MATH ajajoo/math2433

MATH 2433 - 12631

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Lecture : MoWeFre 10-11am in SR 116

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http://www.math.uh.edu/∼ajajoo/Math2433

Parametrized Surfaces;

Surface Area

We can parametrized a surfaceS in space by a vector function

r=r(u,v), where (u,v) ranges over some region Ω of the

uv-plane.

The fundamental vector product

Let the surfaceS be parametrized by a differentiable vector function r=r(u,v) =x(u,v)i+y(u,v)j+z(u,v)k. We denote r_u0 = ∂x ∂ui+ ∂y ∂uj+ ∂z ∂uk, r 0 v = ∂x ∂vi+ ∂y ∂vj+ ∂z ∂vk. Definition

The cross product

N=r0_u×r0_v

The fundamental vector product

The vectorN(u,v) =r0_u(u,v)×r_v0(u,v) is perpendicular to the surfaceS at the tip ofr(u,v).

Example

Find the fundamental vector product for

How to evaluate the area of

S

LetS be a surface parametrized by a continuously differentiable function

r=r(u,v), (u,v)∈Ω.

Definition

If Ω is a basic region in theuv-plane andr is one-to-one on the interior of Ω, we callS asmooth surface.

IfN(u,v) is never zero on the interior of Ω, then we have

area(S) =

Z Z

Ω

||N(u,v)|| dudv.

How to evaluate the area of

S

(particular case)

IfS is the graph ofz =f(x,y) for (x,y)∈Ω, we can parametrize it like this: S : r(x,y) =xi+yj+f(x,y)k, (x,y)∈Ω. Then, we have area(S) = Z Z Ω q [fx(x,y)]2+ [fy(x,y)]2+ 1dxdy.

Example

Find the area of the surfacez =y2, for 0≤x ≤y, 0≤y ≤1.

(continue)

Surface Integrals

Surface integrals

LetH=H(x,y,z) be a scalar field continuous overS :r=r(u,v), with (u,v)∈Ω. The surface integral of H over S is:

Z Z S H(x,y,z)dσ= Z Z Ω H(x(u,v),y(u,v),z(u,v))||N(u,v)||dudv

Notice that ifH(x,y,z) = 1, we have

Z Z S dσ= Z Z Ω ||N(u,v)||dudv =area(S).

Example

Evaluate _{Z Z}

S

2y dσ

whereS is the surface S :z = 1 2y

2_{, 0≤x ≤2, 0≤y≤1.}

(continue)

Evaluate _{Z Z}

S

2y dσ

whereS is the surface S :z = 1 2y

Example

Evaluate _{Z Z}

S

2xy dσ

whereS is the surface x+ 2y+ 2z = 4 in the first octant.

(continue)

Evaluate _{Z Z}

S

2xy dσ

The Vector Differential

Formal definition

The vector differential operator∇is formally defined by

∇= ∂ ∂xi+ ∂ ∂yj+ ∂ ∂zk.

So far, we have seen∇applied to a differentiable scalar field, e.g.

∇f, but it can also be applied to vector fields.

Divergence and curl

Consider vector fieldv=v1(x,y,z)i+v2(x,y,z)j+v3(x,y,z)k.

Divergence ∇ ·v= ∂v1 ∂x + ∂v2 ∂y + ∂v3 ∂z . Curl ∇ ×v=Det   i j k ∂ ∂x ∂ ∂y ∂ ∂z v1 v2 v3  

The curl of a gradient is zero.

Example

(continue)

Example

Given

v(x,y,z) = (3yz)i+ (4xz)j+ (3xy)k,

(continue)

Given

v(x,y,z) = (3yz)i+ (4xz)j+ (3xy)k,

calculate∇ ·v and∇ ×v.

The Divergence Theorem

Divergence theorem

Setv=Qi−Pj. Green’s theorem can be written as:

Z Z Ω (∇ ·v)dxdy = I C (v·)ds,

where is the outer unit normal for Ω.

In 3D, we have:

Divergence theorem

LetT be a solid bounded by a closed oriented surfaceS which is piecewise smooth. If the vector fieldv=v(x,y,z) is continuously differentiable throughoutT, then

Z Z Z T (∇ ·v)dxdydz= Z Z S (v·)dσ.

where is the outer unit normal.

Flux

is calledflux of v out of S.

The divergence theorem tells you that you can evaluate the flux of

vout of S through a triple integral onT, which is the solid bounded byS

flux of vout of S =

Z Z Z

T

Example

Use the divergence theorem to find the total flux out of the solid

x2+y2 ≤4, 0≤z ≤1, given

v(x,y,z) =xi+ 3y2j+ 2z2k.

(continue)

Use the divergence theorem to find the total flux out of the solid

x2+y2 ≤4, 0≤z ≤1, given

Example

Use the divergence theorem to find the total flux out of the solid 0≤x≤1, 0≤y≤1−x, 0≤z ≤1−x−y, given

v(x,y,z) = 2x2i+ 4xyj−4xzk.

(continue)

Use the divergence theorem to find the total flux out of the solid 0≤x≤1, 0≤y≤1−x, 0≤z ≤1−x−y, given