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Lecture 35 : Surface Area; Surface Integrals
In the previous lecture we defined the surface area a(S) of the parametric surface S, defined by
r(u,v) on T, by the double integral RR
a(S) = k r ×r kdudv. (1)
u v
T
Wewill now drive a formula for the area of a surface defined by the graph of a function.
Area of a surface defined by a graph: Suppose a surface S is given by z = f(x,y), (x,y) ∈ T,
that is, S is the graph of the function f(x,y). (For example, S is the unit hemisphere defined by
p 2 2 2 2
z = 1−x −y where(x,y)liesinthecircular region T : x +y ≤ 1.) Then S can be considered
as a parametric surface defined by:
r(x,y) = xi+yj +f(x,y)k, (x,y) ∈ T.
In this case the surface area becomes
RR q 2 2
a(S) = 1+f +f dxdy. (2)
x y
T q
because k r ×r k = k −f i−f j +k k = 1+f2+f2.
u v x y x y
2 2 2 2
Example 1: Let us find the area of the surface of the portion of the sphere x + y + z = 4a
2 2
that lies inside the cylinder x + y = 2ax. Note that the sphere can be considered as a union of
p 2 2 2
two graphs: z = ± 4a −x −y . We will use the formula given in (2) to evaluate the surface
p 2 2 2
area. Let z = f(x,y) = 4a −x −y . Then
√ −x √ −y q 2 2 q 4a2
fx = , fy = and 1+f +f = 2 2 2.
2 2 2 2 2 2 x y 4a −x −y
4a −x −y 4a −x −y
Let T be the projection of the surface z = f(x,y) on the xy-plane (see Figure 1). Then, because
of the symmetry, the surface area is
q π 2acosθ
2
RR 4a2 R R 2ardrdθ
( ) = 2 dxdy × √ .
a S 4a2−x2−y2 =2 2 2 2
T 0 0 4a −r
Remark: Since
2 2 2 2 2 2 2 2 2 2
k r ×r k = kr k kr k sin θ = kr k kr k (1−cos θ) = kr k kr k −(r ·r ) ,
u v u v u v u v u v
the formula given in (1) can be written as RR √
a(S) = EG−F2dudv (3)
T
where E = r ·r ,G = r ·r and F = r ·r .
u u v v u v
2
Example 2: Let us compute the area of the torus
x=(a+bcosφ)cosθ, y =(a+bcosφ)sinθ, z =bsinφ
where 0 ≤ θ ≤ 2π, 0 ≤ φ ≤ 2π, and a and b are constants such that 0 < b < a. Since the surface is
given in the parametric form with the parameters θ and φ, we can either use the formula given in
(1) or (3) and find the surface area. We do not have to know how the surface looks like. However
the surface is given in Figure 2 for understanding. Note that
r =−(a+bcosφ)sinθi+(a+bcosφ)cosθj+0k, rv =−bsinφcosθi−bsinφsinθj+bcosφk.
θ
2 2 √ 2
This implies that E = r ·r = (a+bcosφ) , F = 0, G = b and hence EG−F =b(a+bcosϕ).
u u
Therefore, by (3), the surface area is
a(S) = RR b(a+bcosϕ)dθdφ = R2πR2πb(a+bcosϕ)dθdφ = 4π2ab.
0 0
T
Note that this problem can also be solved using the Pappus theorem : a(S) = 2πρL = 2π ·a·2πb.
Surface Integrals: We will define the concept of integrals, called surface integrals, to the scalar
functions defined on parametric surfaces. Surface integrals are used to define center of mass and
moment of inertia of surfaces, and the surface integrals occur in several applications. We will not
get in to the applications of the surface integrals in this course. We will define the surface integrals
and see how to evaluate them.
Let S be a parametric surface defined by r(u,v),(u,v) ∈ T. Suppose r and r are continuous.
u v
Let g : S → R be bounded. The surface integral of g over S, denoted by RR gdσ, is defined by
S
RR RR RR √ 2
g dσ = g(r(u,v)) k r ×r k dudv = g(r(u,v)) EG−F dudv (4)
u v
S T T
provided the RHS double integral exists. If S is defined by z = f(x,y), then
RR RR q 2 2
g dσ = g[x,y,f(x,y)] 1+f +f dxdy. (5)
x y
S T
where T is the projection of the surface S over the xy-plane.
2 2 2 1/2 RR dσ
Example3: LetSbethehemisphericalsurfacez = (a −x −y ) . Let us evaluate 2 2 2 1/2.
S [x +y +(z+a) ]
Wefirst parameterize the surface S as follows:
S := r(θ,φ) = (asinφcosθ, asinφsinθ, acosφ), 0 ≤ θ ≤ 2π, 0 ≤ φ ≤ π.
√ 2 2 2 2 2 1/2 φ
Simplecalculation shows that EG−F =a sinφand[x +y +(z+a) ] =2acos 2. Therefore,
by equation (4), the surface integral is
RR dσ =R2πRπ/2 a2sinφ dφdθ.
2 2 2 1/2 0 0 φ
S [x +y +(z+a) ] 2acos 2
Example4: LetusevaluatethesurfaceintegralRR g dσ whereg(x,y,z) = x+y+z andthesurface
S
Sisdescribedbyz = 2x+3y, x ≥ 0, y ≥ 0andx+y ≤ 2. Weusetheformulagivenin(5)toevaluate
the surface integral. Note that the projection T of the surface is {(x,y) : x ≥ 0, y ≥ 0, x+y ≤ 2}.
The surface integral is
RR RR q 2 2 R2R2−y √
g dσ = (x+y+z) 1+fx+fy dxdy= 0 0 (x+y+2x+3y) 14dxdy.
S T
Remark: Under certain general conditions (we deal with surfaces satisfying such conditions) the
value of the surface integral is independent of the representation.
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