Calculus 3 Day 24 Notes

8/13/2019 Calculus 3 Day 24 Notes

1/24

Chapter 17

MATH 2433 - 15535

Annalisa Quaini

[email protected]

Office : PGH 662Lecture : MoWe 5:30PM-7:00PM in SEC 102

Office hours : Tu 10AM-12PM and by appointment

FILL IN THE TEACHING EVALUATION ON CASAWEBSITE!!! 5 points will be given to those who fills it inBubble ID and popper number (Popper 19)

http://www.math.uh.edu/quainiA. Quaini, UH MATH 2433 1 / 24


2/24

Chapter 17 Sect. 17.4 Sect. 17.5 Sect. 17.6 Sect. 17.7 Sect. 17.8

Ifh(x, y, z) =h1(x, y, z)i+h2(x, y, z)j+h3(x, y, z)k, the line

integral C

h(r) dr

can be written

C

(h1(x, y, z)dx+h2(x, y, z)dy+h3(x, y, z)dz).

If the curve is parametrized by

C : r(u) =x(u)i+y(u)j+z(u)k, u [a, b],

then we have dx=x(u)du, dy=y(u)du, dz=z(u)du.

A. Quaini, UH MATH 2433 2 / 24


3/24


Definitions

Closed curve

A curve C : r=r(u), u [a, b], is said to be closed if it beginsand ends at the same point

r(a) =r(b).

Simple curveA curve C : r=r(u), u [a, b], is said to be simple if it does notintersect itself

a


4/24


Definition

A Jordan curve is a plane curve that is both closed and simple.

A closed region , the total boundary of which is a Jordan curveC, is called Jordan region.

Greens theorem

Let be a Jordan region with a piecewise-smooth boundary C. IfP and Qare scalar fields such that are continuously differentiableon an open set that contains , then

Qx

(x, y) P

y

(x, y) dxdy= C

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

where the integral on the right is the line integral taken over C inthe counterclockwise direction.



5/24


Finding the area using Greens theorem

IfCis a simple closed curve enclosing the region , then

Area() =

dxdy=

C

ydx=

C

xdy=1

2

C

ydx+xdy


Ch 17 S 17 4 S 17 5 S 17 6 S 17 7 S 17 8


6/24


Parametrized Surfaces;Surface Area


Ch t 17 S t 17 4 S t 17 5 S t 17 6 S t 17 7 S t 17 8


7/24


We can parametrized a surface Sin space by a vector function

r=r(u, v), where (u, v) ranges over some region of theuv-plane.


Chapter 17 Sect 17 4 Sect 17 5 Sect 17 6 Sect 17 7 Sect 17 8


8/24


The fundamental vector product

Let the surface Sbe parametrized by a differentiable vector

function

r=r(u, v) =x(u, v)i+y(u, v)j+z(u, v)k.

We denote

ru=x

ui+

y

uj+

z

uk, rv =

x

vi+

y

vj+

z

vk.

Definition

The cross productN= ru r

v

is called the fundamental vector productof the surface.




9/24


The fundamental vector product

The vector N(u, v) =r

u(u, v) r

v(u, v) is perpendicular to thesurface Sat the tip ofr(u, v).




10/24


Example

Find the fundamental vector product for

r(u, v) =ucos(v)i+usin(v)j+vk.




11/24


How to evaluate the area ofS

Let Sbe a surface parametrized by a continuously differentiablefunctionr=r(u, v), (u, v) .

Definition

If is a basic region in the uv-plane and r is one-to-one on theinterior of , we call S a smooth surface.

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

area(S) =

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




12/24

p

How to evaluate the area ofS(particular case)

IfSis the graph ofz=f(x, y) for (x, y) , we can parametrizeit like this:

S : r(x, y) =xi+yj+f(x, y)k, (x, y) .

Then, we have

area(S) = [fx(x, y)]

2 + [fy(x, y)]2 + 1 dxdy.




13/24

p

Example

Find the area of the surface z=y2, for 0 x y, 0 y 1.




14/24

(continue)

Find the area of the surface z=y2, for 0 x y, 0 y 1.




15/24

Surface Integrals




16/24

Surface integrals

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

S H(x, y, z)d=

H(x(u, v), y(u, v), z(u, v))||N(u, v)|| dudv

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

S

d=

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




17/24

Example

Evaluate

S

2y d

where S is the surface S :z=1

2y2, 0 x 2, 0 y 1.




18/24

(continue)

Evaluate

S

2y d

where S is the surface S :z=1

2y2, 0 x 2, 0 y 1.




19/24

Example

Evaluate S

2xy d

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



( )


20/24

(continue)

Evaluate S

2xy d

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




21/24

The Vector Differential

Operator



F l d fi i i


22/24

Formal definition

The vector differential operator is formally defined by

=

xi+

yj+

zk.

So far, we have seen applied to a differentiable scalarfield, e.g.f, but it can also be applied to vector fields.



Di d l


23/24

Divergence and curl

Consider vector field v=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=

i j k

x

y

z

v1 v2 v3




24/24

The curl of a gradient is zero.


Calculus 3 Day 24 Notes

Documents

Transcript of Calculus 3 Day 24 Notes