Parametric Surfaces and Their Areas

Jim Lambers MAT 280 Spring Semester 2009-10 Lecture 21 Notes

These notes correspond to Section 13.6 in Stewart and Sections 7.3 and 7.4 in Marsden and Tromba.

We have learned that Green’s Theorem can be used to relate a line integral of a two-dimensional vector field F over a closed plane curve C to a double integral of a component of curl F over the region D that is enclosed by C. Our goal is to generalize this result in such a way as to relate the line integral of a three-dimensional vector field F over a closed space curve C to the integral of a component of curl F over a surface enclosed by C. We have also learned that Green’s Theorem relates the integral of the normal component of a two-dimensional vector field over a closed curve C to the double integral of div F over the region D that it encloses. We wish to generalize this result in order to relate the integral of the normal component of a three-dimensional vector field F over a closed surface S to the triple integral of div F over the solid E contained within S. In order to realize either of these generalizations, we need to be able to integrate functions over piecewise smooth surfaces, just as we now know how to integrate functions over piecewise smooth curves. Whereas a smooth curve C, being a curved one-dimensional entity, is most conveniently described by a parameterization r(t), where a ≤ t ≤ b and r(t) is a differentiable function of one vari- able, a smooth surface S, being a curved two-dimensional entity, is most conveniently described by a parametriation r(u, v), where (u, v) lies within a 2-D region, and r(u, v) = ⟨x(u, v), y(y, v), z(u, v)⟩ is a differentiable function of two variables. We say that S is a parametric surface, and x = x(u, v), y = y(u, v), z = z(u, v) are the parametric equations of S. Example The Möbiusstrip is a surface that is famous for being a nonorientable surface; that is, it “has only one side”. It can be parameterized by v u v u v u x(u, v) = 1 + cos cos u, y(u, v) = 1 + cos sin u, z(u, v) = sin , 2 2 2 2 2 2 where 0 ≤ u ≤ 2 and −1 ≤ v ≤ 1. It is shown in Figure 1. □ Example The paraboloid defined by the equation x = 4y2 + 4z2, 0 ≤ x ≤ 4, can also be defined by the parametric equations √ √ x x x = x, y = cos , z = sin , 2 2

1 Figure 1: The M¨obiusstrip where 0 ≤ ≤ 2, since for each x, a point (x, y, z) on the paraboloid must lie on a circle centered at (x, 0, 0) with radius px/4, parallel to the yz-plane. This is an example of a surface of revolution, since the surface is obtained by revolving the curve y = f(x) around the x-axis. □

Let P0 = (x0, y0, z0) = r(u0, v0) be a point on a parametric surface S. A curve deﬁned by g(v) = r(u0, v) that lies within S and passes through P0 has the tangent vector ∂x ∂y ∂z r = g′(v) = (u , v ), (u , v ), (u , v ) v ∂v 0 0 ∂v 0 0 ∂v 0 0 at P0. Similarly, the tangent vector at P0 of the curve h(u) = r(u, v0), that also lies within S and passes through P0, is ∂x ∂y ∂z r = h′(u) = (u , v ), (u , v ), (u , v ) . u ∂u 0 0 ∂u 0 0 ∂u 0 0

If these vectors are not parallel, then together they deﬁne the tangent plane of S at P0. Its normal vector is ru × rv = ⟨a, b, c⟩ which yields the equation a(x − x0) + b(y − y0) + c(z − z0) = 0

2 of the tangent plane. Example (Stewart, Section 13.6, Exercise 30) Consider the surface deﬁned by the parametric equations x = u2, y = v2, z = uv, 0 ≤ u, v ≤ 10.

At the point (x0, y0, z0) = (1, 1, 1), which corresponds to u0 = 1, v0 = 1, the equation of the tangent plane can be obtained by ﬁrst computing the partial derivatives of the coordinate functions. We have xu = 2u, yu = 0, zu = v,

xv = 0, yv = 2v, zv = u.

Evaluating at (u0, v0) yields

ru = ⟨xu, yu, zu⟩ = ⟨2, 0, 1⟩, rv = ⟨xv, yv, zv⟩ = ⟨0, 2, 1⟩.

It follows that the normal to the tangent plane is

n = ru × rv = ⟨2, 0, 1⟩ × ⟨0, 2, 1⟩ = ⟨−2, −2, 4⟩.

We conclude that the equation of the tangent plane is

−2(x − 1) − 2(y − 1) + 4(z − 1) = 0.

□

The vectors ru and rv are helpful for computing the area of a smooth surface S. For simplicity, we assume that S is parametrized by a function r(u, v) with domain D, where D = [a, b] × [c, d] is a rectangle in the uv-plane. We divide [a, b] into n subintervals [ui−1, ui] of width Δu = (b − a)/n, and divide [c, d] into m subintervals [vj−1, vj] of width Δv = (d − c)/m. Then, r approximately maps the rectangle Rij with lower left corner (ui−1, vj−1) into a parallelogram with adjacent edges deﬁned by the vectors

r(ui, vj−1) − r(ui−1, vj−1) ≈ ruΔu and r(ui−1, vj) − r(ui−1, vj−1) ≈ rvΔv. The area of this parallelogram is Aij = ∥ru × rv∥ΔuΔv. Adding all of these areas approximates the area of S, which we denote by A(S). If we let m, n → ∞, we obtain n m X X ZZ A(S) = lim Aij = ∥ru × rv∥ dA. m,n→∞ i=1 j=1 D

3 Example (Stewart, Section 13.6, Exercise 34) We wish to ﬁnd the area of the surface S that is the part of the plane 2x + 5y + z = 10 that lies inside the cylinder x2 + y2 = 9. First, we must ﬁnd parametric equations for this surface. Because x and y are restricted to the circle of radius 3 centered at the origin, it makes sense to use polar coordinates for x and y. We then have the parametric equations

x = u cos v, y = u sin v, z = 10 − u(2 cos v + 5 sin v), where 0 ≤ u ≤ 3 and 0 ≤ v ≤ 2. We then have

ru = ⟨xu, yu, zu⟩ = ⟨cos v, sin v, −2 cos v − 5 sin v⟩,

rv = ⟨xv, yv, zv⟩ = ⟨−u sin v, u cos v, u(2 sin v − 5 cos v)⟩.

We then have √ ∥ru × rv∥ = ∥⟨2u, 5u, u⟩∥ = ∣u∣ 30. It follows that Z 3 Z 2 √ √ Z 3 √ A(S) = u 30 du dv = 2 30 u du = 9 30. 0 0 0

It should be noted that it is to be expected that the direction of ru × rv is parallel to the normal vector of the plane 2x + 5y + z = 10, since it is normal to the surface at every point. □ Often, a surface is deﬁned to be the graph of a function z = f(x, y). Such a surface can be parametrized by x = u, y = v, z = f(u, v), (u, v) ∈ D. It follows that ru = ⟨1, 0, fu⟩, rv = ⟨0, 1, fv⟩.

We then have rv × ru = ⟨fu, fv, 1⟩, which yields the equation of the tangent plane ∂f ∂f (u , v )(x − x ) + (u , v )(y − y ) + (z − z ) = 0, ∂u 0 0 0 ∂v 0 0 0 0 which, using the relations x = u and y = v, can be rewritten as ∂f ∂f (x , y )(x − x ) + (x , y )(y − y ) + (z − z ) = 0. ∂x 0 0 0 ∂y 0 0 0 0 Recall that this is the equation of the tangent plane of a surface deﬁned by an equation of the form z = f(x, y) that had been previously deﬁned. It follows that the area of such a surface is given by the double integral s ZZ df 2 df 2 A(S) = 1 + + dA. D dx dy

4 Example (Stewart, Section 13.6, Exercise 38) To ﬁnd the area A(S) of the surface z = 1+3x+2y2 that lies above the triangle with vertices (0, 0), (0, 1) and (2, 1), we compute ∂z ∂z = 3, = 4y, ∂x ∂y and then evaluate the double integral s Z 1 Z 2y ∂z 2 ∂z 2 A(S) = 1 + + dx dy 0 0 ∂x ∂y Z 1 Z 2y p = 10 + 16y2 dx dy 0 0 Z 1 p = 2y 10 + 16y2 dy 0 1 Z 26 = u1/2 du 16 10 26 1 3/2 = u 24 10 1 = (263/2 − 103/2) 24 ≈ 4.206. □ A surface of revolution S that is obtained by revolving the curve y = f(x), a ≤ x ≤ b, around the x-axis has parametric equations x = u, y = f(u) cos v, z = f(u) sin v, where a ≤ u ≤ b and 0 ≤ v ≤ 2. From these equations, we obtain

p ′ 2 ∥ru × rv∥ = ∣f(u)∣ 1 + [f (u)] , which yields Z b p A(S) = 2 ∣f(u)∣ 1 + [f ′(u)]2 du. a If y = f(x) is revolved around the y-axis instead, then the area is

Z b p A(S) = 2 ∣u∣ 1 + [f ′(u)]2 du, a which can be obtained by considering the case of revolving x = f −1(y) around the y-axis and proceeding with a parametrization similar to the case of revolving around the x-axis.

5 Practice Problems

Practice problems from the recommended textbooks are:

∙ Stewart: Section 13.6, Exercises 29-43 odd

∙ Marsden/Tromba: Section 7.3, Exercises 1, 3, 11; Section 7.4, Exercises 1, 3, 5, 11, 13