Surface Area of Parametrized Surfaces

Transcription

1 Math 3B Discussion ession Week 7 Notes May 1 and 1, 16 In 3A we learned how to parametrize a curve and compute its arc length. More recently we discussed line integrals, or integration along these curves, and realized that this generalizes the computation of arc length. We have a similar story for parametrized surfaces. We ll start by computing their surface areas, and then move onto surface integrals over these surfaces. urface Area of Parametrized urfaces To motivate the computation of surface area for parametrized surfaces, let s first recall how we compute arc length for parametrized curves. Given a parametrization r(t) of a curve, we d like to know the contribution to surface area being made at time t. Notice that this depends on our parametrization, as the figures below indicate: On the left (with the green tangent vectors) we have the usual parametrization of the unit circle: r 1 (t) : (cos t, sin t) t [, π]. On the right (with the red tangent vectors) we re using a parametrization related to the stereographic projection of the circle: ( ) t r (t) : t + 1, t 1 < t <. t + 1 (The stereographic projection parametrization will actually never reach the north pole (, 1), but missing one point won t affect arc length.) We immediately see a difference between these parametrizations: the tangent vectors given by the usual parametrization all have the same length, while the stereographic parametrization tangent vectors have varying lengths. This is related to the speed with which the parametrizations are tracing out the circle. The standard parametrization traces out equal portions of the circle in equal time. For instance, it traces out the upper semicircle as t varies from to π and the lower semicircle as t varies 1

2 from π to π. The stereographic parametrization, on the other hand, traces out the lower semicircle as t varies from 1 to 1 and spends the rest of its infinite time trying to trace out the upper semicircle. o how do we measure arc length when a parametrization is not tracing out equal length in equal time? We need to measure the instantaneous contribution to our arc length (whatever that means) at time t and then integrate these infinitesimal contributions to determine the arc length. For reasons we won t go into now (but which aren t that complicated), it turns out that the magnitude of the velocity vector gives just the right measure of the instantaneous contributions to arc length. We have arc length of C b a r (t) dt, where r is a (regular) parametrization of C on the interval [a, b]. In our examples we have π π arc length sin t, cos t dt sin t + cos tdt and arc length π 1dt π t 4 + 8t + 4t 4 dt (t + 1) 4 (t + 1), 4t (t + 1) dt arctan t 1 + t ) π. ( π π dt (1 + t) 4 (1 + t ) dt 4 4 (1 t ) (t + 1) t (t + 1) 4 dt The situation is very similar in the world of parametrized surfaces. For instance, consider the parametrization of the unit sphere in R 3 given by G(u, v) (cos u sin v, sin u sin v, cos v), for u π and v π. As you re well aware, where curves have tangent lines, surfaces have tangent planes, and it happens that the tangent plane at the point G(u, v ) in our surface is determined by the vectors T u u (u, v ) and T v v (u, v ). For instance, here s a plot of our parametrization of the sphere, along with a few pairs of tangent vectors:

3 Notice that unlike our standard parametrization of the unit circle, the tangent vectors are not constant along the unit sphere. This is indicative of the changing speed with which we re parametrizing the sphere, and of the non-constant nature of the local contribution to the surface area. Just as in the case of arc length, the tangent vectors given by our parametrization provide the perfect measurement of this local contribution. Previously, we used the length of a tangent vector to find the local contribution to a length measurement. Now we have two tangent vectors and we re looking for the local contribution to an area measurement, so it makes good sense to compute a cross product. In particular, we consider the cross product of our tangent vectors. In symbols we have d T u T v da u v dudv. Here is intended to represent the surface area of the surface parametrized by G, so that d is the local contribution to the surface area. ince T u T v is a scalar quantity, we may compute the surface area of as a scalar surface integral: surface area of d d b c a u v dudv, where we assume G parametrizes on the domain u [a, b], v [c, d]. Example. Use the provided parametrization to compute the surface area of the unit sphere. (olution) ince G(u, v) (cos u sin v, sin u sin v, cos v), we have and (u, v) sin u sin v, cos u sin v, u (u, v) cos u cos v, sin u cos v, sin v. v 3

4 Then the cross product is given by u v cos u sin v, sin u sin v, cos u cos v sin v cos v sin u sin v. o we have d u v cos dudv v sin v + sin 4 vdudv Finally, the surface area is given by sin vdudv sin vdudv. A π π sin vdudv π π sin vdv π [ cos v] π π (1 ( 1)) 4π. urface Integrals Now that we have what s known as a volume form (for surfaces this is d, the local surface area contribution; for curves it s the local arc length contribution ds), integration of scalar functions becomes easy, just as with line integrals: Theorem 1. Let G(u, v) be a parametrization of a surface with parameter domain D. Assume that G is continuously differentiable, one-to-one, and regular, except possibly at the boundary of D. Then f(x, y, z)d f(g(u, v)) N(u, v) dudv. D Also in analogue with line integrals, we define vector surface integrals by reducing to a scalar surface integral: Definition. Let R 3 be a surface and suppose F is a vector field whose domain contains. We define the vector surface integral of F along to be F d : (F n)d, where n(p ) is the unit normal vector to the tangent plane of at P, for each point P in. Yet another similarity is given by the way our new integral interacts with parametrizations: Theorem. Let G(u, v) be an oriented parametrization of an oriented surface with parameter domain D. Assume that G is one-to-one and regular, except possibly at the boundary of D. Then F d F(G(u, v)) N(u, v)dudv. D 4

5 We ll use all of this new machinery to solve a problem similar to a problem we solved with line integrals: determining fluid flow through a surface. When we discussed line integrals, we imagined that we had a curve submersed in liquid through which fluid might flow. We then calculated the flow rate (we called it flux) through the curve by dotting the vector field with a positively-oriented unit normal vector to our curve and integrating this scalar quantity along the curve. The situation is very similar for surfaces, except that now our integral is already defined using a normal vector, so in fact the flow rate is given by the usual surface integral: If a fluid is flowing with velocity vector field v, its flow rate across, measured in units of volume per unit time, is given by flow rate across v d. Example. (ection 17.5, Exercise 6) upose we have a net given by y 1 x z, where y, and this net is dipped in a river. If the river has velocity vector field given by v x y, z + y + 4, z, determine the flow rate of water through the net in the positive y-direction. (olution) Here s a plot of the net in question, along with several vectors depicting the flow of the river. Note that these velocity vectors have been given unit length for purposes of the figure, but in fact have varying magnitudes. Now we need a parametrization of this surface. Because the projection of this surface onto the xz-plane is a unit circle, it s very natural to define G(r, θ) (x(r, θ), y(r, θ), z(r, θ)), where x(r, θ) r cos θ, z(r, θ) r sin θ, and then y(r, θ) 1 r, 5

6 for r 1 and θ π. We can now compute our tangent vectors at each point (r, θ): T r (r, θ) cos θ, r, sin θ and T θ (r, θ) r sin θ,, r cos θ. o we have a normal vector is given by i j k T r (r, θ) T θ (r, θ) cos θ r sin θ r sin θ r cos θ r cos θ, r, r sin θ. Notice that since r <, this vector points in the negative y-direction, opposite our orientation. o we choose the normal vector pointing in the opposite direction and have Next, notice that o we have N(r, θ) r cos θ, r, r sin θ. v(g(r, θ)) r cos θ (1 r ), r sin θ + (1 r ) + 4, r sin θ. v(g(r, θ)) N(r, θ) r 3 cos θ r cos θ(1 r ) + r sin θ + 5r r 3 + r 4 sin 3 θ Finally, our flow rate across is given by π 1 v d v(g(r, θ)) N(r, θ)drdθ π 1 r cos θ(r 1 + r cos θ) + r 4 sin 3 θ + r(5 r + r sin θ). r cos θ(r 1 + r cos θ) + r 4 sin 3 θ + r(5 r + r sin θ)drdθ 5π. This last integral isn t hard, it s just really ugly. ince this integral is not very nice, one might expect a spherical parametrization such as G(θ, φ) (cos θ sin φ, cos φ, sin θ sin φ) would work out better. I didn t find the resulting integral to be any nicer. Line Integrals and urface Integrals We ll finish by summarizing the various integrals we ve considered in the last few sections. Throughout, C is a curve parametrized by r(t) with t [a, b] and is a surface parametrized by G(u, v) with (u, v) D. In the flow rate line, we have a fluid with velocity vector field v. We write T for a unit tangent vector to a curve C, and we write n for a unit normal vector to a curve C or a surface. In both cases we assume that the vectors are positively oriented. Given a parametrization G of, we also write N(u, v) T u T v u v. 6

7 arc length/surface area Line Integrals urface Integrals b a r (t) dt D integral of scalar function f u v dudv b a f(r(t)) r (t) dt integral of vector field F f(g(u, v)) D u v dudv C F dr : (F T)ds b F(r(t)) C a r (t)dt F d : (F n)d D fluid flowing through C or F(G(u, v)) N(u, v)dudv C v nds b v(r(t)) a n(t) r (t) dt v d D F(G(u, v)) N(u, v)dudv 7