Surfaces and surface integrals 1 Definition of a surface It is intuitive that a curve is a one dimensional object while a surface has two dimensions. A point can be located on a curve by giving only one coordinate, its distance from the start of the curve, for example, while we need two coordinates to identify a point on a surface. Therefore, while a curve is a vector function of only one parameter, r(t) = f (t)i + g(t)j + h(t)k,
a ≤ t ≤ b,
a surface is a vector function of two parameters, r(u, v) = x(u, v)i + y(u, v)j + z(u, v)k, where u and v vary in a region D. More formally, Definition - A surface is an ordered set of three continuous functions defined in a closed connected region D: x = x(u, v), y = y(u, v), (u, v) ∈ D. z = z(u, v),
v
z D r(u,v) y
u x
Figure 1: As the parameters u and v vary in the domain D the position vector describes a surface in the three dimensional space (x, y, z). The coordinate grid in the domain D (left figure) are transformed in a coordinate grid on the surface (right figure). For example, the coordinate lines u = u0 , where u0 is a constant, (dashed lines) becomes the coordinate curves r(u0 , v).
MA156 - Surfaces and surface integrals
2
z
z
1
1 r(u,v)
r(u,v)
v
u
v y
x
y
v x
Figure 2: Example 2
Figure 3: Example 3
Example 1 - The equations x = x1 + u(x2 − x1 ) + v(x3 − x1 ), y = y1 + u(y2 − y1 ) + v(y3 − y1 ), z = z1 + u(z2 − z1 ) + v(z3 − z1 ),
− ∞ < u, v < +∞
represent a plane that contains the points P1 = (x1 , y1, z1 ),
P2 = (x2 , y2, z2 ),
P3 = (x3 , y3 , z3 ).
Example 2 - The surface r(u, v) = a cos(u)i + a sin(u)j + vk,
0 ≤ u ≤ 2π, 0 ≤ v ≤ 1,
represents a cylinder of radius a and height 1 whose axis is the z axis (see Figure 2). Example 3 - The surface 0 ≤ u ≤ 2π, r(u, v) = v cos(u)i + v sin(u)j + vk, 0 ≤ v ≤ 1, represents a cone of height 1 whose axis is the z axis (see Figure 3). Example 4 - The surface r(u, v) = a sin(u) cos(v)i + a sin(u) sin(v)j + a cos(u)k, represents a sphere of radius a centred at the origin.
0 ≤ u ≤ π, 0 ≤ v ≤ 2π,
MA156 - Surfaces and surface integrals
3 dr dr x du dv
dr dv dv r(u0,v)
dr du du
dS
r(u 0+du,v) r(u,v 0 +dv)
r(u,v0)
Figure 4: Geometrical interpretation of the area element. By changing the coordinate of a point on the surface, r(u0 , v0 ) by small amounts du and dv, we describe a small patch whose area dS is given by the cross product of the partial derivatives of the surface r(u, v).
2 Area of a surface It can be shown that a sensible definition of the area of a surface, r(u, v) where (u, v) ∈ D, is Z ∂r ∂r Area = ∂u × ∂v dudv, D where ∂r/∂u and ∂r/∂v are the partial derivatives of the vector r(u, v) with respect to its argument. The vector n(u, v) =
∂r ∂r × ∂u ∂v
is perpendicular to the surface at the point r(u, v). Its modulus, ∂r ∂r × dS = ∂u ∂v is called the area element: it represents the area of a small patch of the surface obtained by changing the coordinates u and v by small amounts du and dv (see Figure 4). The area of the surface can also be indicated by the symbol Z Area = dS, S
the integral over the surface S of the area element dS. Example 1: Find the area of the cylinder of radius a and height h.
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Using geometry The area of a cylinder is the perimeter of the base times the height of the cylinder: Area = 2πah. Surface Element The surface of the cylinder can be parametrised by r(u, v) = a cos(u)i + a sin(u)j + vk,
0 ≤ u ≤ 2π, 0 ≤ v ≤ h.
To find the surface element we need the partial derivatives of r(u, v) with respect to u and v: ∂r = −a sin(u)i + a cos(u)j, ∂u ∂r = k, ∂v ∂r ∂r × = a cos(u)i + a sin(u)j, ∂u ∂v ∂r ∂r dudv = adudv. dS = × ∂u ∂v
Radial vector in the xy-plane
Therefore the area of the cylinder is Z Z 2π Z h Z ∂r ∂r × dudv = du dv a = 2πah. Area = dS = ∂u ∂v S 0 0 Geometrical construction of the Surface element The most “natural” coordinate system to identify a point on the surface of the cylinder is a cylindrical coordinate system (φ, z), with the implied condition that all the points of the surface have distance r = a from the z-axis (see Figure 5): r(φ, z) = a cos(φ)i + a sin(φ)j + zk,
0 ≤ φ ≤ 2π,
0 ≤ z ≤ h.
The vector normal to the surface is a vector n perpendicular to the z axis and pointing radially outward and the surface element is dS = adzdφ (see Figure 5). The area of the cylinder is Z Z 2π Z h Area = dS = adzdφ = 2πah. S
0
0
MA156 - Surfaces and surface integrals
5
z
z dθ dz z
n a
θ
a
y
a dθ
x
Figure 5: Description of a cylindrical surface and its surface element in cylindrical coordinates. Example 2: Find the area of the cone z =
p x2 + y 2 , 0 ≤ z ≤ 1.
Using geometry The surface of a cone is one half the product of the perimeter of its base times the length of the cone side. The cone described by p z = x2 + y 2 , 0 ≤ z ≤ 1, √ has a base of radius one and the side has length 2. Therefore its area is √ 2 √ Area = 2π = 2π. 2 Surface Element The surface of the cone can be parametrised by p r(x, y) = xi + yj + x2 + y 2k. Since 0 ≤ z ≤ 1 =⇒ 0 ≤
p
x2 + y 2 ≤ 1 =⇒ x2 + y 2 ≤ 1,
the domain where x and y vary is a disk D of radius one and centred at the origin. Finally the surface element is given by ∂r x =i+ p k, ∂x x2 + y 2 ∂r y =j+p k, ∂y x2 + y 2 ∂r ∂r x y × = −p i− p j + k, 2 2 2 ∂x ∂y x +y x + y2 √ ∂r ∂r dxdy = 2dxdy. dS = × ∂x ∂y
MA156 - Surfaces and surface integrals
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Therefore the cone’s area is Z √ Z √ √ 2dxdy = 2 × Area of disk D = 2π. Area = dS = S
D
Exercise - Use the parametrisation 0 ≤ u ≤ 1, 0 ≤ v ≤ 2π,
r(u, v) = u cos(v)i + u sin(v)j + uk, to evaluate the integral.
Example 3: Find the area of a hemisphere of radius a. Using geometry Area = 2πa2 . Surface Element The surface can be parametrised by r(u, v) = a cos(u) sin(v)i + a sin(u) sin(v)j + a cos(v)k,
0 ≤ u ≤ 2π, 0 ≤ v ≤ π/2.
The surface element is given by: ∂r = −a sin(u) sin(v)i + a cos(u) sin(v)j, ∂u ∂r = a cos(u) cos(v)i + a sin(u) cos(v)j − a sin(v)k, ∂v ∂r ∂r × = −a2 cos(u) sin2 (v)i − a2 sin(u) sin2 (v)j − a2 sin(v) cos(v)k, ∂u ∂v ∂r ∂r dudv = sin(v)a2 dudv. × dS = ∂u ∂v
Opposite of the radial vector
The area of the hemisphere is Z
Z Area =
dS = S
Z
2π
0
π/2
dv sin(v)a2 = 2πa2 .
du 0
Geometrical construction of the surface element The most “natural” coordinate system to describe a sphere is the spherical one, (φ, θ), with the implied condition that the radial coordinate is fixed, r = a, where a is the radius of the sphere: r(φ, θ) = a cos(φ) sin(θ)i + a sin(φ) sin(θ)j + a cos(θ)k,
0 ≤ φ ≤ 2π,
0 ≤ θ ≤ π/2.
MA156 - Surfaces and surface integrals
7 z
n
z
a dθ
θ
φ
a sinθ d φ y
a
x
y
x
Figure 6: Description of a hemispherical surface and its area element in spherical coordinates. The surface element (see Figure 6) is dS = a2 sin(θ)dθdφ, while the vector normal to the hemispherical surface points in the radial direction. The area of the hemisphere is Z π/2 Z 2π Z π/2 Z 2 2 Area = dS = dθ dφ a sin(θ) = 2πa sin(θ)dθ = 2πa2 . S
0
0
0
3 Surface integral of a function It is possible to extend the concept of integral of a function of two variables, f (x, y), on a region D of the plane, Z f (x, y) dA, D
to the integral of a function of three variables, f (x, y, z), on a surface r(u, v), Z f dS, S
the surface integral of the function f . If we assume that the surface r(u, v) is a sheet of material of surface density f , then we can interpret the surface integral as the mass of the sheet. Suppose that the surface r(u, v) is defined on a domain D. The surface integral of f (x, y, z) is defined as Z Z ∂r ∂r dudv. f dS = f [r(u, v)] × ∂u ∂v S
D
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Notice that the function f is evaluated only on the points of the surface, i.e. f [r(u, v)] = f [x(u, v), y(u, v), z(u, v)]. Example - Find the mass of the cylinder r(u, v) = a cos(u)i + a sin(u)j + vk,
0 ≤ u ≤ 2π, 0 ≤ v ≤ h,
if its surface density is f (x, y, z) = z(x2 + y 2). The area element is ∂r ∂r dudv = adudv. dS = × ∂u ∂v The function evaluated on the surface is f [r(u, v)] = f [a cos(u), a sin(u), v] = [a cos(u)]2 + [a sin(u)]2 v = a2 v. The mass of the surface is given by Z Z 2π Z M = f dS = du S
0
Z
h 2
3
dv a a v = a 0
Z
2π
v dv = πa3 h2 .
du 0
h
0
