Section Parametrized Surfaces and Surface Integrals. (I) Parametrizing Surfaces (II) Surface Area (III) Scalar Surface Integrals

Transcription

1 Section 16.4 Parametrized Surfaces and Surface Integrals (I) Parametrizing Surfaces (II) Surface Area (III) Scalar Surface Integrals MATH 127 (Section 16.4) Parametrized Surfaces and Surface Integrals The University of Kansas 1 / 41

2 In chapter 13 we parametrized curves using a single parameter. More parameters are not required because a curve is a one dimensional object. In this section, we explore the parametrization of surfaces. Consider a vector function r (u, v) of two variables r (u, v) = f (u, v), g(u, v), h(u, v) with points (u, v) in the domain D which lie in a uv-plane. The set of points satisfying the equation is the surface parametrized by r. MATH 127 (Section 16.4) Parametrized Surfaces and Surface Integrals The University of Kansas 2 / 41

3 If a surface S is parametrized by r (u, v), then the surface can be sketched using the grid lines of the domain. Grid lines along a parametrized surface are the image of the vertical and horizontal lines in the domain, the uv-plane. For a constant a, r (a, v) is a parametrization of the vertical grid lines. For a constant b, r (u, b) is a parametrization of the horizontal grid lines. MATH 127 (Section 16.4) Parametrized Surfaces and Surface Integrals The University of Kansas 3 / 41

4 Example 1: Find a parametrization of the plane containing the point P and the vectors a and b. Solution: Points in the plane can be found by starting at P and moving along the vectors a then b. r (u, v) = P + u a + v b Example 1b: The plane containing the points (3, 0, 0), (0, 2, 0), and (0, 0, 1) contains the vectors 3, 0, 1 and 3, 2, 0 and has a normal vector n = 2, 3, 6. Scalar Equation: 2x + 3y + 6z = 6 Parametrization: r (u, v) = 3u 3v, 2v, u + 1 MATH 127 (Section 16.4) Parametrized Surfaces and Surface Integrals The University of Kansas 4 / 41

5 Example 2: Find a parametrization for the cylinder x 2 + y 2 = r 2. Solution: We use the cylindrical coordinates transformation. r (θ, z) = r cos(θ), r sin(θ), z [0, 2π) [, ] MATH 127 (Section 16.4) Parametrized Surfaces and Surface Integrals The University of Kansas 5 / 41

6 Example 3: Find a parametrization for the sphere x 2 + y 2 + z 2 = r 2. Solution: We use the spherical coordinates transformation. r (θ, φ) = r cos(θ) sin(φ), r sin(θ) sin(φ), r cos(φ) [0, 2π) [0, π] MATH 127 (Section 16.4) Parametrized Surfaces and Surface Integrals The University of Kansas 6 / 41

7 Example 4: Identify the surface parametrized by r (u, v) = u cos(v), u sin(v), u Solution: Eliminate u and v to obtain an equation in (x, y, z): x 2 + y 2 = ( u cos(v) ) 2 ( ) 2 + u sin(v) = u 2 = z 2 The parametric surface is a cone! Two-variable functions z = f (x, y) define a surface and can be easily parametrized as r (x, y) = x, y, f (x, y) MATH 127 (Section 16.4) Parametrized Surfaces and Surface Integrals The University of Kansas 7 / 41

8 Example 5: Let r (u, v) = u cos(v), u sin(v), v. What are the horizontal grid lines, when v is constant? What are the vertical grid lines, when u is constant? Solution: MATH 127 (Section 16.4) Parametrized Surfaces and Surface Integrals The University of Kansas 8 / 41

9 Example 6: Find a parametrization for the hyperboloid of one sheet x 2 + y 2 z 2 = 1 Solution: Don t forget that there are an infinite number of parametrizations for any surface! We will find two for this surface: (I) Set z = u, then x 2 + y 2 = 1 + c 2. We can then let x = 1 + u 2 cos(v) y = 1 + u 2 sin(v) (II) Let z = tan(u), then x 2 + y 2 = sec 2 (u). We can then let x = tan(u) cos(v) y = tan(u) sin(v) MATH 127 (Section 16.4) Parametrized Surfaces and Surface Integrals The University of Kansas 9 / 41

10 Surface Area of a Parametrized Surface In section 13.3 we calculated the arclength of a parametrized curve and then we used the resulting formula to define line integrals. In a smiliar fashion, we will calculate the surface area of a surface parametrized by r (u, v) over the domain R. We partition R into m n subregions R ij. Each subregion corresponds to a portion of the surface S ij. S = S ij S ij = r (D ij ), the Image of D ij under r MATH 127 (Section 16.4) Parametrized Surfaces and Surface Integrals The University of Kansas 10 / 41

11 Similar to the Jacobian calculations of section 15.6, we approximate S ij as a rectangle of width r u (u i, v j ) u and length r v (u i, v j ) v. Using the parallelogram property of the cross product: Area(S ij ) r u (u i, v j ) u r v (u i, v j ) v = r u (u i, v j ) r v (u i, v j ) u v m n The area of S is approximately r u (u i, v j ) r v (u i, v j ) u v. i=1 j=1 Taking the limit as (m, n) (, ) yields Area(S) = r u (u, v) r v (u, v) da D MATH 127 (Section 16.4) Parametrized Surfaces and Surface Integrals The University of Kansas 11 / 41

12 Area(S) = r u (u, v) r v (u, v) da D Example 7: Find the surface area of the cylinder x 2 + y 2 = 4 between z = 0 and z = 5. Solution: First, parameterize the cylinder as Second, calculate r u r v. r (u, v) = 2 cos(u), 2 sin(u), v [0, 2π] [0, 5] r u (u, v) = 2 sin(u), 2 cos(u), 0 r v (u, v) = 0, 0, 1 Area(S) = 2π r u r v = 2 cos(u), 2 sin(u), 0 = 2 2 dv du = 20π MATH 127 (Section 16.4) Parametrized Surfaces and Surface Integrals The University of Kansas 12 / 41

13 Surface Area of a 2-Variable Function The surface area of the graph of z = f (x, y) defined on D has a natural parametrization r (x, y) = x, y, f (x, y). r x (x, y) = 1, 0, f x (x, y) r y (x, y) = 0, 1, f y (x, y) Since r x r y = f x, f y, 1, we have r x r y = The area of the surface is given by 1 + f x (x, y) 2 + f y (x, y) 2 da D 1 + f 2 x + f 2 y. MATH 127 (Section 16.4) Parametrized Surfaces and Surface Integrals The University of Kansas 13 / 41

14 Example 8: Calculate the surface area of the sphere of radius a. Solution: Half of the sphere is the graph z = f (x, y) = a 2 x 2 y 2 where x 2 + y 2 a 2. f x (x, y) = x a 2 x 2 y 2 f y (x, y) = y a 2 x 2 y 2 Thus the area of the sphere is ( ) 2 ( ) 2 2 x y da D a 2 x 2 y 2 a 2 x 2 y 2 2π a a = 2 r dr dθ = 0 0 a 2 4πa2 r 2 MATH 127 (Section 16.4) Parametrized Surfaces and Surface Integrals The University of Kansas 14 / 41

15 Example 9: A torus (surface of a donut) can be obtained by rotating the circle on the yz-plane centered at (0, b, 0) with radius a (b > a > 0) about the z-axis. What is the surface area of the torus? The surface area is calculated 2π 2π 0 0 Solution: The torus has a parametrization x(θ, α) = b cos(θ) + a cos(α) cos(θ) y(θ, α) = b sin(θ) + a cos(α) sin(θ) z(θ, α) = a sin(α) where 0 θ 2π and 0 α 2π. r θ r α = a(b + a cos(α)) a(b + a cos(α)) dθ dα = 2πa 2π 0 (b + a cos(α)) dα = 4π 2 ab MATH 127 (Section 16.4) Parametrized Surfaces and Surface Integrals The University of Kansas 15 / 41

16 Surface Integrals - Scalar Surface Integral Suppose that f (x, y, z) gives the density of some quantity per unit area at each point on a surface S parametrized by r (u, v) over the domain R. Then the total amount of the quantity is calculated as f (x, y, z) ds = f ( r (u, v)) r u r v da S R The symbol ds is the surface element or the surface area differential. ds = r u r v da uv MATH 127 (Section 16.4) Parametrized Surfaces and Surface Integrals The University of Kansas 16 / 41

17 Example 10: Suppose that the mass density of a surface is given by f (x, y, z) = x 2. Compute the total mass of the sphere x 2 + y 2 + z 2 = 1. Solution: First, parametrize the sphere using spherical coordinates. r (θ, φ) = sin(φ) cos(θ), sin(φ) sin(θ), cos(φ) r u r v = sin(φ) Total Mass: S x 2 ds = = 2π π 0 0 2π π 0 0 = cos3 (φ) 3 (sin(φ) cos(θ)) 2 sin(φ) dφ dθ sin(φ) 3 cos(θ) dφ dθ cos(φ) π 0 sin(θ) 2π 0 = 4π 3 MATH 127 (Section 16.4) Parametrized Surfaces and Surface Integrals The University of Kansas 17 / 41

18 Example 1 MATH 127 (Section 16.4) Parametrized Surfaces and Surface Integrals The University of Kansas 18 / 41

19 IClicker 1 (A) 26 2π (B) 27 2π (C) 26π (D) 26 2 MATH 127 (Section 16.4) Parametrized Surfaces and Surface Integrals The University of Kansas 18 / 41

20 IClicker 2 (A) (B) (C) MATH 127 (Section 16.4) Parametrized Surfaces and Surface Integrals The University of Kansas 19 / 41

21 Section 16.5 Surface Integrals of Vector Fields (I) Tangent Lines and Planes of Parametrized Surfaces (II) Oriented Surfaces (III) Vector Surface Integrals (IV) Flux, Fluid Flow, Electric and Magnetic Fields MATH 127 (Section 16.4) Parametrized Surfaces and Surface Integrals The University of Kansas 20 / 41

22 Let S be parametrized by r (u, v) on the domain R. If r u is nonzero at (a, b), then the tangent line to the horizontal grid line at r (a, b) has a directional vector r u (a, b). If r v is nonzero at (a, b), then the tangent line to the vertical grid line at r (a, b) has a directional vector r v (a, b). If r u r v is nonzero at (a, b), then the tangent plane to S at r (a, b) has a normal vector ( r u r v ) (a, b). MATH 127 (Section 16.4) Parametrized Surfaces and Surface Integrals The University of Kansas 21 / 41

23 If it is possible to choose a unit normal vector n at every points so that n varies continuously over the surface, then the surface is called an oriented surface and the choice of n provides an orientation for the surface. S is smooth if r u r v is nonzero for all points in R. If S is a smooth orientable surface which is parametrized by r (u, v), then the surface is automatically supplied with the orientation of the unit normal vector n = r u r v r u r v MATH 127 (Section 16.4) Parametrized Surfaces and Surface Integrals The University of Kansas 22 / 41

24 Not all surfaces are can be oriented! Discovered in 1858 by August Möbius, the Möbius strip is a surface which cannot be oriented. If an ant were to crawl along the strip starting at a point P, it would eventually end up on the other side of the strip. If the any continued to crawl in the same direction, then it would end up back at the same point P without ever crossing an edge. The Möbius strip has only one side! For r [ 0.5, 0.5] and θ [0, 2π], the Möbius strip can be parametrized as x = 2 cos(θ) + r cos(θ/2) y = 2 sin(θ) + r cos(θ/2) z = r sin(θ/2) MATH 127 (Section 16.4) Parametrized Surfaces and Surface Integrals The University of Kansas 23 / 41

25 A closed surface is the boundary of a solid region. By convention, the positive orientation of a closed surface is the one for which the normal vectors point outward from the solid region. Example 1: The unit sphere parametrized using the spherical coordinate transformation has positive orientation. r (φ, θ) = sin(φ) cos(θ), sin(φ) sin(θ), cos(φ) r φ r θ = sin 2 (φ) cos(θ), sin 2 (φ) sin(θ), sin(φ) cos(φ) MATH 127 (Section 16.4) Parametrized Surfaces and Surface Integrals The University of Kansas 24 / 41

26 The normal component of F~ to S is F~ n~. If F~ is a continuous vector field defined on an oriented surface S with unit normal vector n~, then the vector surface integral of F~ over S is ZZ ZZ ~ ~ F ds = F~ n~ ds S MATH 127 (Section 16.4) S Parametrized Surfaces and Surface Integrals The University of Kansas 25 / 41

27 If F is a continuous vector field defined on an oriented surface S with unit normal vector n, then the vector surface integral of F over S is F ds = F n ds The integral is also called the flux of F across S. Let S be parametrized by r (u, v) over R. F ds = F n ds S S = = S R R S F ( r (u, v)) r u r v r u r v r u r v da F ( r (u, v)) ( r u r v ) da MATH 127 (Section 16.4) Parametrized Surfaces and Surface Integrals The University of Kansas 26 / 41

28 Example 2: Find the flux of F (x, y, z) = 0, yz, z 2 outward through the surface S cut from the cylinder y 2 + z 2 = 1, z 0, by the planes x = 0 and x = 1. Solution: The surface can be parametrized with an outward orientation as r (x, θ) = x, sin(θ), cos(θ) x [0, 1] θ [ π/2, π/2] r x r θ = 1, 0, 0 0, cos(θ), sin(θ) = 0, sin(θ), cos(θ) S F ( r (x, θ)) = 0, sin(θ) cos(θ), cos 2 (θ) F ds = = = R π/2 1 π/2 π/2 π/2 F ( r φ r θ ) da 0 sin 2 (θ) cos(θ) + cos 3 (θ) dx dθ sin 2 (θ) cos(θ) + cos 3 (θ) dθ = 2 MATH 127 (Section 16.4) Parametrized Surfaces and Surface Integrals The University of Kansas 27 / 41

29 Example 3: Find the flux of the vector field F (x, y, z) = z, y, x across the unit sphere x 2 + y 2 + z 2 = 1 Solution: The unit sphere can be parametrized as r (φ, θ) = sin(φ) cos(θ), sin(φ) sin(θ), cos(φ) φ [0, π] θ [0, 2π] r φ r θ = sin 2 (φ) cos(θ), sin 2 (φ) sin(θ), sin(φ) cos(φ) F ( r (φ, θ)) = cos(φ), sin(φ) sin(θ), sin(φ) cos(θ) S F ds = F ( r φ r θ ) da = R 2π π sin 2 (φ) cos(φ) cos(θ) + sin 3 (φ) sin 2 (θ) dφ dθ = 4π 3 MATH 127 (Section 16.4) Parametrized Surfaces and Surface Integrals The University of Kansas 28 / 41

30 Fluid Flux For a fluid with velocity vector field V, the flow rate across the surface S (volume per unit time) is V ds S MATH 127 (Section 16.4) Parametrized Surfaces and Surface Integrals The University of Kansas 29 / 41

31 Example 4: A fluid has density ρ kg/m 3 and flows with velocity V (x, y, z) = z, y 2, x 2 where x, y, z are measured in meters and the components of V in meters per second. Find the rate of flow outward through the cylinder x 2 + y 2 = 4, 0 z 1. Solution: First, parametrize the surface: r (θ, z) = 2 cos(θ), 2 sin(θ), z θ [0, 2π] z [0, 1] r θ r z = 2 sin(θ), 2 cos(θ), 0 0, 0, 1 = 2 cos(θ), 2 sin(θ), 0 Note that r has positive orientation as it points out from the cylinder. ρ V ds = ρ z, 4 sin 2 (θ), 4 cos 2 (θ) 2 cos(θ), 2 sin(θ), 0 da S R 2π = ρ z cos(θ) + 8 sin 3 (θ) dz dθ ( = ρ sin(θ) + 1 ) 2π 3 cos3 (θ) cos(θ) 0 = 0 kg s MATH 127 (Section 16.4) Parametrized Surfaces and Surface Integrals The University of Kansas 30 / 41

32 Example 5: Calculate the flux of F (x, y, z) = 0, y, z through the surface S consisting of the paraboloid y = x 2 + z 2, y 1 and the disk x 2 + z 2 1, y = 1. Solution: S is the closed surface consisting of S 1 + S 2. S 1 is parametrized by S 2 is parametrized by r (x, z) = x, x 2 + z 2, z s (r, θ) = r sin(θ), 1, r cos(θ) on x 2 + z 2 1. on r [0, 1], θ [0, 2π]. r x r z = 2x, 1, 2z s r s θ = 0, r, 0 The parametrization of S points outward from the surface. F ds S = F ds + S 1 F ds S 2 = x 2 z 2 2z 2 da + R 1 1 r 2 cos 2 (θ) da R 2 = π + 5π/3 = 2π/3 MATH 127 (Section 16.4) Parametrized Surfaces and Surface Integrals The University of Kansas 31 / 41

33 Types of Integral Let f be a scalar function and F a vector field. Scalar Line Integral along a curve C parametrized by r (t) on [a, b]. b f ds = f ( r (t)) r (t) dt C Vector Line Integral along an oriented curve C: b F d r = F ( r (t)) r (t) dt a C a Scalar Surface Integral over a surface S parametrized by r (u, v) on R: f ds = f ( r (u, v)) r u r v da S Vector Surface Integral over an oriented surface S: F ds = F ( r (u, v)) ( r u r v ) da R S R MATH 127 (Section 16.4) Parametrized Surfaces and Surface Integrals The University of Kansas 32 / 41