Section 12.6
Surface Area
MATH 122 (Section 12.6)
Surface Area
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Surface Area
For a curve C , we can straighten it out and measure its length.
For a surface, we cannot make it flat without changing its area. What is
the intuition of the area of a surface?
The answer lies in the concept of limits. Let’s revisit the reasoning which
led to our formulation of the arclength:
Z
b
0 ~r (t) dt
a
Similar reasoning will allow us to calculate the surface area as a double
integral.
MATH 122 (Section 12.6)
Surface Area
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Recall: Arclength
Let ~r (t) = hf (t), g (t), h(t)i be a smooth curve for t ∈ [a, b].
Partition [a, b] into n subintervals of equal size
∆t = b−a
n .
a = t0 < . . . < ti < ti+1 < . . . < tn = b
Over each interval [ti , ti+1 ], then length of the
curve is approximately
|~r (ti+1 ) − ~r (ti )| ≈ ~r 0 (ti ) ∆t
The total length is then lim
n→∞
The arclength is
Z
Z b
0 ~r (t) dt =
L=
a
MATH 122 (Section 12.6)
b
n
X
0 ~r (ti ) ∆t.
i=1
q
[f 0 (t)]2 + [g 0 (t)]2 + [h0 (t)]2 dt.
a
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We use a similar approach to obtain the area of a surface S with
parametrization
~r (u, v ) = hx(u, v ), y (u, v ), z(u, v )i
Domain D
We partition the domain D into m × n subregions Dij where 1 ≤ i ≤ m
and 1 ≤ j ≤ n. Each subregion corresponds to a portion of the surface,
call this part Sij .
S = ∪Sij
MATH 122 (Section 12.6)
Sij = ~r (Dij ), the Image of Dij under ~r
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Approximate Sij as a rectangle of width ~ru (ui , vj ) ∆u and length
~rv (ui , vj ) ∆v .
Using the parallelogram property of the cross product:
Area(Sij ) ≈ |~ru (ui , vj ) ∆u × ~rv (ui , vj ) ∆v | = |~ru (ui , vj ) × ~rv (ui , vj ) | ∆u ∆v
The area of S is approximately
m X
n
X
|~ru (ui , vj ) × ~rv (ui , vj ) | ∆u ∆v .
i=1 j=1
Taking the limit as (m, n) → (∞, ∞) yields
Z Z
Area(S) =
|~ru (u, v ) × ~rv (u, v ) | dA
D
MATH 122 (Section 12.6)
Surface Area
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A torus (surface of a donut) can be obtained by rotating the circle on the
yz-plane centered at (0, b, 0) with radius a (b > a > 0) about the z-axis.
What is the surface area of the torus?
Solution: The torus has a parametrization
x(θ, α) = b cos(θ) + a cos(α) cos(θ)
y (θ, α) = b sin(θ) + a cos(α) sin(θ)
z(θ, α) = a sin(α)
where 0 ≤ θ ≤ 2π and 0 ≤ α ≤ 2π.
|~rθ × ~rα | = a(b + a cos(α))
The surface area is calculated
Z
Z 2π Z 2π
a(b + a cos(α)) dθ dα = 2πa
0
(b + a cos(α)) dα = 4π 2 ab
0
0
MATH 122 (Section 12.6)
2π
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Surface Area of a 2-Variable Function
The surface area of the graph of z = f (x, y ) defined on D has a natural
parametrization ~r (x, y ) = hx, y , f (x, y )i.
~rx (x, y ) = h1, 0, fx (x, y )i
~ry (x, y ) = h0, 1, fy (x, y )i
Since ~rx × ~ry = h−fx , −fy , 1i, we have |~rx × ~ry | =
q
1 + fx2 + fy2 .
The area of the surface is given by
Z Z q
1 + fx (x, y )2 + fy (x, y )2 dA
D
MATH 122 (Section 12.6)
Surface Area
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Calculate the surface area of the sphere of radius a.
Solution: Half of the sphere is the graph z = f (x, y ) =
where x 2 + y 2 ≤ a2 .
−x
fx (x, y ) = p
a2 − x 2 − y 2
−y
fy (x, y ) = p
a2 − x 2 − y 2
Thus the area of the sphere is
v
!2
Z Z u
u
−x
t
2
1+ p
+
a2 − x 2 − y 2
D
Z
2π
Z
a
=2
0
0
MATH 122 (Section 12.6)
√
a
r dr dθ
− r2
a2
p
a2 − x 2 − y 2
=
Surface Area
−y
p
a2 − x 2 − y 2
!2
dA
4πa2
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