Parametrized Surfaces


 Γερασιμος Σπανός
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1 Parametrized Surfaces Recall from our unit on vectorvalued functions at the beginning of the semester that an R 3 valued function c(t) in one parameter is a mapping of the form c : I R 3 where I is some subinterval of the real line. If c is differentiable, we can think of c as smoothly embedding the interval I as a closed curve C in R 3, a parametrized curve.
2 Parametrized Surfaces Recall from our unit on vectorvalued functions at the beginning of the semester that an R 3 valued function c(t) in one parameter is a mapping of the form c : I R 3 where I is some subinterval of the real line. If c is differentiable, we can think of c as smoothly embedding the interval I as a closed curve C in R 3, a parametrized curve. Our goal now is to study parametrized surfaces, which are a natural next step from parametrized curves. Now instead of embedding a 1dimensional domain I into R 3, we will embed 2dimensional domain D into R 3 and do calculus on the embedded surface.
3 Definition We consider a continuously differentiable R 3 valued function of two variables G : R 2 R 3 : G(u, v) = (x(u, v), y(u, v), z(u, v))
4 Definition We consider a continuously differentiable R 3 valued function of two variables G : R 2 R 3 : The graph of G is the set G(u, v) = (x(u, v), y(u, v), z(u, v)) {(x, y, z) R 3 : there exists (u, v) with G(u, v) = (x, y, z)}. (This definition is analogous to that of the graph of a vectorvalued function.) The graph of G is also called a parametrized surface. The domain D of G is called the parameter domain.
5 Example (Parametrization of a Sphere) Fix any R > 0. Let D = [0, 2π] [0, π], a rectangle in R 2. Define a mapping G on D by G(θ, φ) = (R cos θ sin φ, R sin θ sin φ, R cos φ). Then G maps D onto the sphere of radius R in R 3.
6 Example (Parametrization of the Graph of a RealValued Function of Two Variables) Let f (x, y) be any realvalued function of two variables with domain D. Define mapping G on D by G(x, y) = (x, y, f (x, y)). Then G maps D onto the graph of f. (In particular the graph of G equals the graph of f.)
7 Definition Let G(u, v) = (x, y, z) be a continuously differentiable R 3 valued function of two variables. Define the utangent vector and the vtangent vector functions of G, respectively, to be T u = (x u, y u, z u ) and T v = (x v, y v, z v ). T u and T v are both functions from R 2 into R 3, and they output nonparallel tangent vectors to the graph of G at input (u, v).
8 Definition Let G(u, v) = (x, y, z) be a continuously differentiable R 3 valued function of two variables. Define the utangent vector and the vtangent vector functions of G, respectively, to be T u = (x u, y u, z u ) and T v = (x v, y v, z v ). T u and T v are both functions from R 2 into R 3, and they output nonparallel tangent vectors to the graph of G at input (u, v). Assume both T u and T v are nonzero. Define the normal vector associated to G to be the function n = T u T v. Then n is always orthogonal to the tangent plane of the graph of G at a point G(u, v).
9 Example Let G(θ, z) = (2 cos θ, 2 sin θ, z) be a parametrization of the cylinder described by x 2 + y 2 = Compute T θ, T z, and n. 2. Find an equation for the tangent plane to the graph of G at G( π 4, 5).
10 Solution to (1) Known: G(θ, z) = (2 cos θ, 2 sin θ, z)
11 Solution to (1) Known: G(θ, z) = (2 cos θ, 2 sin θ, z) Taking derivatives, we have and hence T θ (θ, z) = ( 2 sin θ, 2 cos θ, 0) and T z = (0, 0, 1), n(θ, z) = T θ T z = det [ i j k 2 sin θ 2 cos θ ] = (2 cos θ, 2 sin θ, 0).
12 Solution to (2) Known: G(θ, z) = (2 cos θ, 2 sin θ, z) n = (2 cos θ, 2 sin θ, 0)
13 Solution to (2) Known: G(θ, z) = (2 cos θ, 2 sin θ, z) n = (2 cos θ, 2 sin θ, 0) To find the tangent plane, compute the normal at input ( π 4, 5): n( π 4, 5) = (2 cos π 4, 2 sin π 4, 0) = ( 2, 2, 0). The tangent plane passes through the point G( π 4, 5) = ( 2, 2, 5), and so its equation [using 0 = n ((x, y, z) (x 0, y 0, z 0 ))] is given by 0 = 2(x 2) + 2(y 2).
14 Surface Integrals Definition Let G(u, v) = (x, y, z) be a onetoone continuously differentiable parametrization of a surface S in R 3 with domain D, with nonzero tangent vectors T u and T v. Let f (x, y, z) be a realvalued function of three variables. Define the surface integral of f over S to be S f (x, y, z)ds = f (G(u, v)) n(u, v) d(u, v). D
15 Surface Integrals Definition Let G(u, v) = (x, y, z) be a onetoone continuously differentiable parametrization of a surface S in R 3 with domain D, with nonzero tangent vectors T u and T v. Let f (x, y, z) be a realvalued function of three variables. Define the surface integral of f over S to be S f (x, y, z)ds = D f (G(u, v)) n(u, v) d(u, v). Fact If G is as above, then the surface integral 1dS is exactly the surface S area of the graph of G.
16 Example 1. Calculate the surface area of the portion S of the cone x 2 + y 2 = z 2 within the cylinder x 2 + y 2 = For the same S, calculate S x 2 zds.
17 Surface Area of S First we parametrize the cone by G using a variant of spherical coordinates: G(ρ, θ) = (x, y, z) ( = ρ cos θ sin π 4, ρ sin θ sin π 4, ρ cos π ) 4 ( ) = ρ cos θ, sin θ, ρ on the domain D = [0, 2 2] [0, 2π] in the (ρ, θ)plane.
18 Surface Area of S First we parametrize the cone by G using a variant of spherical coordinates: G(ρ, θ) = (x, y, z) ( = ρ cos θ sin π 4, ρ sin θ sin π 4, ρ cos π ) 4 ( ) = ρ cos θ, sin θ, ρ on the domain D = [0, 2 2] [0, 2π] in the (ρ, θ)plane. Now compute tangent and normal vectors: ( 2 T ρ = cos θ, 2 sin θ, )
19 Surface Area of S First we parametrize the cone by G using a variant of spherical coordinates: G(ρ, θ) = (x, y, z) ( = ρ cos θ sin π 4, ρ sin θ sin π 4, ρ cos π ) 4 ( ) = ρ cos θ, sin θ, ρ on the domain D = [0, 2 2] [0, 2π] in the (ρ, θ)plane. Now compute tangent and normal vectors: ( 2 T ρ = cos θ, 2 sin θ, ) ( T θ = 2 ρ sin θ, ) 2 ρ cos θ, 0 2 2
20 Surface Area of S First we parametrize the cone by G using a variant of spherical coordinates: G(ρ, θ) = (x, y, z) ( = ρ cos θ sin π 4, ρ sin θ sin π 4, ρ cos π ) 4 ( ) = ρ cos θ, sin θ, ρ on the domain D = [0, 2 2] [0, 2π] in the (ρ, θ)plane. Now compute tangent and normal vectors: ( 2 T ρ = cos θ, 2 sin θ, ) ( T θ = 2 ρ sin θ, ) 2 ρ cos θ, n = T ρ T θ = det = i j k 2 cos θ 2 sin θ ρ sin θ 2 ρ cos θ ( 12 ρ cos θ, 12 ρ sin θ, 12 ρ ).
21 Surface Area of S, contd. Known: n = ( 1 2 ρ cos θ, 1 2 ρ sin θ, 1 2 ρ) Now the length of n is n = 1 4 ρ2 cos 2 θ ρ2 sin 2 θ ρ2 = 2 2 ρ,
22 Surface Area of S, contd. Known: n = ( 1 2 ρ cos θ, 1 2 ρ sin θ, 1 2 ρ) Now the length of n is n = 1 4 ρ2 cos 2 θ ρ2 sin 2 θ ρ2 = 2 2 ρ, and so we are able to compute the area surface integral: S 1dS = = = D 2π n(u, v) d(u, v) [θ]2π 0 = 4 2π. 2 2 ρdρdθ [ ] ρ2 0
23 Surface Integral of x 2 z Known: n = 2 f (x, y, z) = x 2 z G(ρ, θ) = ( 2 2 ρ 2 ρ cos θ, 2 2 sin θ, 2 2 ρ )
24 Surface Integral of x 2 z Known: n = 2 f (x, y, z) = x 2 z G(ρ, θ) = ( 2 2 ρ 2 ρ cos θ, 2 2 sin θ, 2 2 ρ ) Compute the composition: f (G(ρ, θ)) = 2 4 ρ3 cos 2 θ,
25 Surface Integral of x 2 z Known: n = 2 f (x, y, z) = x 2 z G(ρ, θ) = ( 2 2 ρ 2 ρ cos θ, 2 2 sin θ, 2 2 ρ ) Compute the composition: and hence S f (G(ρ, θ)) = 2 4 ρ3 cos 2 θ, fds = = D 2π 2 2 = 1 4 = 1 4 f (G(u, v)) n(u, v) d(u, v) 0 0 2π [ 1 5 ρ5 = 32 2π. 5 0 ] ρ3 cos 2 2 θ 2 ρdρdθ ρ 4 cos 2 θdρdθ [ 1 2 θ + 1 ] 2π sin 2θ 4 0
26 Fact Suppose g(x, y) is a function of two variables let S be the graph of g. Suppose G(x, y) is a parametrization of S. Then T x = (1, 0, g x) and T y = (0, 1, g y ); n = 1 + g 2 x + g 2 y ;
27 Fact Suppose g(x, y) is a function of two variables let S be the graph of g. Suppose G(x, y) is a parametrization of S. Then T x = (1, 0, g x) and T y = (0, 1, g y ); n = 1 + g 2 x + g 2 y ; and therefore for any function f with domain S we have S f (x, y, z)ds = D f (x, y, g(x, y)) 1 + g 2 x + g 2 y d(x, y).
28 Proof. Let g(x, y) be a function. Then the graph of g is parametrized by G(u, v) = (x, y, z) = (u, v, g(u, v)).
29 Proof. Let g(x, y) be a function. Then the graph of g is parametrized by Computing tangent vectors, we have G(u, v) = (x, y, z) = (u, v, g(u, v)). T u = (1, 0, g x(u, v)) and T v (0, 1, g y (u, v))
30 Proof. Let g(x, y) be a function. Then the graph of g is parametrized by Computing tangent vectors, we have G(u, v) = (x, y, z) = (u, v, g(u, v)). T u = (1, 0, g x(u, v)) and T v (0, 1, g y (u, v)) and hence n = det i j k 1 0 g x(u, v) 0 1 g y (u, v) n = ( 1 + g 2 x + g 2 y ). = ( g x, g y, 1) and
31 Proof. Let g(x, y) be a function. Then the graph of g is parametrized by Computing tangent vectors, we have G(u, v) = (x, y, z) = (u, v, g(u, v)). T u = (1, 0, g x(u, v)) and T v (0, 1, g y (u, v)) and hence n = det i j k 1 0 g x(u, v) 0 1 g y (u, v) n = ( 1 + g 2 x + g 2 y ). = ( g x, g y, 1) and Therefore if we want to compute the surface integral of f over the graph of g, we get S f (x, y, z)ds = D f (x, y, g(x, y)) 1 + g 2 x + g 2 y d(x, y).
32 Example Compute S (z x)ds, where S is the graph of z = x + y 2, 0 x y, 0 y 1.
33 Example Compute S (z x)ds, where S is the graph of z = x + y 2, 0 x y, 0 y 1. Solution. Take g(x, y) = x + y 2, so n = (2y) 2 = 2 + 4y 2 )
34 Example Compute S (z x)ds, where S is the graph of z = x + y 2, 0 x y, 0 y 1. Solution. Take g(x, y) = x + y 2, so n = (2y) 2 = 2 + 4y 2 ) and f (x, y, g(x, y) = g(x, y) x = x + y 2 x = y 2.
35 Example Compute S (z x)ds, where S is the graph of z = x + y 2, 0 x y, 0 y 1. Solution. Take g(x, y) = x + y 2, so n = (2y) 2 = 2 + 4y 2 ) and f (x, y, g(x, y) = g(x, y) x = x + y 2 x = y 2. Now compute (using substitution rule with u = 2 + 4y 2 ): S (z x)ds = = 1 y y y 2 dxdy y y 2 dy = 1 30 ( ).
36 Flux Integrals Our goal now is develop an integral which computes the amount of flow of a given vector field in R 3 through a given oriented surface S, which we call a flux integral.
37 Flux Integrals Our goal now is develop an integral which computes the amount of flow of a given vector field in R 3 through a given oriented surface S, which we call a flux integral. Definition Let G(u, v) be a parametrization of the surface S in R 3, and let n be its associated normal vector. Let F be a vector field over R 3.
38 Flux Integrals Our goal now is develop an integral which computes the amount of flow of a given vector field in R 3 through a given oriented surface S, which we call a flux integral. Definition Let G(u, v) be a parametrization of the surface S in R 3, and let n be its associated normal vector. Let F be a vector field over R 3. The normal component of F with respect to G is F n. Informally speaking, the normal component of F is how much of the vector field F is orthogonal to the surface S.
39 Flux Integrals Our goal now is develop an integral which computes the amount of flow of a given vector field in R 3 through a given oriented surface S, which we call a flux integral. Definition Let G(u, v) be a parametrization of the surface S in R 3, and let n be its associated normal vector. Let F be a vector field over R 3. The normal component of F with respect to G is F n. Informally speaking, the normal component of F is how much of the vector field F is orthogonal to the surface S. The flux integral or vector surface integral S F ds of F across S is the surface integral of the normal component of F, i.e. S F ds = S ( F n)ds.
40 Fact Let G, S, n, F be as in the previous definition, and let D be the domain of G. Then S F ds = D F (G(u, v)) n(u, v)d(u, v). To help remember the formula, note the analogy here: computationally, flux integrals are to surface integrals as vector field line integrals are to scalar line integrals.
41 Example Compute S F ds, where F = (0, 0, x) and S is the graph of G(u, v) = (u 2, v, u 3 v 2 ) with domain D = [0, 1] [0, 1].
42 Example Compute S F ds, where F = (0, 0, x) and S is the graph of G(u, v) = (u 2, v, u 3 v 2 ) with domain D = [0, 1] [0, 1]. Solution. Compute the tangents and normal of G: T u = (2u, 0, 3u 2 ) and T v = (0, 1, 2v), and
43 Example Compute S F ds, where F = (0, 0, x) and S is the graph of G(u, v) = (u 2, v, u 3 v 2 ) with domain D = [0, 1] [0, 1]. Solution. Compute the tangents and normal of G: T u = (2u, 0, 3u 2 ) and T v = (0, 1, 2v), and i j k n = det 2u 0 3u 2 = ( 3u 2, 4uv, 2u) v
44 Example Compute S F ds, where F = (0, 0, x) and S is the graph of G(u, v) = (u 2, v, u 3 v 2 ) with domain D = [0, 1] [0, 1]. Solution. Compute the tangents and normal of G: T u = (2u, 0, 3u 2 ) and T v = (0, 1, 2v), and i j k n = det 2u 0 3u 2 = ( 3u 2, 4uv, 2u) v Now compute the normal component of F : F (G(u, v)) n = (0, 0, u 2 ) ( 3u 2, 4uv, 2u) = 2u 3.
45 Example Compute S F ds, where F = (0, 0, x) and S is the graph of G(u, v) = (u 2, v, u 3 v 2 ) with domain D = [0, 1] [0, 1]. Solution. Compute the tangents and normal of G: T u = (2u, 0, 3u 2 ) and T v = (0, 1, 2v), and i j k n = det 2u 0 3u 2 = ( 3u 2, 4uv, 2u) v Now compute the normal component of F : F (G(u, v)) n = (0, 0, u 2 ) ( 3u 2, 4uv, 2u) = 2u 3. Finally compute the integral: 1 1 F ds = 2u 3 dudv S 0 0 [ ] 1 1 = 2 u4 [v] = 1 2.
46 Example Let v = (x 2 + y 2, 0, z 2 ) model the velocity (in cm/s) of a threedimensional body of water. Compute the the volume of water passing each second through the upper hemisphere S of the unit sphere centered at the origin. (Hint: Compute the corresponding flux integral.)
47 Example Let v = (x 2 + y 2, 0, z 2 ) model the velocity (in cm/s) of a threedimensional body of water. Compute the the volume of water passing each second through the upper hemisphere S of the unit sphere centered at the origin. (Hint: Compute the corresponding flux integral.) Solution. First parametrize the upper hemisphere using spherical coordinates: G(θ, φ) = (cos θ sin φ, sin θ sin φ, cos φ) on [0, 2π] [0, π 2 ] in the (θ, φ)plane.
48 Example Let v = (x 2 + y 2, 0, z 2 ) model the velocity (in cm/s) of a threedimensional body of water. Compute the the volume of water passing each second through the upper hemisphere S of the unit sphere centered at the origin. (Hint: Compute the corresponding flux integral.) Solution. First parametrize the upper hemisphere using spherical coordinates: G(θ, φ) = (cos θ sin φ, sin θ sin φ, cos φ) on [0, 2π] [0, π 2 ] in the (θ, φ)plane. Now compute tangents and normals. T θ = ( sin θ sin φ, cos θ sin φ, 0) and T φ = (cos θ cos φ, sin θ cos φ, sin φ); n = T θ T φ = (cos θ sin 2 φ, sin θ sin 2 φ, cos φ).
49 Example Let v = (x 2 + y 2, 0, z 2 ) model the velocity (in cm/s) of a threedimensional body of water. Compute the the volume of water passing each second through the upper hemisphere S of the unit sphere centered at the origin. (Hint: Compute the corresponding flux integral.) Solution. First parametrize the upper hemisphere using spherical coordinates: G(θ, φ) = (cos θ sin φ, sin θ sin φ, cos φ) on [0, 2π] [0, π 2 ] in the (θ, φ)plane. Now compute tangents and normals. T θ = ( sin θ sin φ, cos θ sin φ, 0) and T φ = (cos θ cos φ, sin θ cos φ, sin φ); n = T θ T φ = (cos θ sin 2 φ, sin θ sin 2 φ, cos φ). Now v(g(θ, φ)) = (sin 2 φ, 0, cos 2 φ), and so
50 Example Let v = (x 2 + y 2, 0, z 2 ) model the velocity (in cm/s) of a threedimensional body of water. Compute the the volume of water passing each second through the upper hemisphere S of the unit sphere centered at the origin. (Hint: Compute the corresponding flux integral.) Solution. First parametrize the upper hemisphere using spherical coordinates: G(θ, φ) = (cos θ sin φ, sin θ sin φ, cos φ) on [0, 2π] [0, π 2 ] in the (θ, φ)plane. Now compute tangents and normals. T θ = ( sin θ sin φ, cos θ sin φ, 0) and T φ = (cos θ cos φ, sin θ cos φ, sin φ); n = T θ T φ = (cos θ sin 2 φ, sin θ sin 2 φ, cos φ). Now v(g(θ, φ)) = (sin 2 φ, 0, cos 2 φ), and so v(g(θ, φ)) n(θ, φ) = sin 4 φ cos θ + sin φ cos 3 φ.
51 Example Let v = (x 2 + y 2, 0, z 2 ) model the velocity (in cm/s) of a threedimensional body of water. Compute the the volume of water passing each second through the upper hemisphere S of the unit sphere centered at the origin. (Hint: Compute the corresponding flux integral.) Solution. First parametrize the upper hemisphere using spherical coordinates: G(θ, φ) = (cos θ sin φ, sin θ sin φ, cos φ) on [0, 2π] [0, π 2 ] in the (θ, φ)plane. Now compute tangents and normals. T θ = ( sin θ sin φ, cos θ sin φ, 0) and T φ = (cos θ cos φ, sin θ cos φ, sin φ); n = T θ T φ = (cos θ sin 2 φ, sin θ sin 2 φ, cos φ). Now v(g(θ, φ)) = (sin 2 φ, 0, cos 2 φ), and so v(g(θ, φ)) n(θ, φ) = sin 4 φ cos θ + sin φ cos 3 φ. Then we compute the surface integral (using symmetry on the first half): π/2 2π v ds = (sin 4 φ cos θ + sin φ cos 3 φ)dθdφ S 0 0 = 0 + π 2.
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