. . . . . .

Section 12.6Surface Area

Math 21a

March 31, 2008

Announcements

◮ Office hours Tuesday, Wednesday 2–4pm SC 323◮ Problem Sessions: Mon, 8:30; Thur, 7:30; SC 103b

..Image: Flickr user Tracy O

. . . . . .

Announcements

◮ Office hours Tuesday, Wednesday 2–4pm SC 323◮ Problem Sessions: Mon, 8:30; Thur, 7:30; SC 103b

. . . . . .

Outline

Last time: Integration over polar regionsIntegration in Polar Coordinates

Easy AreaRectangles and Parallelograms in the planeParallelograms in space

Area of curved surfacesDivideApproximateTake the Limit

Special casesGraphsSurfaces of Revolution

Worksheet

Next Time

. . . . . .

Polar slicing

.

Here the boundaries of x and yare complicated

.

Here the boundary r is afunction of θ

. . . . . .

Integration in Polar Coordinates

FactIf f is continuous on a polar region of the form

D = { (r, θ) | α ≤ θ ≤ β, h1(θ) ≤ r ≤ h2(θ) }

then ∫∫D

f(x, y) dA =

∫ β

α

∫ h2(θ)

h1(θ)f(r cos θ, r sin θ) r dr dθ.

◮ Notice the “area element” is r dr dθ, not dr dθ!

. . . . . .

Outline

Last time: Integration over polar regionsIntegration in Polar Coordinates

Easy AreaRectangles and Parallelograms in the planeParallelograms in space

Area of curved surfacesDivideApproximateTake the Limit

Special casesGraphsSurfaces of Revolution

Worksheet

Next Time

. . . . . .

Rectangles and Parallelograms in the plane

.

.

.ℓ

.w

.

.

.b

.a .h = a sin θ

. .θ

A = ℓw A = bh = ab sin θ

. . . . . .

Parallelograms in space

The magnitude of the cross product a× b is the area of theparallelogram with sides a and b.

. .a

.b .|b| sin θ

.θ

A = |a| |b| sin θ = |a× b|

. . . . . .

Outline

Last time: Integration over polar regionsIntegration in Polar Coordinates

Easy AreaRectangles and Parallelograms in the planeParallelograms in space

Area of curved surfacesDivideApproximateTake the Limit

Special casesGraphsSurfaces of Revolution

Worksheet

Next Time

. . . . . .

Parametrizing a surface

A parametric surface S is defined by a vector-valued function of twoparameters

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

where (u, v) varies throughout a region D in the plane.

..u

.v

.D .r

.x .y

.z

.S

. . . . . .

Divide and Conquer

Suppose D is a rectangle [a, b] × [c, d]. Divide [a, b] into m pieces and[c, d] into n pieces.The area of the surface is the sum of areas of pieces of the surface:

A =∑

i,j

∆Ai,j

where ∆Aij is the area of the surface restricted to the subdomain[ui, ui+1] × [vi, vi+1]

. . . . . .

ApproximateThe vectors representing the sides of a small rectangle “pushforward” to vectors tangent to the surface.

..u

.v

.∆u i

.∆v j.D

.x .y

.z

.∆u ru.∆v rv .S

They span a parallelogram approximating this piece of the surface. Sothe area of that piece is

∆A ≈ |∆u ru × ∆v rv| = |ru × rv| ∆v ∆u

. . . . . .

Take the Limit

So

A ≈m∑

i=1

n∑j=1

∣∣ru(uij, vij) × rv(uij, vij)∣∣ ∆v ∆u

Taking the limit we get

A =

∫ b

a

∫ d

c|ru × rv| dv du

More generally, if D is any region (not necessarily a rectangle), thesurface area is ∫∫

D

|ru × rv| dA

Remembering the dA is over the domain (u, v) space.

. . . . . .

Outline

Last time: Integration over polar regionsIntegration in Polar Coordinates

Easy AreaRectangles and Parallelograms in the planeParallelograms in space

Area of curved surfacesDivideApproximateTake the Limit

Special casesGraphsSurfaces of Revolution

Worksheet

Next Time

. . . . . .

Surface areas of GraphsNotation as in Stewart, p. 869ff.

Suppose f(x, y) is a function oftwo variables with domain D.The graph of f over D is thesurface in space given by

S = { (x, y, z) | (x, y) ∈ D, z = f(x, y) } .

.D.x

.y

.z

.S

. . . . . .

To find the surface area of S, use the parametrization

r(x, y) = ⟨x, y, f(x, y)⟩

Then

rx =

⟨1, 0,

∂f∂x

⟩ry =

⟨0, 1,

∂f∂y

⟩rx × ry =

⟨− ∂f

∂x,− ∂f

∂y, 1

⟩So

A =

∫∫D

√1 +

(∂f∂x

)2

+

(∂f∂y

)2

dA

. . . . . .

Surfaces of Revolution

A surface of revolution can be described by rotating the graph ofy = f(x) over the interval [a, b] around the x-axis.

..x

.y

.z

. . . . . .

Choose the parametrization

r(x, θ) = ⟨u, f(u) cos θ, f(u) sin θ⟩

where a ≤ x ≤ b, 0 ≤ θ ≤ 2π. Then

rx =⟨1, f′(x) cos θ, f′(x) sin θ

⟩rθ = ⟨0,−f(x) sin θ, f(x) cos θ⟩

rx × rθ =⟨f′(x)f(x),−f(x) cos θ,−f(x) sin θ

⟩|ru × rv| = f(x)

√1 + f′(x)2

So

A =

∫ 2π

0

∫ b

af(x)

√1 + f′(x)2 dx dθ = 2π

∫ b

af(x)

√1 + f′(x)2 dx

. . . . . .

Outline

Last time: Integration over polar regionsIntegration in Polar Coordinates

Easy AreaRectangles and Parallelograms in the planeParallelograms in space

Area of curved surfacesDivideApproximateTake the Limit

Special casesGraphsSurfaces of Revolution

Worksheet

Next Time

. . . . . .

Worksheet #1

ProblemFind the surface area of the part of the plane z = 2 + 3x + 4y that liesabove the rectangle [0, 5] × [1, 4].

SolutionThis is a graph. So

A =

∫ 5

0

∫ 4

1

√1 + 9 + 16 dy dx = 15

√26

. . . . . .

Worksheet #1

ProblemFind the surface area of the part of the plane z = 2 + 3x + 4y that liesabove the rectangle [0, 5] × [1, 4].

SolutionThis is a graph. So

A =

∫ 5

0

∫ 4

1

√1 + 9 + 16 dy dx = 15

√26

. . . . . .

Worksheet #2

ProblemFind the surface area of theparameterized surface

x(u, v) = u2

y(u, v) = uv

z(u, v) =12

v2

where 0 ≤ u ≤ 1 and0 ≤ v ≤ 2.

0.0

0.5

1.0

0.0

0.5

1.0

1.5

2.0

0.0

0.5

1.0

1.5

2.0

. . . . . .

SolutionWe have

ru = ⟨2u, v, 0⟩rv = ⟨0, u, v⟩

ru × rv =⟨

v2,−2uv, 2u2⟩

|ru × rv| = 2u2 + v2

So

A =

∫ 1

0

∫ 2

0(2u2 + v2) dv du = 4

. . . . . .

Worksheet #3

ProblemFind the area of the part of thesurface y = 4x + z2 that liesbetween the planes x = 0,x = 1, z = 0, and z = 1.

0.0

0.5

1.0

0

2

4

0.0

0.5

1.0

. . . . . .

SolutionUse the parametrization

r(u, v) =⟨

u, 4u + v2, v⟩

where 0 ≤ u ≤ 1, 0 ≤ v ≤ 1. Then

ru = ⟨1, 4, 0⟩rv = ⟨0, 2v, 1⟩

ru × rv = ⟨4,−1, 2v⟩

So

A =

∫ 1

0

∫ 1

0

√17 + 4v2 dv du =

∫ 1

0

√17 + 4v2 dv

. . . . . .

About the integral

To find∫ 1

0

√17 + 4v2 dv, use the substitution 2v =

√17 tan θ. Then

2 dv =√

17 sec2 θ dθ and√

17 + 4v2 =√

17 sec θ. So∫ 1

0

√17 + 4v2 dv =

172

∫ arctan(2/√

17)

0sec3 θ dθ

=174

[sec θ tan θ + ln | sec θ + tan θ|]θ=arctan(2/√

17)θ=0

=174

[sec

(arctan

(2√17

))tan

(arctan

(2√17

))+ ln

∣∣∣∣sec(

arctan(

2√17

))+ tan

(arctan

(2√17

))∣∣∣∣]

. . . . . .

Now sec(

arctan(

2√17

))=

√2117

. So this simplifies (a little) to

174

[√21√17

2√17

+ ln

∣∣∣∣∣√

2117

+2√17

∣∣∣∣∣]

or+ √212

+174

ln

(2 +

√21√

17

)

. . . . . .

Worksheet #4

ProblemFind the area of the part of the sphere x2 + y2 + z2 = 4z that lies insidethe paraboloid z = x2 + y2.

First find a better descriptionof the surface. The twosurfaces intersect when

z + z2 = 4z =⇒ z = 0, 3

So we want the portion of thesphere where z ≥ 3.

..x

.z

. . . . . .

Solving the sphere equation gives

z = 2 +√

4 − x2 − y2

So we want

A =

∫∫D

√1 +

(∂z∂x

)2

+

(∂z∂y

)2

dA

where D is the circle of radius√

3 in the xy-plane.

. . . . . .

The integrand becomes2√

4 − x2 − y2, which makes it good for

integration in polar coordinates!

A =

∫ 2π

0

∫ √3

0

2√4 − x2 − y2

r dr dθ = 4π

∫ √3

0

r dr√4 − r2

Regular u-substitution u = 4 − r2, du = −2r dr gives

A = 4π

. . . . . .

Second parametrization

Use a version of spherical coordinates to get

r(u, v) = ⟨2 sin u cos v, 2 sin u sin v, 2 cos u + 2⟩

where 0 ≤ u ≤ π

3, 0 ≤ v ≤ 2π. Then

ru = ⟨2 cos u cos v, 2 cos u sin v,−2 sin u⟩rv = ⟨−2 sin u sin v, 2 sin u cos v, 0⟩

ru × rv =⟨

4 sin2 u cos v, 4 sin2 u sin v, 4 sin u cos 4⟩

|ru × rv| = 4 sin u.

So

A =

∫ π/3

0

∫ 2π

04 sin u dv du = 4 · 2π [− cos u]u=π/3

u=0 = 4π

. . . . . .

Outline

Last time: Integration over polar regionsIntegration in Polar Coordinates

Easy AreaRectangles and Parallelograms in the planeParallelograms in space

Area of curved surfacesDivideApproximateTake the Limit

Special casesGraphsSurfaces of Revolution

Worksheet

Next Time

. . . . . .

Next time:Triple Integrals