lesson 21: surface area
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TRANSCRIPT
. . . . . .
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