V2 Gauss From Math 2220 Class 39pi.math.cornell.edu/~back/m222_f14/slides/dec1_v2.pdfFrom Math 2220...

From Math2220 Class 39

V2

Stokes andGauss

Why Green’sand Gauss’

ConservativeVector Fields

SystematicMethod ofFinding aPotential

dTheta

IntegralTheoremProblems

Surface ofRevolutionCase

Graph Case

SurfaceIntegrals

SurfaceParametriza-tion

From Math 2220 Class 39

Dr. Allen Back

Dec. 1, 2014


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Stokes and Gauss

Integration of a conservative vector field cartoon.


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Stokes and Gauss

Green’s Theorem cartoon.


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Stokes and Gauss

Stokes’ Theorem cartoon.


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Stokes and Gauss

Both sides of Stokes involve integrals whose signs depend onthe orientation, so to have a chance at being true, there needsto 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 thecurve has the inside of the surface to the left.

As with all orientations, this can be expressed in terms of thesign of some determinant. (Or in many cases in terms of thesign of some combination of dot and cross products.)


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Stokes and Gauss

Problem: Let S be the portion of the unit spherex2 + y2 + z2 = 1 with z ≥ 0. Orient the hemisphere with anupward unit normal. Let ~F (x , y , z) = (y ,−x , ez2). Calculatethe value of the surface integral∫∫

S∇× ~F · n̂ dS .


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Stokes and Gauss

Gauss’ Theorem field cartoon.


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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 outwardfrom the inside of the region.

(For the 2d analogue of Gauss (really an application of Green’s)∫C~F · n̂ =

∫∫inside

(Px + Qy ) dx dy

we also use an outward normal, where here C must of course bea closed curve.


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Stokes and Gauss

Problem: Let W be the solid cylinder x2 + y2 ≤ 3 with1 ≤ z ≤ 5. Let ~F (x , y , z) = (x , y , z). Find the value of thesurface integral ∫∫

∂W~F · n̂ dS .


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Why Green’s and Gauss’

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

CP(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

on the set R and C is given its usual “inside to the left”orientation.


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For a y -simple region∫∫R−Py dy dx =

∫C=∂R

P dx .

is fairly easily justified.


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For an x-simple region∫∫R

Qx dx dy =

∫C=∂R

Q dy .


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So for a region that is both y -simple and x-simple we haveGreen:∫

∂RP(x , y) dx + Q(x , y) dy =

∫∫R

Qy − Px dx dy .


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


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

Think about the line integral around a small rectangle withsides ∆x and ∆y .

If you assume (or justify) that evaluating the vector field in themiddle of each edge gives a good approximation in the lineintegral, then Qx − Py emerges quite naturally.


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


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


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For what might be called a z-simple region W,

g(x , y) ≤ z ≤ h(x , y) with (x , y) ∈ D ⊂ R2

a very similar argument shows that∫∫∂W

(0, 0,R) · n̂ dS =∫∫∫

WRz dz dx dy

as Gauss says about the z-part of any C 1 vector field.


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As with Green, the other components can be handled similarlyfor appropriately shaped elementary regions.If you can cut a region into two pieces where we know Gaussholds on each piece (e.g. a doughnut shaped region), thenGauss also holds for the entire region.


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While the proof of Stokes’ shares many elements with theproofs of Green’s and Gauss’, the best “classical style” proof ofStokes’ involves using a parametrization to reduce Stokes’ toGreen’s in the parameterizing (i.e. uv plane)


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The differential forms point of view introduced in section 8.5makes all these theorems one theorem, usually called Stokes’,and the proof becomes a combination of more advanced linearalgebra constructions (differential forms) together with the onevariable Fundamental Theorem of Calculus.Time permitting, we’ll talk a bit about this on Monday.


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Conservative Vector Fields

A vector field ~F (x , y , z) which can be written as

~F = ∇f

is called conservative. We already know∫C~F · d~s = 0

for any closed curve.


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The origin of the term is physics (I think) where in the case of~F a force, it does no work (and so saps/adds no energy) as aparticle traverses the closed curve.


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A vector field ~F (x , y , z) which can be written as

~F = ∇f

is called conservative. We already know∫C~F · d~s = 0

for any closed curve.In physics, the convention is to choose φ so that

~F = −∇φ

and φ is referred to as the potential energy.


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Conservation of energy (in e.g mechanics) becomes a theoremin multivariable calculus combining the definition of a flow linewith the computation of a line integral.Newton’s 2nd law (~F = m~a and other versions) is also key . . . .


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The concept of voltage arises here too; it is just a potentialenergy per unit charge.


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The theorem (vector identity) curl(∇f ) = 0 means the

curl(~F ) = 0

is a necessary condition for the existence of a function fsatisfying

∇f = ~F .


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It turns out that for vector fields defined on e.g. all of R2 orR3, the converse of the theorem curl(∇f ) = 0 is true. (For R2,we’re thinking of the scalar curl.)In other words, in such a case, if curl(~F ) = 0, (for a C 1 vectorfield), there is guaranteed to be a function f (x , y , z) such that

∇f = ~F .


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While this hold for vector fields with domains R2, R3, or moregenerally any “simply connected” region, the example dθ belowshows this converse does not hold in general.


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Time permitting, we’ll talk about simple connectivity nextweek.


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Systematic Method of Finding a Potential

Finding a potential by inspection is fine when you can, but it isnot systematic.I often ask on a final exams for this.


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Problem: Use a systematic method to find a function f (x , y , z)for which

∇f = (2xy , x2 + z2, 2yz + 1).


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∇f = (y2zexyz + 1y, (1 + xyz)exyz − x

y2,

−(cos2 (xyz))ez + xy2exyx − ez sin2 (xyz)).


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∇f = (2xz , 2y , x2 + 6ez).


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dθ

∫C−y dx + x dy

for C the unit circle traversed counterclockwise.


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dθ

∫C

x dx + y dy

for C the unit circle traversed counterclockwise.


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dθ

∫C

−y dx + x dyx2 + y2

is still nonzero for C a circle of radius R centered at the origintraversed counterclockwise.

This is remarkable since

∇ tan−1 yx

=1

x2 + y2(−y , x)) .


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dθ

Why does this not contradict∫C ∇f · d~s = 0 for a closed curve

(i.e. start=end) C ?


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Integral Theorem Problems

Problem: Let

~F (x , y , z) = (y2 + z2, x2 + z2, x2).

Find ∫C~F · d~s

where C is the boundary of the plane x + 2y + 2z = 2intersected with the first octant, oriented counterclockwisefrom above.


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Problem: Find the flux of the vector field

~F (x , y , z) = (xy , yz , xz)

through the boundary of the unit cube0 ≤ x ≤ 1, 0 ≤ y ≤ 1, 0 ≤ z ≤ 1 where the boundary of thecube has its usual outward normal.


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Problem: Find ∫∫S~F · n̂ dS

for ~F (x , y , z) = (0, yz , z2) and S the portion of the cylindery2 + z2 = 1 with 0 ≤ x ≤ 1, z ≥ 0, and the positiveorientation chosen to be a radial outward (from the axis of thecylinder) normal.


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12

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

x2

32+

y2

42= 1

oriented counterclockwise.


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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 saywhether one of

∫C1

~F · d~s and∫C2

~F · d~s is bigger than theother?


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Surface of Revolution Case

This is not worth memorizing!If one rotates about the z-axis the path (curve) z = f (x) in thexz-plane for 0 ≤ a ≤ x ≤ b, one obtains a surface of revolutionwith a parametrization

Φ(u, v) = (u cos v , u sin v , f (u))

and dS =?


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Surface of Revolution Case

dS = u√

1 + (f ′(u))2 du dv .


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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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Graph Case

dS =√

1 + f 2u + f2v du dv .


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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 2x + f2y

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 γdu dv .

(Note that du dv is essentially the same as dx dy here.)


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Surface Integrals

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


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Surface Integrals

Basic Parametrization Picture


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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 n̂ = ±~Tu × ~Tv |‖~Tu × ~Tv‖

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


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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) · n̂ dS .


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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) · n̂ 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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Surface Integrals

3d Flux Picture


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Surface Integrals

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

Sρ~F · n̂ dS

(with associated Riemann Sum∑ρ(x∗i , y

∗j , z∗k )~F (x∗i , y

∗j , z∗k ) · n̂(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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Surface Integrals

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


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Surface Integrals

2d Flux Picture

There’s an analagous 2d Riemann sum and interp of∫C~F · n̂ ds.


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Surface Integrals


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Surface Integrals

Problem: Calculate ∫∫S~F (x , y , z) · n̂ dS

for the vector field ~F (x , y , z) = (x , y , z) and S the part of theparaboloid z = 1− x2 − y2 above the xy -plane. Choose thepositive orientation of the paraboloid to be the one with normalpointing downward.


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Surface Integrals

Problem: Calculate the surface area of the above paraboloid.


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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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Φ(u, v) =< u, v ,F (u, v) > .

Parameters u and v just different names for x and y resp.


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Φ(u, v) =< u, v ,F (u, v) > .

Use this idea if you can’t think of something better.


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Φ(u, v) =< u, v ,F (u, v) > .


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Φ(u, v) =< u, v ,F (u, v) > .

Note the curves where u and v are constant are visible in thewireframe.


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


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Φ(u, v) =< 2u cos v , u sin v , 4u2 > .


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Φ(u, v) =< 2u cos v , u sin v , 4u2 > .

Algebraically, we are rescaling the algebra behind polarcoordinates where

x = r cos θ

y = r sin θ

leads to r2 = x2 + y2.


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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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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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Parabolic Cylinder z = x2

Graph parametrizations are often optimal for paraboliccylinders.


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Φ(u, v) =< u, v , u2 >


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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 thecurve z = x2 in the zx plane.


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Elliptic Cylinder x2 + 2z2 = 6

The trigonometric trick is often good for elliptic cylinders


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Φ(u, v) =<√

3·√

2 cos v , u,√

3 sin v >=<√

6 cos v , u,√

3 sin v >


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Φ(u, v) =<√

6 cos v , u,√

3 sin v >


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Φ(u, v) =<√

6 cos v , u,√

3 sin v >

What happened here is we started with the polar coordinateidea

x = r cos θ

z = r sin θ

but noted that the algebra wasn’t right for x2 + 2z2 so shiftedto

x =√

2r cos θ

z = r sin θ


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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 whenr =√

3.


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Ellipsoid x2 + 2y 2 + 3z2 = 4

A similar trick occurs for using spherical coordinate ideas inparameterizing ellipsoids.


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A similar trick occurs for using spherical coordinate ideas inparameterizing ellipsoids.

Φ(u, v) =< 2 sin u cos v ,√

2 sin u sin v ,

√4

3cos u >


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Φ(u, v) =< 2 sin u cos v ,√

2 sin u sin v ,

√4

3cos u >


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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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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 hyperbolicversion cosh2 θ − sinh2 θ = 1 leads to the nicest hyperbolaparameterizations.


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Just as cos2 θ + sin2 θ = 1 helps with ellipses, the hyperbolicversion cosh2 θ − sinh2 θ = 1 leads to the nicest hyperbolaparameterizations.

Φ(u, v) =< 2 sinh v , u, 2 cosh v >


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Φ(u, v) =< 2 sinh v , u, 2 cosh v >


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Saddle z = x2 − y 2

The hyperbolic trick also works with saddles


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Saddle z = x2 − y 2

Φ(u, v) =< u cosh v , u sinh v , u2 >


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

The spherical coordinate idea for ellipsoids with sinφ replacedby cosh u works well here.


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


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


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Top Part of Cone z2 = x2 + y 2

So z =√

x2 + y2.


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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 >


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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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ctionN

Stokes and GaussWhy Green's and Gauss'Conservative Vector FieldsSystematic Method of Finding a PotentialdThetaIntegral Theorem ProblemsSurface of Revolution CaseGraph CaseSurface IntegralsSurface Parametrization

V2 Gauss From Math 2220 Class 39pi.math.cornell.edu/~back/m222_f14/slides/dec1_v2.pdfFrom Math 2220...

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