Area and Jacobians. Outline

Transcription

1 Area and Jacobians Outline 1. Jacobians Let f : 2 2 be a smooth map from the uv-plane to the xy-plane. The Jacobian of f is the absolute value of the determinant of the derivative matrix: Jf = det(df) = x y u v x y v u If is any region in the uv-plane and f is one-to-one on, then the area of the region f() in the xy-plane can be determined by integrating the Jacobian: area of f() = Jf(u, v) dudv. More generally, if g(x, y) is a real-valued function, then we can use the Jacobian to integrate g: g(x, y) dxdy = g(f(u, v)) Jf(u, v) dudv. f() This is known as the change of variables formula for double integrals. See section 12.9 of Stewart for more discussion and examples using the Jacobian. 2. Surface Area If σ: U S is a surface patch, the Jacobian of σ is the quantity Jσ = σ u σ v. In terms of the first fundamental form I, the Jacobian is given by the formula Jσ = det(i). A surface patch σ is called equiareal if its Jacobian is the constant function 1. If is any region in U, then the surface area of the region σ() S can be determined by integrating the Jacobian: surface area of σ() = Jσ(u, v) dudv. More generally, if g is a real-valued function on S, then the surface integral of g on σ() is defined by the formula g da = g(σ(u, v)) Jσ(u, v) dudv. σ()

2 3. Maps Between Surfaces If f : S 1 S 2 is a map between surfaces, then the Jacobian of f is a certain real-valued function Jf : S 1 [, ). It can be defined in several different ways: If t and u are linearly independent tangent vectors at a point p S 1, then Jf(p) = D pf(t) D p f(u) t u Let σ: U S 1 be a surface patch, and suppose that f is regular on σ(u). Then Jf(σ(u, v)) = J(f σ)(u, v). Jσ(u, v) Let σ: U S 1 be a surface patch, and suppose that f is regular on σ(u). Let I 1 be the first fundamental form for σ, and let I 2 be the first fundamental form for f σ. Then det(i 2 ) Jf = det(i 1 ) A map f is called equiareal if its Jacobian is the constant function 1. Note that the Jacobian is always at a critical point, so an equiareal map must be regular. If is any region on S 1, then the surface area of the image f() on S 2 can be computed using a surface integral of the Jacobian: area of f() = Jf da. More generally, if g is a real-valued function on S 2, then the surface integral of g on f() can be computed using the formula g da = (g f)jf da. f()

3 Practice Problems 1. Let f : (, ) (, 2π) 2 be the function f(r, θ) = (r cos θ, r sin θ). (a) Find the Jacobian of f. (b) Use your answer to part (a) to find the area of the region in the plane defined by r < 1 + cos θ. 2. Use the substitutions x = u + v and y = u v to find the area of the region in the plane defined by the inequality (x + y) 2 < x y < Let S be the portion of the paraboloid z = x 2 + y 2 lying below the plane z = 1. Find the surface area of S. 4. Compute y da, where S is the portion of the helicoid z = θ satisfying < r < 1 and S < θ < π. 5. Compute S z2 da, where S is the portion of the surface z = e x sin y satisfying < x < 1 and < y < π. 6. Compute S z da, where S is the portion of the cone z2 = x 2 + y 2 for which < z < Let S 1 be the cylinder x 2 + y 2 = 1, let S 2 be the catenoid r = cosh z, and let f : S 1 S 2 be the map f(x, y, z) = (x cosh z, y cosh z, z). Compute the Jacobian of f.

4 Solutions [ ] cos θ r sin θ 1. (a) We have Jf = det sin θ r cos θ = r. (b) We can parameterize the given region using the transformation f(r, θ), where < θ < 2π and < r < 1 + cos θ. Since Jf = r, the area is 2π +cos θ 2π r drdθ = 1 2 (1 + cos θ) 2 dθ = 3π 2 Incidentally, the curve r = 1+cos θ in this problem is a cardioid, which is a type of epicycloid. Here is a plot of the region in Mathematica: r Sqrt x^2 y^2 ; cos x r; egionplot r 1 cos, x,.5, 2.2, y, 1.35, Substituting x = u + v and y = u v into the given inequalities, yields 4u 2 < 2v < 2, which simplifies to 2u 2 < v < 1. Therefore, we can parameterize the given region using the transformation f(u, v) = (u + v, u v), where 1/ 2 < u < 1/ 2 and 2u 2 < v < 1. Since Jf = [ det 1 1 ] 1 1 = 2, the area of the region is / 2 1/ 2 2u 2 2 dvdu = / 2 1/ 2 (1 2u 2 )du = Incidentally, the region (x + y) 2 < x y < 2 in this problem is a parabolic segment. Here is

5 a Mathematica plot of the region: egionplot x y ^2 x y 2, x,.2, 1.75, y,.2, We can parameterize almost all of the surface using the surface patch σ(r, θ) = ( r cos θ, r sin θ, r 2) where < r < 1 and < θ < 2π. Since σ r and σ θ are orthogonal, we have Jσ = σ r σ θ = σ r σ θ = r 1 + 4r 2, so the surface area is 2π r 1 + 4r 2 dθdr = 2π r 1 + 4r 2 dr = π( ) 6 4. We can parameterize this surface using the surface patch σ(r, θ) = ( r cos θ, r sin θ, θ ) where < r < 1 and < θ < π. Since σ r and σ θ are orthogonal, we have Jσ = σ r σ θ = σ r σ θ = 1 + r 2, so S y da = π r sin θ 1 + r 2 dθdr = π sin θ dθ r 1 + r 2 dr = We can parameterize this surface using the surface patch σ(x, y) = (x, y, e x sin y), where < x < 1 and < y < π. Then Jσ = σ x σ y = (1,, e x sin y) (, 1, e x cos y) = ( e x sin y, e x cos y, 1) = 1 + e 2x. Since z 2 = e 2x sin 2 y, the integral is π e 2x sin 2 y 1 + e 2x dydx = e 2x 1 + e 2x dx π sin 2 y dy = (1 + e2 ) 3/2 2 3/2 π 6

6 6. We can parameterize almost all of this surface using the surface patch σ(r, θ) = (r cos θ, r sin θ, r), where < r < 1 and < θ < 2π. Since σ r and σ θ are orthogonal, we have Since z = r, the integral is 2π Jσ = σ r σ θ = σ r σ θ = r 2. r ( r 2 ) dθdr = 2π 2 r 2 dr = 2π Let t 1 = ( y, x, ) and t 2 = (,, 1) be tangent vectors to the cylinder S 1 at the point (x, y, z). Then Df(t 1 ) = yf x + xf y = ( y cosh z, x cosh z, ) and Df(t 2 ) = f z = (x sinh z, y sinh z, 1). Since t 1 and t 2 are orthonormal, the Jacobian is just Df(t 1 ) Df(t 2 ). Since Df(t 1 ) and Df(t 2 ) are orthogonal, it follows that Jf = Df(t 1 ) Df(t 2 ) = ( cosh 2 z )( 1 + sinh 2 z ) = cosh 4 z.