens Problems Flow Lines Div and Curl Surface of Revolution Case Graph Case Surface Parametriza- tion

Why Surface Integrals

The Vector Case

1 Electromagnetism. (Two of Maxwell’s equations involve

surface integrals.)

2 Partial differential equations. (Integration by parts is an

important technique here, and is based on the chapter 8 integral theorems.)

3 Heat, fluid flow and difffusion. 4 (Related to Markov chains in probability.) 5 At an advanced level, financial mathematics. (e.g.

Black-Scholes is similar to the heat equation.)

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From Math 2220 Class 36 V2c Surface Integrals Why Surface Integrals Other Problems Why Green’s Greens Problems Flow Lines Div and Curl Surface of Revolution Case Graph Case Surface Parametriza- tion

Why Surface Integrals

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From Math 2220 Class 36 V2c Surface Integrals Why Surface Integrals Other Problems Why Green’s Greens Problems Flow Lines Div and Curl Surface of Revolution Case Graph Case Surface Parametriza- tion

Why Surface Integrals

The Scalar Case

1 As with multiple integrals, going from a density of

something (this time per unit surface area) to a total.

2 Vector surface integrals are surface integrals of the scalar

function “normal component.”

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From Math 2220 Class 36 V2c Surface Integrals Why Surface Integrals Other Problems Why Green’s Greens Problems Flow Lines Div and Curl Surface of Revolution Case Graph Case Surface Parametriza- tion

Why Surface Integrals

1d-Heat Equation ∂T ∂t = c ∂2T ∂x2 (T(x, t) is the temperature at position x and time t along a rod.) Black-Scholes ∂V ∂t + cS2 ∂2V ∂S2 + rS ∂V ∂S − rV = 0. (V (S, t) is the price of a derivative as a function

  • f an underlying stock price S at time t.)

(See e.g. “Black-Scholes model” in Wikipedia.)

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From Math 2220 Class 36 V2c Surface Integrals Why Surface Integrals Other Problems Why Green’s Greens Problems Flow Lines Div and Curl Surface of Revolution Case Graph Case Surface Parametriza- tion

Other Problems

Problem: Find

  • S
  • F · ˆ

n dS for F(x, y, z) = (0, yz, z2) and S the portion of the cylinder y2 + z2 = 1 with 0 ≤ x ≤ 1, z ≥ 0, and the positive

  • rientation chosen to be a radial outward (from the axis of the

cylinder) normal.

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From Math 2220 Class 36 V2c Surface Integrals Why Surface Integrals Other Problems Why Green’s Greens Problems Flow Lines Div and Curl Surface of Revolution Case Graph Case Surface Parametriza- tion

Other Problems

Problem: Letting r denote the vector field (x, y, z), find the value of the surface integral

  • S
  • r
  • r3 · ˆ

n dS where S is the sphere of radius R about the origin, oriented with an outward normal.

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From Math 2220 Class 36 V2c Surface Integrals Why Surface Integrals Other Problems Why Green’s Greens Problems Flow Lines Div and Curl Surface of Revolution Case Graph Case Surface Parametriza- tion

Other Problems

Spherical coordinates (letting ρ = R be constant) gives a natural parametrization of the sphere of radius R centered at the origin. It is geometrically to be expected (and you’ve done some verifications like this at least in look at problems) that the Jacobian result for spherical coordinates translates to the area element on such a sphere to be dS = R2 sin φ dφ dθ.

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From Math 2220 Class 36 V2c Surface Integrals Why Surface Integrals Other Problems Why Green’s Greens Problems Flow Lines Div and Curl Surface of Revolution Case Graph Case Surface Parametriza- tion

Why Green’s

Green’s theorem says that for simple closed (piecewise smooth) curve C whose inside is a region R, we have

  • C

P(x, y) dx + Q(x, y) dy =

  • R

∂Q ∂x − ∂P ∂y dx dy as long as the vector field F(x, y) = (P(x, y), Q(x, y)) is C 1

  • n the set R and C is given its usual “inside to the left”
  • rientation.
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From Math 2220 Class 36 V2c Surface Integrals Why Surface Integrals Other Problems Why Green’s Greens Problems Flow Lines Div and Curl Surface of Revolution Case Graph Case Surface Parametriza- tion

Why Green’s

For a y-simple region

  • R

−Py dy dx =

  • C=∂R

P dx. is fairly easily justified.

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From Math 2220 Class 36 V2c Surface Integrals Why Surface Integrals Other Problems Why Green’s Greens Problems Flow Lines Div and Curl Surface of Revolution Case Graph Case Surface Parametriza- tion

Why Green’s

For an x-simple region

  • R

Qx dx dy =

  • C=∂R

Q dy.

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From Math 2220 Class 36 V2c Surface Integrals Why Surface Integrals Other Problems Why Green’s Greens Problems Flow Lines Div and Curl Surface of Revolution Case Graph Case Surface Parametriza- tion

Why Green’s

So for a region that is both y-simple and x-simple we have Green:

  • ∂R

P(x, y) dx + Q(x, y) dy =

  • R

Qy − Px dx dy.

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From Math 2220 Class 36 V2c Surface Integrals Why Surface Integrals Other Problems Why Green’s Greens Problems Flow Lines Div and Curl Surface of Revolution Case Graph Case Surface Parametriza- tion

Why Green’s

Intuitively, why the different signs? +Qx yet −Py. And why this combination of Qx − Py?

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From Math 2220 Class 36 V2c Surface Integrals Why Surface Integrals Other Problems Why Green’s Greens Problems Flow Lines Div and Curl Surface of Revolution Case Graph Case Surface Parametriza- tion

Why Green’s

Intuitively, why the different signs? +Qx yet −Py. And why this combination of Qx − Py? Think about the line integral around a small rectangle with sides ∆x and ∆y. If you assume (or justify) that evaluating the vector field in the middle of each edge gives a good approximation in the line integral, then Qx − Py emerges quite naturally.

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From Math 2220 Class 36 V2c Surface Integrals Why Surface Integrals Other Problems Why Green’s Greens Problems Flow Lines Div and Curl Surface of Revolution Case Graph Case Surface Parametriza- tion

Why Green’s

If you can cut a region into two pieces where we know Green’s holds on each piece (e.g. a ring shaped region), then Green also holds for the entire region. (Because the line integrals over the “cuts” show up twice with opposite signs (think “inside to the left”) and cancel.

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From Math 2220 Class 36 V2c Surface Integrals Why Surface Integrals Other Problems Why Green’s Greens Problems Flow Lines Div and Curl Surface of Revolution Case Graph Case Surface Parametriza- tion

Why Green’s

So in the end, Green’s theorem holds for regions whose boundaries include several closed curves (multiply-connected regions) as long as we orient each boundar curve according to the inside to the left rule.

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From Math 2220 Class 36 V2c Surface Integrals Why Surface Integrals Other Problems Why Green’s Greens Problems Flow Lines Div and Curl Surface of Revolution Case Graph Case Surface Parametriza- tion

Greens Problems

1 2

  • C x dy − y dx for C the boundary of the ellipse

x2 32 + y2 42 = 1

  • riented counterclockwise.
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From Math 2220 Class 36 V2c Surface Integrals Why Surface Integrals Other Problems Why Green’s Greens Problems Flow Lines Div and Curl Surface of Revolution Case Graph Case Surface Parametriza- tion

Greens Problems

Let

  • F =

1 x2 + y2 (−y, x). If C1 and C2 are two simple closed curves enclosing the origin (and oriented with the usual inside to the left), can you say whether one of

  • C1

F · d s and

  • C2

F · d s is bigger than the

  • ther?
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From Math 2220 Class 36 V2c Surface Integrals Why Surface Integrals Other Problems Why Green’s Greens Problems Flow Lines Div and Curl Surface of Revolution Case Graph Case Surface Parametriza- tion

Flow Lines

Definition: A path c : [a, b] → R3 is said to be a flow line (or integral curve) of a vector field F(x, y, z) if

  • c′(t) =

F(c(t)) for all t ∈ [a, b].

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From Math 2220 Class 36 V2c Surface Integrals Why Surface Integrals Other Problems Why Green’s Greens Problems Flow Lines Div and Curl Surface of Revolution Case Graph Case Surface Parametriza- tion

Flow Lines

Flow lines always exist at least locally because of existence and uniqueness theorems for ordinary differential equations.

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From Math 2220 Class 36 V2c Surface Integrals Why Surface Integrals Other Problems Why Green’s Greens Problems Flow Lines Div and Curl Surface of Revolution Case Graph Case Surface Parametriza- tion

Flow Lines

At the end of Math 2210, you studied how to use eigenvalues/eigenvectors/diagonalization/(Jordan Normal form) to solve linear differential equations.

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From Math 2220 Class 36 V2c Surface Integrals Why Surface Integrals Other Problems Why Green’s Greens Problems Flow Lines Div and Curl Surface of Revolution Case Graph Case Surface Parametriza- tion

Flow Lines

Show that c(t) = (x0et, y0et) is a flow line of F(x, y) = (x, y).

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From Math 2220 Class 36 V2c Surface Integrals Why Surface Integrals Other Problems Why Green’s Greens Problems Flow Lines Div and Curl Surface of Revolution Case Graph Case Surface Parametriza- tion

Flow Lines

Show that c(t) = x(t) y(t)

  • =

cos t − sin t sin t cos t x0 y0

  • is a flow line of

F(x, y) = (−y, x).

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From Math 2220 Class 36 V2c Surface Integrals Why Surface Integrals Other Problems Why Green’s Greens Problems Flow Lines Div and Curl Surface of Revolution Case Graph Case Surface Parametriza- tion

Div and Curl

We will discuss the meaning of these when we get to the 3d integral theorems (8.2 and 8.4.) For a vector field F(x, y, z) = (P, Q, R) div( F) is a (scalar) function; i.e. a number at each point. curl( F) is another vector field; i.e. a vector at each point.

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From Math 2220 Class 36 V2c Surface Integrals Why Surface Integrals Other Problems Why Green’s Greens Problems Flow Lines Div and Curl Surface of Revolution Case Graph Case Surface Parametriza- tion

Div and Curl

div( F) = Px + Qy + Rz for F(x, y, z) = (P, Q, R). (∇ · F).

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From Math 2220 Class 36 V2c Surface Integrals Why Surface Integrals Other Problems Why Green’s Greens Problems Flow Lines Div and Curl Surface of Revolution Case Graph Case Surface Parametriza- tion

Div and Curl

curl( F) =   ˆ i ˆ j ˆ k

∂ ∂x ∂ ∂y ∂ ∂z

P Q R   for F(x, y, z) = (P, Q, R). (∇ × F).

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From Math 2220 Class 36 V2c Surface Integrals Why Surface Integrals Other Problems Why Green’s Greens Problems Flow Lines Div and Curl Surface of Revolution Case Graph Case Surface Parametriza- tion

Div and Curl

Please take a look at the table of vector analysis identities on page 255. Many are just the 1-variable product or chain rules.

1 ∇(fg) 2 div(f

F)

3 curl(f

F)

4 curl(

G × F) would be more complicated.

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From Math 2220 Class 36 V2c Surface Integrals Why Surface Integrals Other Problems Why Green’s Greens Problems Flow Lines Div and Curl Surface of Revolution Case Graph Case Surface Parametriza- tion

Surface of Revolution Case

This is not worth memorizing! If one rotates about the z-axis the path (curve) z = f (x) in the xz-plane for 0 ≤ a ≤ x ≤ b, one obtains a surface of revolution with a parametrization Φ(u, v) = (u cos v, u sin v, f (u)) and dS =?

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From Math 2220 Class 36 V2c Surface Integrals Why Surface Integrals Other Problems Why Green’s Greens Problems Flow Lines Div and Curl Surface of Revolution Case Graph Case Surface Parametriza- tion

Surface of Revolution Case

dS = u

  • 1 + (f ′(u))2 du dv.
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From Math 2220 Class 36 V2c Surface Integrals Why Surface Integrals Other Problems Why Green’s Greens Problems Flow Lines Div and Curl Surface of Revolution Case Graph Case Surface Parametriza- tion

Graph Case

This is not worth memorizing! For the graph parametrization of z = f (x, y), Φ(u, v) = (u, v, f (u, v)) and dS =?

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From Math 2220 Class 36 V2c Surface Integrals Why Surface Integrals Other Problems Why Green’s Greens Problems Flow Lines Div and Curl Surface of Revolution Case Graph Case Surface Parametriza- tion

Graph Case

dS =

  • 1 + f 2

u + f 2 v du dv.

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From Math 2220 Class 36 V2c Surface Integrals Why Surface Integrals Other Problems Why Green’s Greens Problems Flow Lines Div and Curl Surface of Revolution Case Graph Case Surface Parametriza- tion

Graph Case

For such a graph, the normal to the surface at a point (x, y, f (x, y)) (this is the level set z − f (x, y) = 0) is (−fx, −fy, 1) so we can see that cos γ = 1

  • 1 + f 2

x + f 2 y

determines the angle γ of the normal with the z-axis. And so at the point (u, v, f (u, v)) on a graph, dS = 1 cos γ du dv. (Note that du dv is essentially the same as dx dy here.)

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From Math 2220 Class 36 V2c Surface Integrals Why Surface Integrals Other Problems Why Green’s Greens Problems Flow Lines Div and Curl Surface of Revolution Case Graph Case Surface Parametriza- tion

Paraboloid z = x2 + 4y 2

The graph z = F(x, y) can always be parameterized by Φ(u, v) =< u, v, F(u, v) > .

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From Math 2220 Class 36 V2c Surface Integrals Why Surface Integrals Other Problems Why Green’s Greens Problems Flow Lines Div and Curl Surface of Revolution Case Graph Case Surface Parametriza- tion

Paraboloid z = x2 + 4y 2

The graph z = F(x, y) can always be parameterized by Φ(u, v) =< u, v, F(u, v) > . Parameters u and v just different names for x and y resp.

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From Math 2220 Class 36 V2c Surface Integrals Why Surface Integrals Other Problems Why Green’s Greens Problems Flow Lines Div and Curl Surface of Revolution Case Graph Case Surface Parametriza- tion

Paraboloid z = x2 + 4y 2

The graph z = F(x, y) can always be parameterized by Φ(u, v) =< u, v, F(u, v) > . Use this idea if you can’t think of something better.

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From Math 2220 Class 36 V2c Surface Integrals Why Surface Integrals Other Problems Why Green’s Greens Problems Flow Lines Div and Curl Surface of Revolution Case Graph Case Surface Parametriza- tion

Paraboloid z = x2 + 4y 2

The graph z = F(x, y) can always be parameterized by Φ(u, v) =< u, v, F(u, v) > .

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From Math 2220 Class 36 V2c Surface Integrals Why Surface Integrals Other Problems Why Green’s Greens Problems Flow Lines Div and Curl Surface of Revolution Case Graph Case Surface Parametriza- tion

Paraboloid z = x2 + 4y 2

The graph z = F(x, y) can always be parameterized by Φ(u, v) =< u, v, F(u, v) > . Note the curves where u and v are constant are visible in the wireframe.

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From Math 2220 Class 36 V2c Surface Integrals Why Surface Integrals Other Problems Why Green’s Greens Problems Flow Lines Div and Curl Surface of Revolution Case Graph Case Surface Parametriza- tion

Paraboloid z = x2 + 4y 2

A trigonometric parametrization will often be better if you have to calculate a surface integral.

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From Math 2220 Class 36 V2c Surface Integrals Why Surface Integrals Other Problems Why Green’s Greens Problems Flow Lines Div and Curl Surface of Revolution Case Graph Case Surface Parametriza- tion

Paraboloid z = x2 + 4y 2

A trigonometric parametrization will often be better if you have to calculate a surface integral. Φ(u, v) =< 2u cos v, u sin v, 4u2 > .

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From Math 2220 Class 36 V2c Surface Integrals Why Surface Integrals Other Problems Why Green’s Greens Problems Flow Lines Div and Curl Surface of Revolution Case Graph Case Surface Parametriza- tion

Paraboloid z = x2 + 4y 2

A trigonometric parametrization will often be better if you have to calculate a surface integral. Φ(u, v) =< 2u cos v, u sin v, 4u2 > .

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From Math 2220 Class 36 V2c Surface Integrals Why Surface Integrals Other Problems Why Green’s Greens Problems Flow Lines Div and Curl Surface of Revolution Case Graph Case Surface Parametriza- tion

Paraboloid z = x2 + 4y 2

A trigonometric parametrization will often be better if you have to calculate a surface integral. Φ(u, v) =< 2u cos v, u sin v, 4u2 > . Algebraically, we are rescaling the algebra behind polar coordinates where x = r cos θ y = r sin θ leads to r2 = x2 + y2.

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From Math 2220 Class 36 V2c Surface Integrals Why Surface Integrals Other Problems Why Green’s Greens Problems Flow Lines Div and Curl Surface of Revolution Case Graph Case Surface Parametriza- tion

Paraboloid z = x2 + 4y 2

A trigonometric parametrization will often be better if you have to calculate a surface integral. Φ(u, v) =< 2u cos v, u sin v, 4u2 > . Here we want x2 + 4y2 to be simple. So x = 2r cos θ y = r sin θ will do better.

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From Math 2220 Class 36 V2c Surface Integrals Why Surface Integrals Other Problems Why Green’s Greens Problems Flow Lines Div and Curl Surface of Revolution Case Graph Case Surface Parametriza- tion

Paraboloid z = x2 + 4y 2

A trigonometric parametrization will often be better if you have to calculate a surface integral. Φ(u, v) =< 2u cos v, u sin v, 4u2 > . Here we want x2 + 4y2 to be simple. So x = 2r cos θ y = r sin θ will do better. Plug x and y into z = x2 + 4y2 to get the z-component.

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From Math 2220 Class 36 V2c Surface Integrals Why Surface Integrals Other Problems Why Green’s Greens Problems Flow Lines Div and Curl Surface of Revolution Case Graph Case Surface Parametriza- tion

Parabolic Cylinder z = x2

Graph parametrizations are often optimal for parabolic cylinders.

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From Math 2220 Class 36 V2c Surface Integrals Why Surface Integrals Other Problems Why Green’s Greens Problems Flow Lines Div and Curl Surface of Revolution Case Graph Case Surface Parametriza- tion

Parabolic Cylinder z = x2

Φ(u, v) =< u, v, u2 >

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From Math 2220 Class 36 V2c Surface Integrals Why Surface Integrals Other Problems Why Green’s Greens Problems Flow Lines Div and Curl Surface of Revolution Case Graph Case Surface Parametriza- tion

Parabolic Cylinder z = x2

Φ(u, v) =< u, v, u2 >

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From Math 2220 Class 36 V2c Surface Integrals Why Surface Integrals Other Problems Why Green’s Greens Problems Flow Lines Div and Curl Surface of Revolution Case Graph Case Surface Parametriza- tion

Parabolic Cylinder z = x2

Φ(u, v) =< u, v, u2 > One of the parameters (v) is giving us the “extrusion”

  • direction. The parameter u is just being used to describe the

curve z = x2 in the zx plane.

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From Math 2220 Class 36 V2c Surface Integrals Why Surface Integrals Other Problems Why Green’s Greens Problems Flow Lines Div and Curl Surface of Revolution Case Graph Case Surface Parametriza- tion

Elliptic Cylinder x2 + 2z2 = 6

The trigonometric trick is often good for elliptic cylinders

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From Math 2220 Class 36 V2c Surface Integrals Why Surface Integrals Other Problems Why Green’s Greens Problems Flow Lines Div and Curl Surface of Revolution Case Graph Case Surface Parametriza- tion

Elliptic Cylinder x2 + 2z2 = 6

Φ(u, v) =< √ 3· √ 2 cos v, u, √ 3 sin v >=< √ 6 cos v, u, √ 3 sin v >

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From Math 2220 Class 36 V2c Surface Integrals Why Surface Integrals Other Problems Why Green’s Greens Problems Flow Lines Div and Curl Surface of Revolution Case Graph Case Surface Parametriza- tion

Elliptic Cylinder x2 + 2z2 = 6

Φ(u, v) =< √ 6 cos v, u, √ 3 sin v >

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From Math 2220 Class 36 V2c Surface Integrals Why Surface Integrals Other Problems Why Green’s Greens Problems Flow Lines Div and Curl Surface of Revolution Case Graph Case Surface Parametriza- tion

Elliptic Cylinder x2 + 2z2 = 6

Φ(u, v) =< √ 6 cos v, u, √ 3 sin v >

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From Math 2220 Class 36 V2c Surface Integrals Why Surface Integrals Other Problems Why Green’s Greens Problems Flow Lines Div and Curl Surface of Revolution Case Graph Case Surface Parametriza- tion

Elliptic Cylinder x2 + 2z2 = 6

Φ(u, v) =< √ 6 cos v, u, √ 3 sin v >

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From Math 2220 Class 36 V2c Surface Integrals Why Surface Integrals Other Problems Why Green’s Greens Problems Flow Lines Div and Curl Surface of Revolution Case Graph Case Surface Parametriza- tion

Elliptic Cylinder x2 + 2z2 = 6

Φ(u, v) =< √ 6 cos v, u, √ 3 sin v > What happened here is we started with the polar coordinate idea x = r cos θ z = r sin θ but noted that the algebra wasn’t right for x2 + 2z2 so shifted to x = √ 2r cos θ z = r sin θ

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From Math 2220 Class 36 V2c Surface Integrals Why Surface Integrals Other Problems Why Green’s Greens Problems Flow Lines Div and Curl Surface of Revolution Case Graph Case Surface Parametriza- tion

Elliptic Cylinder x2 + 2z2 = 6

Φ(u, v) =< √ 6 cos v, u, √ 3 sin v > x = √ 2r cos θ z = r sin θ makes the left hand side work out to 2r2 which will be 6 when r = √ 3.

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SLIDE 67

From Math 2220 Class 36 V2c Surface Integrals Why Surface Integrals Other Problems Why Green’s Greens Problems Flow Lines Div and Curl Surface of Revolution Case Graph Case Surface Parametriza- tion

Ellipsoid x2 + 2y 2 + 3z2 = 4

A similar trick occurs for using spherical coordinate ideas in parameterizing ellipsoids.

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SLIDE 68

From Math 2220 Class 36 V2c Surface Integrals Why Surface Integrals Other Problems Why Green’s Greens Problems Flow Lines Div and Curl Surface of Revolution Case Graph Case Surface Parametriza- tion

Ellipsoid x2 + 2y 2 + 3z2 = 4

A similar trick occurs for using spherical coordinate ideas in parameterizing ellipsoids. Φ(u, v) =< 2 sin u cos v, √ 2 sin u sin v,

  • 4

3 cos u >

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SLIDE 69

From Math 2220 Class 36 V2c Surface Integrals Why Surface Integrals Other Problems Why Green’s Greens Problems Flow Lines Div and Curl Surface of Revolution Case Graph Case Surface Parametriza- tion

Ellipsoid x2 + 2y 2 + 3z2 = 4

Φ(u, v) =< 2 sin u cos v, √ 2 sin u sin v,

  • 4

3 cos u >

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SLIDE 70

From Math 2220 Class 36 V2c Surface Integrals Why Surface Integrals Other Problems Why Green’s Greens Problems Flow Lines Div and Curl Surface of Revolution Case Graph Case Surface Parametriza- tion

Hyperbolic Cylinder x2 − z2 = −4

You may have run into the hyperbolic functions cosh x = ex + e−x 2 sinh x = ex − e−x 2

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SLIDE 71

From Math 2220 Class 36 V2c Surface Integrals Why Surface Integrals Other Problems Why Green’s Greens Problems Flow Lines Div and Curl Surface of Revolution Case Graph Case Surface Parametriza- tion

Hyperbolic Cylinder x2 − z2 = −4

You may have run into the hyperbolic functions cosh x = ex + e−x 2 sinh x = ex − e−x 2 Just as cos2 θ + sin2 θ = 1 helps with ellipses, the hyperbolic version cosh2 θ − sinh2 θ = 1 leads to the nicest hyperbola parameterizations.

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SLIDE 72

From Math 2220 Class 36 V2c Surface Integrals Why Surface Integrals Other Problems Why Green’s Greens Problems Flow Lines Div and Curl Surface of Revolution Case Graph Case Surface Parametriza- tion

Hyperbolic Cylinder x2 − z2 = −4

Just as cos2 θ + sin2 θ = 1 helps with ellipses, the hyperbolic version cosh2 θ − sinh2 θ = 1 leads to the nicest hyperbola parameterizations. Φ(u, v) =< 2 sinh v, u, 2 cosh v >

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SLIDE 73

From Math 2220 Class 36 V2c Surface Integrals Why Surface Integrals Other Problems Why Green’s Greens Problems Flow Lines Div and Curl Surface of Revolution Case Graph Case Surface Parametriza- tion

Hyperbolic Cylinder x2 − z2 = −4

Φ(u, v) =< 2 sinh v, u, 2 cosh v >

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SLIDE 74

From Math 2220 Class 36 V2c Surface Integrals Why Surface Integrals Other Problems Why Green’s Greens Problems Flow Lines Div and Curl Surface of Revolution Case Graph Case Surface Parametriza- tion

Saddle z = x2 − y 2

The hyperbolic trick also works with saddles

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SLIDE 75

From Math 2220 Class 36 V2c Surface Integrals Why Surface Integrals Other Problems Why Green’s Greens Problems Flow Lines Div and Curl Surface of Revolution Case Graph Case Surface Parametriza- tion

Saddle z = x2 − y 2

Φ(u, v) =< u cosh v, u sinh v, u2 >

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SLIDE 76

From Math 2220 Class 36 V2c Surface Integrals Why Surface Integrals Other Problems Why Green’s Greens Problems Flow Lines Div and Curl Surface of Revolution Case Graph Case Surface Parametriza- tion

Saddle z = x2 − y 2

Φ(u, v) =< u cosh v, u sinh v, u2 >

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SLIDE 77

From Math 2220 Class 36 V2c Surface Integrals Why Surface Integrals Other Problems Why Green’s Greens Problems Flow Lines Div and Curl Surface of Revolution Case Graph Case Surface Parametriza- tion

Hyperboloid of 1 Sheet x2 + y 2 − z2 = 1

The spherical coordinate idea for ellipsoids with sin φ replaced by cosh u works well here.

slide-78
SLIDE 78

From Math 2220 Class 36 V2c Surface Integrals Why Surface Integrals Other Problems Why Green’s Greens Problems Flow Lines Div and Curl Surface of Revolution Case Graph Case Surface Parametriza- tion

Hyperboloid of 1 Sheet x2 + y 2 − z2 = 1

Φ(u, v) =< cosh u cos v, cosh u sin v, sinh u >

slide-79
SLIDE 79

From Math 2220 Class 36 V2c Surface Integrals Why Surface Integrals Other Problems Why Green’s Greens Problems Flow Lines Div and Curl Surface of Revolution Case Graph Case Surface Parametriza- tion

Hyperboloid of 1 Sheet x2 + y 2 − z2 = 1

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SLIDE 80

From Math 2220 Class 36 V2c Surface Integrals Why Surface Integrals Other Problems Why Green’s Greens Problems Flow Lines Div and Curl Surface of Revolution Case Graph Case Surface Parametriza- tion

Hyperboloid of 2-Sheets x2 + y 2 − z2 = −1

Φ(u, v) =< sinh u cos v, sinh u sin v, cosh u >

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SLIDE 81

From Math 2220 Class 36 V2c Surface Integrals Why Surface Integrals Other Problems Why Green’s Greens Problems Flow Lines Div and Curl Surface of Revolution Case Graph Case Surface Parametriza- tion

Hyperboloid of 2-Sheets x2 + y 2 − z2 = −1

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SLIDE 82

From Math 2220 Class 36 V2c Surface Integrals Why Surface Integrals Other Problems Why Green’s Greens Problems Flow Lines Div and Curl Surface of Revolution Case Graph Case Surface Parametriza- tion

Top Part of Cone z2 = x2 + y 2

So z =

  • x2 + y2.
slide-83
SLIDE 83

From Math 2220 Class 36 V2c Surface Integrals Why Surface Integrals Other Problems Why Green’s Greens Problems Flow Lines Div and Curl Surface of Revolution Case Graph Case Surface Parametriza- tion

Top Part of Cone z2 = x2 + y 2

So z =

  • x2 + y2.

The polar coordinate idea leads to Φ(u, v) =< u cos v, u sin v, u >

slide-84
SLIDE 84

From Math 2220 Class 36 V2c Surface Integrals Why Surface Integrals Other Problems Why Green’s Greens Problems Flow Lines Div and Curl Surface of Revolution Case Graph Case Surface Parametriza- tion

Top Part of Cone z2 = x2 + y 2

So z =

  • x2 + y2.

The polar coordinate idea leads to Φ(u, v) =< u cos v, u sin v, u >

slide-85
SLIDE 85

From Math 2220 Class 36 V2c Surface Integrals Why Surface Integrals Other Problems Why Green’s Greens Problems Flow Lines Div and Curl Surface of Revolution Case Graph Case Surface Parametriza- tion

Mercator Parametrization of the Sphere

For 0 ≤ v ≤ ∞, 0 ≤ u ≤ 2π Φ(u, v) = (sech(v) cos u, sech(v) sin u, tanh(v)). (Note tanh2(v) + sech2(v) = 1)

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SLIDE 86

From Math 2220 Class 36 V2c Surface Integrals Why Surface Integrals Other Problems Why Green’s Greens Problems Flow Lines Div and Curl Surface of Revolution Case Graph Case Surface Parametriza- tion

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