From Math 2220 Class 37 Div and Curl Why Greens Greens Dr. Allen - - PowerPoint PPT Presentation

From Math 2220 Class 37 V1c Symmetries Can Help Flow Lines Div and Curl Why Green’s Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface

From Math 2220 Class 37

• Dr. Allen Back
• Nov. 21, 2014

From Math 2220 Class 37 V1c Symmetries Can Help Flow Lines Div and Curl Why Green’s Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface

Symmetries Can Help

The most important thing you should know (e.g. for exams and homework) is how to setup (and perhaps compute if not too hard) surface integrals, triple integrals, etc. But occasionally symmetries can save a considerable amount of work.

From Math 2220 Class 37 V1c Symmetries Can Help Flow Lines Div and Curl Why Green’s Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface

Symmetries Can Help

Problem: Calculate the following surface integral:

• S

y2 − z2 dS for S the part of the paraboloid x = y2 + z2 with 1 ≤ x ≤ 5.

From Math 2220 Class 37 V1c Symmetries Can Help Flow Lines Div and Curl Why Green’s Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface

Symmetries Can Help

In this case the symmetry of interchanging y and z is an

• rthogonal (and so distance/angle preserving transformation of

R3.) (Algebraically this is the transformation y → z, z → y, x → x.) And this transformation preserves the region we are integrating

• ver.

At the Riemann sum level, for every little area ∆S near (x, y, z) ∈ S there is a matching area near (x, z, y) whose contributions at the Riemann sum level cancel. So the value of the surface integral is 0.

From Math 2220 Class 37 V1c Symmetries Can Help Flow Lines Div and Curl Why Green’s Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface

Symmetries Can Help

If you use this argument on homework or an exam, please

1 Describe the symmetry you are using. 2 Mention that the region of integration is symmetric under

this symmetry.

3 State clearly what cancellation or simplification results.

(In other words, don’t just say “ by symmetry” this is 0 unless it really is obvious.)

From Math 2220 Class 37 V1c Symmetries Can Help Flow Lines Div and Curl Why Green’s Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface

Symmetries Can Help

Using the same potion of a paraboloid, here are two more examples where we can use symmetry to argue for an answer of 0 :

1

• S (0, y, −z) · ˆ

n dS

2

• S yz dS

From Math 2220 Class 37 V1c Symmetries Can Help Flow Lines Div and Curl Why Green’s Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface

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].

From Math 2220 Class 37 V1c Symmetries Can Help Flow Lines Div and Curl Why Green’s Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface

Flow Lines

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

From Math 2220 Class 37 V1c Symmetries Can Help Flow Lines Div and Curl Why Green’s Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface

Flow Lines

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

From Math 2220 Class 37 V1c Symmetries Can Help Flow Lines Div and Curl Why Green’s Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface

Flow Lines

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

From Math 2220 Class 37 V1c Symmetries Can Help Flow Lines Div and Curl Why Green’s Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface

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).

From Math 2220 Class 37 V1c Symmetries Can Help Flow Lines Div and Curl Why Green’s Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface

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.

From Math 2220 Class 37 V1c Symmetries Can Help Flow Lines Div and Curl Why Green’s Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface

Div and Curl

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

From Math 2220 Class 37 V1c Symmetries Can Help Flow Lines Div and Curl Why Green’s Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface

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).

From Math 2220 Class 37 V1c Symmetries Can Help Flow Lines Div and Curl Why Green’s Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface

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.

From Math 2220 Class 37 V1c Symmetries Can Help Flow Lines Div and Curl Why Green’s Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface

Div and Curl

(There is a more complicated identity

• a × (

b × c) = ( a · c) b − ( a · b) c useful in E & M, but the related curl identity (useful in some proofs of Stokes’ theorem is tricky to get right.)

From Math 2220 Class 37 V1c Symmetries Can Help Flow Lines Div and Curl Why Green’s Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface

Div and Curl

Other very important ones are

1 div(curl

F) = 0.

2 curl(∇f ) = 0.

(for C 2 functions and vector fields.) They all are consequences

• f

fxy = fyx for such functions.

From Math 2220 Class 37 V1c Symmetries Can Help Flow Lines Div and Curl Why Green’s Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface

Div and Curl

Problem: Let r = (x, y, z), and so r = r =

• x2 + y2 + z2.

Show that a vector field of the form f (r) r satisfies ∇ × F = 0. (In physics, this is called a central force field.)

From Math 2220 Class 37 V1c Symmetries Can Help Flow Lines Div and Curl Why Green’s Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface

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.

From Math 2220 Class 37 V1c Symmetries Can Help Flow Lines Div and Curl Why Green’s Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface

Why Green’s

For a y-simple region

• R

−Py dy dx =

• C=∂R

P dx. is fairly easily justified.

From Math 2220 Class 37 V1c Symmetries Can Help Flow Lines Div and Curl Why Green’s Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface

Why Green’s

For an x-simple region

• R

Qx dx dy =

• C=∂R

Q dy.

From Math 2220 Class 37 V1c Symmetries Can Help Flow Lines Div and Curl Why Green’s Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface

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.

From Math 2220 Class 37 V1c Symmetries Can Help Flow Lines Div and Curl Why Green’s Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface

Why Green’s

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

From Math 2220 Class 37 V1c Symmetries Can Help Flow Lines Div and Curl Why Green’s Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface

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.

From Math 2220 Class 37 V1c Symmetries Can Help Flow Lines Div and Curl Why Green’s Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface

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.

From Math 2220 Class 37 V1c Symmetries Can Help Flow Lines Div and Curl Why Green’s Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface

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.

From Math 2220 Class 37 V1c Symmetries Can Help Flow Lines Div and Curl Why Green’s Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface

Greens Problems

1 2

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

x2 32 + y2 42 = 1

• riented counterclockwise.

From Math 2220 Class 37 V1c Symmetries Can Help Flow Lines Div and Curl Why Green’s Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface

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?

From Math 2220 Class 37 V1c Symmetries Can Help Flow Lines Div and Curl Why Green’s Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface

Stokes and Gauss

Integration of a conservative vector field cartoon.

From Math 2220 Class 37 V1c Symmetries Can Help Flow Lines Div and Curl Why Green’s Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface

Stokes and Gauss

Green’s Theorem cartoon.

From Math 2220 Class 37 V1c Symmetries Can Help Flow Lines Div and Curl Why Green’s Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface

Stokes and Gauss

Stokes’ Theorem cartoon.

From Math 2220 Class 37 V1c Symmetries Can Help Flow Lines Div and Curl Why Green’s Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface

Stokes and Gauss

Both sides of Stokes involve integrals whose signs depend on the orientation, so to have a chance at being true, there needs to be some compatibility between the choices. The rule is that, from the “positive” side of the surface, (i.e. the side chosen by the orientation), the positive direction of the curve has the inside of the surface to the left. As with all orientations, this can be expressed in terms of the sign of some determinant. (Or in many cases in terms of the sign of some combination of dot and cross products.)

From Math 2220 Class 37 V1c Symmetries Can Help Flow Lines Div and Curl Why Green’s Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface

Stokes and Gauss

Problem: Let S be the portion of the unit sphere x2 + y2 + z2 = 1 with z ≥ 0. Orient the hemisphere with an upward unit normal. Let F(x, y, z) = (y, −x, z). Calculate the value of the surface integral

• S

∇ × F · ˆ n dS.

From Math 2220 Class 37 V1c Symmetries Can Help Flow Lines Div and Curl Why Green’s Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface

Stokes and Gauss

Gauss’ Theorem field cartoon.

From Math 2220 Class 37 V1c Symmetries Can Help Flow Lines Div and Curl Why Green’s Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface

Stokes and Gauss

The surface integral side of Gauss depends on the orientation, so there needs to be a choice making the theorem true. The rule is that the normal to the surface should point outward from the inside of the region. (For the 2d analogue of Gauss (really an application of Greens)

• C
• F · ˆ

n =

• inside

(Px + Qy) dx dy we also use an outward normal, where here C must of course be a closed curve.

From Math 2220 Class 37 V1c Symmetries Can Help Flow Lines Div and Curl Why Green’s Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface

Stokes and Gauss

Problem: Let W be the solid cylinder x2 + y2 ≤ 3 with 1 ≤ z ≤ 5. Let F(x, y, z) = (x, y, z). Find the value of the surface integral

• ∂W
• F · ˆ

n dS.

From Math 2220 Class 37 V1c Symmetries Can Help Flow Lines Div and Curl Why Green’s Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface

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 =?

From Math 2220 Class 37 V1c Symmetries Can Help Flow Lines Div and Curl Why Green’s Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface

Surface of Revolution Case

dS = u

• 1 + (f ′(u))2 du dv.

From Math 2220 Class 37 V1c Symmetries Can Help Flow Lines Div and Curl Why Green’s Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface

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 =?

From Math 2220 Class 37 V1c Symmetries Can Help Flow Lines Div and Curl Why Green’s Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface

Graph Case

dS =

• 1 + f 2

u + f 2 v du dv.

From Math 2220 Class 37 V1c Symmetries Can Help Flow Lines Div and Curl Why Green’s Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface

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.)

From Math 2220 Class 37 V1c Symmetries Can Help Flow Lines Div and Curl Why Green’s Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface

Surface Integrals

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

From Math 2220 Class 37 V1c Symmetries Can Help Flow Lines Div and Curl Why Green’s Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface

Surface Integrals

Basic Parametrization Picture

From Math 2220 Class 37 V1c Symmetries Can Help Flow Lines Div and Curl Why Green’s Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface

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.)

From Math 2220 Class 37 V1c Symmetries Can Help Flow Lines Div and Curl Why Green’s Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface

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.

From Math 2220 Class 37 V1c Symmetries Can Help Flow Lines Div and Curl Why Green’s Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface

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.

From Math 2220 Class 37 V1c Symmetries Can Help Flow Lines Div and Curl Why Green’s Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface

Surface Integrals

3d Flux Picture

From Math 2220 Class 37 V1c Symmetries Can Help Flow Lines Div and Curl Why Green’s Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface

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.

From Math 2220 Class 37 V1c Symmetries Can Help Flow Lines Div and Curl Why Green’s Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface

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.

From Math 2220 Class 37 V1c Symmetries Can Help Flow Lines Div and Curl Why Green’s Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface

Surface Integrals

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

• C
• F · ˆ

n ds.

From Math 2220 Class 37 V1c Symmetries Can Help Flow Lines Div and Curl Why Green’s Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface

Surface Integrals

From Math 2220 Class 37 V1c Symmetries Can Help Flow Lines Div and Curl Why Green’s Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface

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.

From Math 2220 Class 37 V1c Symmetries Can Help Flow Lines Div and Curl Why Green’s Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface

Surface Integrals

Problem: Calculate the surface area of the above paraboloid.

From Math 2220 Class 37 V1c Symmetries Can Help Flow Lines Div and Curl Why Green’s Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface

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.

From Math 2220 Class 37 V1c Symmetries Can Help Flow Lines Div and Curl Why Green’s Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface

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.

From Math 2220 Class 37 V1c Symmetries Can Help Flow Lines Div and Curl Why Green’s Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface

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.

From Math 2220 Class 37 V1c Symmetries Can Help Flow Lines Div and Curl Why Green’s Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface

Paraboloid z = x2 + 4y 2

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

From Math 2220 Class 37 V1c Symmetries Can Help Flow Lines Div and Curl Why Green’s Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface

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.

From Math 2220 Class 37 V1c Symmetries Can Help Flow Lines Div and Curl Why Green’s Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface

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.

From Math 2220 Class 37 V1c Symmetries Can Help Flow Lines Div and Curl Why Green’s Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface

Paraboloid z = x2 + 4y 2

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

From Math 2220 Class 37 V1c Symmetries Can Help Flow Lines Div and Curl Why Green’s Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface

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.

From Math 2220 Class 37 V1c Symmetries Can Help Flow Lines Div and Curl Why Green’s Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface

Paraboloid z = x2 + 4y 2

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

From Math 2220 Class 37 V1c Symmetries Can Help Flow Lines Div and Curl Why Green’s Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface

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 > .

From Math 2220 Class 37 V1c Symmetries Can Help Flow Lines Div and Curl Why Green’s Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface

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 > .

From Math 2220 Class 37 V1c Symmetries Can Help Flow Lines Div and Curl Why Green’s Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface

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.

From Math 2220 Class 37 V1c Symmetries Can Help Flow Lines Div and Curl Why Green’s Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface

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.

From Math 2220 Class 37 V1c Symmetries Can Help Flow Lines Div and Curl Why Green’s Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface

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.

From Math 2220 Class 37 V1c Symmetries Can Help Flow Lines Div and Curl Why Green’s Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface

Parabolic Cylinder z = x2

Graph parametrizations are often optimal for parabolic cylinders.

From Math 2220 Class 37 V1c Symmetries Can Help Flow Lines Div and Curl Why Green’s Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface

Parabolic Cylinder z = x2

Φ(u, v) =< u, v, u2 >

From Math 2220 Class 37 V1c Symmetries Can Help Flow Lines Div and Curl Why Green’s Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface

Parabolic Cylinder z = x2

Φ(u, v) =< u, v, u2 >

From Math 2220 Class 37 V1c Symmetries Can Help Flow Lines Div and Curl Why Green’s Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface

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.

From Math 2220 Class 37 V1c Symmetries Can Help Flow Lines Div and Curl Why Green’s Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface

Elliptic Cylinder x2 + 2z2 = 6

The trigonometric trick is often good for elliptic cylinders

From Math 2220 Class 37 V1c Symmetries Can Help Flow Lines Div and Curl Why Green’s Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface

Elliptic Cylinder x2 + 2z2 = 6

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

From Math 2220 Class 37 V1c Symmetries Can Help Flow Lines Div and Curl Why Green’s Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface

Elliptic Cylinder x2 + 2z2 = 6

Φ(u, v) =< √ 6 cos v, u, √ 3 sin v >

From Math 2220 Class 37 V1c Symmetries Can Help Flow Lines Div and Curl Why Green’s Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface

Elliptic Cylinder x2 + 2z2 = 6

Φ(u, v) =< √ 6 cos v, u, √ 3 sin v >

From Math 2220 Class 37 V1c Symmetries Can Help Flow Lines Div and Curl Why Green’s Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface

Elliptic Cylinder x2 + 2z2 = 6

Φ(u, v) =< √ 6 cos v, u, √ 3 sin v >

From Math 2220 Class 37 V1c Symmetries Can Help Flow Lines Div and Curl Why Green’s Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface

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

From Math 2220 Class 37 V1c Symmetries Can Help Flow Lines Div and Curl Why Green’s Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface

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.

From Math 2220 Class 37 V1c Symmetries Can Help Flow Lines Div and Curl Why Green’s Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface

Ellipsoid x2 + 2y 2 + 3z2 = 4

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

From Math 2220 Class 37 V1c Symmetries Can Help Flow Lines Div and Curl Why Green’s Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface

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 >

From Math 2220 Class 37 V1c Symmetries Can Help Flow Lines Div and Curl Why Green’s Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface

Ellipsoid x2 + 2y 2 + 3z2 = 4

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

• 4

3 cos u >

From Math 2220 Class 37 V1c Symmetries Can Help Flow Lines Div and Curl Why Green’s Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface

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

From Math 2220 Class 37 V1c Symmetries Can Help Flow Lines Div and Curl Why Green’s Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface

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.

From Math 2220 Class 37 V1c Symmetries Can Help Flow Lines Div and Curl Why Green’s Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface

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 >

From Math 2220 Class 37 V1c Symmetries Can Help Flow Lines Div and Curl Why Green’s Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface

Hyperbolic Cylinder x2 − z2 = −4

Φ(u, v) =< 2 sinh v, u, 2 cosh v >

From Math 2220 Class 37 V1c Symmetries Can Help Flow Lines Div and Curl Why Green’s Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface

Saddle z = x2 − y 2

The hyperbolic trick also works with saddles

From Math 2220 Class 37 V1c Symmetries Can Help Flow Lines Div and Curl Why Green’s Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface

Saddle z = x2 − y 2

Φ(u, v) =< u cosh v, u sinh v, u2 >

From Math 2220 Class 37 V1c Symmetries Can Help Flow Lines Div and Curl Why Green’s Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface

Saddle z = x2 − y 2

Φ(u, v) =< u cosh v, u sinh v, u2 >

From Math 2220 Class 37 V1c Symmetries Can Help Flow Lines Div and Curl Why Green’s Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface

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.

From Math 2220 Class 37 V1c Symmetries Can Help Flow Lines Div and Curl Why Green’s Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface

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

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

From Math 2220 Class 37 V1c Symmetries Can Help Flow Lines Div and Curl Why Green’s Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface

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

From Math 2220 Class 37 V1c Symmetries Can Help Flow Lines Div and Curl Why Green’s Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface

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

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

From Math 2220 Class 37 V1c Symmetries Can Help Flow Lines Div and Curl Why Green’s Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface

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

From Math 2220 Class 37 V1c Symmetries Can Help Flow Lines Div and Curl Why Green’s Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface

Top Part of Cone z2 = x2 + y 2

So z =

• x2 + y2.

From Math 2220 Class 37 V1c Symmetries Can Help Flow Lines Div and Curl Why Green’s Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface

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 >

From Math 2220 Class 37 V1c Symmetries Can Help Flow Lines Div and Curl Why Green’s Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface

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 >

From Math 2220 Class 37 V1c Symmetries Can Help Flow Lines Div and Curl Why Green’s Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface

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)

From Math 2220 Class 37 V1c Symmetries Can Help Flow Lines Div and Curl Why Green’s Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface

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