of water in the oceans of the earth. At each point of the surface the depth of the ocean is given by a function f.
To measure the total volume we divide up the surfaceS into smaller surface areas 4Sij, each of which is so small that we can think of it as approximately flat under
which the depth of the ocean is approximately constant. The volume of water below 4Sij is approximatelyf(xij, yij, zij)4Sij, where (xij, yij, zij) is any point in4Sij.
The total volume of water in the ocean is approximatelyP
i,jf(xij, yij, zij)4Sij.
We therefore define thesurface integral of a function f over the surface S to be
Z Z S f dS = lim 4Sij→0 X i,j f(xij, yij, zij)4Sij
where the sum is over a partition ofS into smaller surface areas4Sij, (xij, yij, zij)
is any point in4Sij and we take the limit as the partition becomes finer.
Suppose thatS is a parameterized surface: X(u, v) =x(u, v)i+y(u, v)j+z(u, v)k, where (u, v) ∈ D. Let Rij = {(u, v); ui ≤ u ≤ ui +4u, vj ≤ v ≤ vj + 4v}
be a small rectangle in the u-v plane and let Sij be the image of Rij under the
map (u, v)→X(u, v). Then the area of Sij is approximately the area in of the
parallelogram in the tangent plane spanned by the vectors Xu4u and Xv4v:
4Sij ∼
Xu ×Xv(ui, vj)
4u4v
Substituting this we get a Riemann sum for a double integral in theu-v plane. We therefore define the surface integral of a function f over a surfaceS:
Z Z S
f dS =
Z Z D
f x(u, v), y(u, v), z(u, v)Xu ×Xv(u, v) dudv
We can symbolically write
dS =Xu×Xv(u, v) dudv
Ex. FindRR Sz
2dS, whereS ={(x, y, z); x2+y2+z2 = 1} is the unit sphere.
Sol. 1A parametrization is X(φ, θ) = sinφcosθi+sinφsinθj+cosφk, 0≤φ≤π, 0≤θ <2π, and we showed before that
dS =Xφ×Xθ(φ, θ) dφ dθ= sinφ dφ dθ Hence Z Z S z2dS = Z 2π 0 Z π 0 cos2φ sinφ dφ dθ = Z 2π 0 −cos 3φ 3 π 0dθ = Z 2π 0 2 3 dθ= 4π 3 Sol. 2 A parametrization of the northern hemisphere S+ is X(x, y) = xi+yj+
f(x, y)k, where f(x, y) =p1−x2−y2 and dS =kXx×Xykdxdy= q 1 +f2 x +fy2dxdy= dxdy p 1−x2−y2
Since the integral of z2 over the southern and northern hemispheres are the same
Z S z2dS = 2 Z S+ z2dS = 2 Z Z x2+y2≤1 (1−x2−y2)p dxdy 1−x2−y2 = 2 Z 2π 0 Z 1 0 (1−r2)1/2rd dθ= 2 Z 2π 0 1 3(1−r 2 )3/2 1 0dθ= 2 Z 2π 0 1 3dθ= 4π 3 Ex. Find Z Z S
x dS, whereS is the triangle with vertices (1,0,0), (0,1,0) and (0,0,1). Sol. The surface is a piece of a plane ax+by+cz=d and putting in the 3 points we get a=d, b=d and c=d, e.g. a=b=c=d= 1. The surface is therefore given by h(x, y, z) =x+y+z = 1 and (x, y) ∈ D = {(x, y);x ≥ 0, y ≥ 0, x+y ≤ 1}. The normal to the surface is therefore n=∇h/|∇h|= (i+j+k)/√3. We have
dS = dxdy |cosγ| = dxdy |n·k| = √ 3dxdy
If we rewrite D={(x, y); 0≤x ≤1, 0≤y ≤1−x} we get
Z Z S x dS = Z Z D x√3dxdy= Z 1 0 Z 1−x 0 x√3dy dx = Z 1 0 xy√3 1−x y=0 dx= Z 1 0 x(1−x)√3dx= √ 3 6
A surface is calledclosed if it has no boundary.
The sphere is closed but the upper hemisphere is not since its boundary is the equator. A closed surface has an inside and an outside.
Anoriented surfaceis a two-sided surface with one side specified as theoutside. Theoutward normal points away from from the outside of the surface.
A orientation can hence be specified by giving an outward normal everywhere. The upper hemisphere is orientable.
A M¨obius strip is not orientable since it only has one side.
A surface is called regular if it has a tangent plane everywhere. A parameterized surface is regular ifXu×Xv is nonvanishing everywhere.
A sphere is a regular surface but a cone is not regular at the tip.
There are a couple of alternative ways to express the surface area element: dS =kXu×Xvkdudv =
p
kXuk2kXvk2−(Xu·Xv)2 dudv
and sinceX=xi+yj+zkwe can also write (after some work)
dS =kXu ×Xvkdudv= s ∂(x, y) ∂(u, v) 2 + ∂(y, z) ∂(u, v) 2 + ∂(x, z) ∂(u, v) 2 dudv
For a surface of revolution of a function y=f(x) about the x-axis we have
Area = 2π
Z b a
|f(x)|p1 + (f0(x))2dx
and revolved about they axis
Area = 2π
Z b a
Surface Integrals of vector functions.
The flow rate of fluid out of the total surface S, or the flux of the velocity vector fieldF out of the surface S, with outward unit normal n, is given by
Z Z S
F·n dS
That only the normal component of F matters is clear since a tangential velocity would not contribute to the flow of fluid out from the surface.
The question of how to calculate the flux reduces how to calculate surface integrals. In a parametrization X=X(u, v) we have
dS =kXu×Xvkdudv, n =±Xu ×Xv/kXu ×Xvk. Hence Z Z S F·n dS =± Z Z F·(Xu×Xv)dudv
Here the sign is positive if Xu×Xv points out from the surface.
Ex. Find the flux of F = xi+yj−2zk out of the surface S of the cube C = {(x, y, z); 0≤x≤1, 0≤y≤1, 0≤z ≤1}.
Sol. The Cube has six sides S1 with x= 0, S2 withx = 1,S3 with y = 0,S4 with
y = 1, S5 with z = 0 and S6 with z = 1. On S1, the outward normal is −i and
F·n= (yj−2zk)·(−i) = 0, on S2,F·n= (i+yj−2zk)·i= 1, on S3, F·n= 0,
on S4, F·n = 1, on S5, F·n = 0, and on S6, F·n = −2. Since the area of each
side is one it follows that
Z Z S F·ndS = Z Z S1 F·ndS+...+ Z Z S6 F·ndS = 0 + 1 + 0 + 1 + 0−2 = 0
Ex. Find the flux of the vector field F = xi+ yj −2zk out of the sphere S ={(x, y, z); x2+y2+z2 = 1}.
Sol. The surface can be written h(x, y, z) = x2+y2 +z2 = 1. The outward unit
normal to the unit sphere is n = ∇h/|∇h| = (xi+ yj +zk)/px2+y2+z2 =
xi+yj+zk, when x2+y2+z2 = 1. Therefore Z Z S F·ndS = Z Z S x2+y2−2z2dS
There are several ways to proceed:
(1) In Spherical coordinates, X(φ, θ) = sinφcosθi+ sinφsinθj+ cosφk. Then dS =|Xφ×Xθ|dφdθ= sinφ dφdθ. Hence Z Z S F·ndS = Z 2π 0 Z π 0
sin2φcos2θ+ sin2φsin2θ−2 cos2φsinφ dφ dθ
=
Z 2π
0
Z π
0
(sin2φ−2 cos2φ) sinφ dφdθ =
Z 2π 0 Z π 0 (1−3 cos2φ) sinφ dφdθ = Z 2π 0 −cosφ+ cos3φ π φ=0dθ= 0
(2) The sphere S can be written as the union of the northern hemisphere S+ and
southern hemisphere S+ and each of these can be viewed as a graph over the x-y
planeS± ={(x, y, z); z =p1−x2−y2, (x, y)∈D}, whereD={(x, y); x2+y2≤1}.
Since the integrand and the sphere are symmetric under changingz to−z we have
Z Z S x2+y2−2z2dS = 2 Z Z S+ x2+y2−2z2dS
We can now instead writedS = dxdy |cosγ| = dxdy |n·k| = dxdy z = dxdy p 1−x2−y2 so Z Z S+ x2+y2−2z2dS = Z Z D 3(x2+y2)−2p dxdy 1−x2−y2
Introducing polar coordinates we get
Z Z D dxdy p 1−x2−y2 = Z 2π 0 Z 1 0 (3r2−2)(1−r2)−1/2r dr dθ = Z 2π 0 Z 1 0 −3(1−r2)1/2r+(1−r2)−1/2dr dθ= Z 2π 0 (1−r2)3/2−(1−r2)1/2 1 r=0 dθ= 0.
(3) Finally, one can also use symmetry to see that
Z Z S x2dS = Z Z S y2dS = Z Z S z2dS.
