Section 16.6 Parametric Surfaces and Their Areas

Section 16.6 Parametric Surfaces and Their Areas

Xin Li

[email protected]

MAC2313 Summer 2020

Xin Li (FSU) Section 16.6 MAC2313 Summer 2020 1 / 20

parametric curve and surface

Recall: When we parameterized a curve, we took value of t from someinterval [a, b] and plug them into

~r(t) = x(t)~i + y(t)~j + z(t)~k

and the resulting set of vectors will be the position vectors for the pointson the curve.To parameterize a surface, we take points (u, v) on region D in theuv-plane and plug them into

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

The resulting set of vectors will be the position vectors for the points onthe surface S that we are trying to parameterize. This is often called theparameteric representation of the parametric surface S


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parametric surface

The parameteric equation for a surface isx = x(u, v), y = y(u, v), z = (u, v)


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Example 1

Identify and sketch the surface with vector equation

~r(u, v) = 2 cos u~i + v~j + 2 sin u~k


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{Fz÷g xz=4as2ut4sin2u= I

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Parametric surface

If a parametric surface S is given by a vector function ~r(u, v), thenthere are two useful families of curves that lie on S , one family with uconstant and the other with v constant.

These families correspond to vertical and horizontal lines in theuv-plane.

If we keep u constant by putting u = u0, then ~r(u0, v) becomes avector function of the single parameter v and defines a curve C1 lyingon S .


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Grid Curves

Similarly, if we keep v constant by putting v = v0, then ~r(u, v0)becomes a vector function of the single parameter u and defines acurve C2 lying on S

We call these ~r(u0, v) and ~r(u, v0) Grid Curves

In fact, when a computer graphs a parametric surface, it usuallydepicts the surface by plotting these grid curves


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Example 2

Give parametric representations for the sphere x2 + y2 + z2 = 30

Solution: Equation of sphere is x2 + y2 + z2 = ⇢2, so ⇢ =p30

Recall the parametric equation of the spherical

x = ⇢ sin� cos ✓

y = ⇢ sin� sin ✓

z = ⇢ cos�

We have the corresponding vector equation

~r(✓,�) =p30 sin� cos ✓~i +

p30 sin� sin ✓~j +

p30 cos�~k

where 0 ✓ 2⇡, 0 � ⇡


--

✓ ( 8,0) → (x. y , Z)

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Example 2

~r(✓,�) =p30 sin� cos ✓~i +

p30 sin� sin ✓~j +

p30 cos�~k

The grid curves for a sphere are curves of constant latitude or constantlongitude.


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Tangent Plane

Recall the equation of a plane

a(x � x0) + b(y � y0) + c(z � z0) = 0

where the point(x0, y0, z0)

is on the plane and the normal vector of the plane is

~n =< a, b, c >


# PT. pot = o

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Tangent Plane

We now find the tangent plane to the parametric surface S at a point P0

with position vector ~r(u0, v0). The surface S is given by a vector function

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

Definition

~ru =@x

@u(u0, v0)~i +

@y

@u(u0, v0)~j +

@z

@u(u0, v0)~k

~rv =@x

@v(u0, v0)~i +

@y

@v(u0, v0)~j +

@z

@v(u0, v0)~k


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Example 3

Find the tangent plane to the surface with parameteric equationsx = u2, y = v2, z = u + 2v at the point (1, 1, 3)

Solution:


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Tcu, v) = u' Itv' j t cut ZHI

Fu = zu it O j t I = C ZU,o, I >

Fu = OF t zvjt 2. I = C O,

2V,27

n' = Fux I =/ zig ! 1=4%4 - incult- ( (zu )C2)- (1)Co)) Tt ( @a) ( 24 - G) co) ) I

= - 2 V 'T - 4Uj t 4kV

Example 3

Find the tangent plane to the surface with parameteric equationsx = u2, y = v2, z = u + 2v at the point (1, 1, 3)


I = - 2 Vj - 4 Uj t Kurtat point qq.az ) u2=I → a- It

soYy, y zV'=/ → V = -11 V -

- I

ur'

fr "utzu

Ut2V= 3 utzv =3 -

ICI, 1) = - 2T -4T -14kt =L -2 , -4,43

Tangent plane-2 (x - l ) - 4cg - l ) -14 (2--3)=0

Surface Area

Now we define the surface area of a general parametric surface. Let’schoose (u⇤i , v

⇤j ) to be the lower left color of small region Rij


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Surface Area

Let ~r⇤u = ~ru(u⇤i , v⇤j ) and ~r⇤v = ~rv (u⇤i , v

⇤j ) be the tangent vectors at point

Pij . The vector of the two edges of the region can be approximated by thevectors �u~r⇤u and �v~r⇤v


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Surface Area

So we approximate Sij by the parallelogram determined by the vectors�u~r⇤u and �v~r⇤v .

The area of this parallelogram is the magnitude of the cross product.

|(�u~r⇤u )⇥ (�v~r⇤v )| = |~r⇤u ⇥ ~r⇤v |�u�v

So the Area of S can be approximated bymX

i=1

nX

j=1

|~r⇤u ⇥ ~r⇤v |�u�v


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Surface Area Definition

DefinitionIf a smooth parametric surface S is given by the equation

~r(u, v) = x(u, v)~i + y(u, v)~j + z(u, v)~k , (u, v) 2 D

and S is covered just once as (u, v) ranges throughout the parameterdomain D, then the surface area of S is

A(S) =

ZZ

D|~ru ⇥ ~rv |dA

where ~ru = @x@u~i + @y

@u~j + @z

@u~k and ~rv = @x

@v~i + @y

@v~j + @z

@v~k


-ao•⇒#① define region D

.

② compute Fa.

I.

Fax RT. truant

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Example 4

Find the surface area of a sphere of radius a.

Solution:


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1- a asf I

D= { ( 0,0 ) ) off E.ae,

#this.

Tp =F-

acosy asOTt a asf snot- a Singh

To = - A Sino since i t a sing asOTT OIG

Example 4



*xri=l÷ -

÷:X= a'sing asso it a's.io/smoj-a2smo/oso/k

g-

hipxro f-#(a'snip a)'

t (a's.no/smoj4-@smoeos8T=-a" Saito as 'O ta 't Sm 44 sin 'o ta 't s.io/cos2oT--

=#a " ssh 401 + a't

sink cos 201= FIE - so

Example 4



try x Fo I = fatso = AZ sin of

azotfsm.co/3o for#EK.

ALS) -- Sf, try xtro IDA= fo"

[ a 's.no/do1dI-

= a' (f."do ) ( S? sing doll

= a'

(off"

) fast to" )= a

'

( 22 ) ( - asset cos o ) = a'

( 22) ( 2) =4T⑦

Example 4



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Section 16.6 Parametric Surfaces and Their Areas