Parametric Surfaces and their AreaPart II

Parametric Surfaces – Tangent Plane

The line u = u0 is mapped to the gridline C2 =r(u0,v)

r

( , ) ( , ), ( , ), ( , ) r u v x u v y u v z u vConsider the parametric surface

defined over some domain D.

The line v = v0 is mapped to the gridline C1=r(u,v0)

The vector is the tangent vector to C1 at

The vector is the tangent vector to C2 at

Thus, the vector is the Normal vector to the surface at .

𝑷

𝐫𝑢

𝐫𝒗

𝐧=𝐫𝑢×𝐫𝑣

Parametric surfaces – Tangent PlaneWe can use this fact to find the equation of the tangent plane to a parametric surface at a given point.

( , ) cos , sin , r u v u v u v u

Thus the normal vector to the tangent plane is

Cartesian coordinates of the point corresponding to u = 2,

0, 2,2u v n r r

Example 6: Find the equation of the tangent plane to

at

:2

v

( , ) cos ,sin ,1 2, = 0,1,12

u uu v v v

r r

( , ) sin , cos ,0 2, = 2,0,02

v vu v u v u v

r r

22, 0 ,2,2 P

r

Equation of tangent plane: 0(x−0) − 2(y − 2)+2(z − 2)=0

which simplifies to

Note that the surface is a cone. Can you see that from the parametric equations?

Parametric Surfaces – Surface Area

Divide the domain D into sub-rectangles . Let’s choose to be the lower left corner of .

r

( , ) ( , ), ( , ), ( , ) r u v x u v y u v z u vConsider the parametric surface S described by the equation

defined over some domain D.

Let and be the tangent vectors at

The rectangle is mappedto the patch . maps to the point

Area of the surface S is

Parametric Surfaces and their area

Let and be the tangent vectors at

The area of the patch can be approximated by the area of the parallelogram determined by the vectors and

Area of parallelogram: is approximately the area of .

( ) r r u vD

A S dudv

Parametric Surfaces and their area – Example 7

Find the surface area of the helicoid

cos ,sin ,0u v vr

2 2sin , cos , cos sinu v v v u v u v r r

( , ) cos , sin ,u v u v u v vr

1 5 2

0 01( )

Du v u dvduA S dA

r r

0 5 , 0 1v u

sin ,cos ,v v u

2 2 2 2sin cos 1u v v v u u r r

1 2

05 1 u du 5

2 ln 2 1 5.742

Parametric Surfaces and their area – Example 8

Find the surface area of a sphere of radius a

( , ) cos sin , sin sin , cosa a a r

0 2 , 0 The sphere can be parameterized by

sin sin , cos sin ,0a a r cos cos , sin cos , sina a a r

sin sin cos sin 0

cos cos sin cos sin

a a

a a a

i j k

r r

2 2 2 2 2 2 2 2cos sin , sin sin , sin sin cos cos sin cosa a a a

2 2 2cos sin , sin sin , sin cosa

4 2 4 2 4 2 2cos sin sin sin sin cosa r r

2 4 2 2sin sin cosa 2 2 2 2sin sin cos a 2 sina

2 sina r r22 2 2

0 0 0 A(S) = sin 2 sin 4a d d a d a

We then have:

=1

2 2 2 2 2 2 2cos sin , sin sin , sin cos sin cosa a a =1

Then and

Parametric Surfaces and their area

If the surface is of the form , for in some domain D, then it can be parameterized by

222 21 1( )

x yD D

z zf f dA dA

x yA S

2 21r r x y x yf f

Surface Area:

Parametric Surfaces and their area – Example 9

Find the surface area of the part of the sphere that lies above the cone

Since z is positive, the surface can be written in explicit form as

2 216z x y

2 2 2 216 x y x y

Find the curve C of intersection:

2 22( ) 16x y 2 2 8x y

2 2( , ) 8D x y x y

22

( ) 1D

z zA S dA

x y

2 2

2 2 2 21

16 16D

x ydA

x y x y

Parametric Surfaces and their area – Example 9 continued

2 2( , ) 8D x y x y 2 2

2 2 2 2( ) 1

16 16D

x yA S dA

x y x y

2 2

2 2 2 21

16 16D

x ydA

x y x y

2 2 2 2

2 2

16

16D

x y x ydA

x y

2 2

4

16D

dAx y

Changing to polar coordinates:

2

8 2

0 0

4

16

rd drr

8

208

16

r

drr

8 4 2 2 29.44

on the domain

Parametric Surfaces and their area – Example 10Find the surface area cut out of the cylinder by the cylinder

We can use symmetry and find twice the area of the surface

2 2( , ) 81D x y x y

22

1( ) 2D

z z

x yA S dA

2

22 1

81D

xdA

x

2

22 1

81

D

xdA

x

2

2

9 81

29 81

118

81

x

xdydx

x648

9

918 2 dx