Lesson 21: Surface Area

Slide 1: Section 12.6 Surface Area Math 21a March 31, 2008 Announcements ◮ Office hours Tuesday, Wednesday 2–4pm SC 323 ◮ Problem Sessions: Mon, 8:30; Thur, 7:30; SC 103b . . Image: Flickr user Tracy O . . . . . .

Slide 2: Announcements ◮ Office hours Tuesday, Wednesday 2–4pm SC 323 ◮ Problem Sessions: Mon, 8:30; Thur, 7:30; SC 103b . . . . . .

Slide 3: Outline Last time: Integration over polar regions Integration in Polar Coordinates Easy Area Rectangles and Parallelograms in the plane Parallelograms in space Area of curved surfaces Divide Approximate Take the Limit Special cases Graphs Surfaces of Revolution Worksheet Next Time . . . . . .

Slide 4: Polar slicing . . Here the boundaries of x and y Here the boundary r is a are complicated function of θ . . . . . .

Slide 5: Integration in Polar Coordinates Fact If f is continuous on a polar region of the form D = { (r, θ) | α ≤ θ ≤ β, h1 (θ) ≤ r ≤ h2 (θ) } then ∫∫ ∫ ∫ β h2 (θ) f(x, y) dA = f(r cos θ, r sin θ) r dr dθ. α h1 (θ) D ◮ Notice the “area element” is r dr dθ, not dr dθ! . . . . . .

Slide 6: Outline Last time: Integration over polar regions Integration in Polar Coordinates Easy Area Rectangles and Parallelograms in the plane Parallelograms in space Area of curved surfaces Divide Approximate Take the Limit Special cases Graphs Surfaces of Revolution Worksheet Next Time . . . . . .

Slide 7: Rectangles and Parallelograms in the plane w . . a . . = a. θ h sin θ . . . ℓ . b . A = ℓw A = bh = ab sin θ . . . . . .

Slide 8: Parallelograms in space The magnitude of the cross product a × b is the area of the parallelogram with sides a and b. . b | . b| sin θ θ . . . a A = |a| |b| sin θ = |a × b| . . . . . .

Slide 9: Outline Last time: Integration over polar regions Integration in Polar Coordinates Easy Area Rectangles and Parallelograms in the plane Parallelograms in space Area of curved surfaces Divide Approximate Take the Limit Special cases Graphs Surfaces of Revolution Worksheet Next Time . . . . . .

Slide 10: Parametrizing a surface A parametric surface S is defined by a vector-valued function of two parameters r(u, v) = x(u, v)i + y(u, v)j + z(u, v)k where (u, v) varies throughout a region D in the plane. . z v . D . . r . S . u . x . . y . . . . . .

Slide 11: Divide and Conquer Suppose D is a rectangle [a, b] × [c, d]. Divide [a, b] into m pieces and [c, d] into n pieces. The area of the surface is the sum of areas of pieces of the surface: ∑ A= ∆Ai,j i,j where ∆Aij is the area of the surface restricted to the subdomain [ui , ui+1 ] × [vi , vi+1 ] . . . . . .

Slide 12: Approximate The vectors representing the sides of a small rectangle “push forward” to vectors tangent to the surface. z . v . ∆ . vj D . ∆ . v rv ∆ . u ru . S ∆ . ui . u . x . y . They span a parallelogram approximating this piece of the surface. So the area of that piece is ∆A ≈ |∆u ru × ∆v rv | = |ru × rv | ∆v ∆u . . . . . .

Slide 13: Take the Limit So ∑∑ m n A≈ ru (uij , vij ) × rv (uij , vij ) ∆v ∆u i=1 j=1 Taking the limit we get ∫ b∫ d A= |ru × rv | dv du a c More generally, if D is any region (not necessarily a rectangle), the surface area is ∫∫ |ru × rv | dA D Remembering the dA is over the domain (u, v) space. . . . . . .

Slide 14: Outline Last time: Integration over polar regions Integration in Polar Coordinates Easy Area Rectangles and Parallelograms in the plane Parallelograms in space Area of curved surfaces Divide Approximate Take the Limit Special cases Graphs Surfaces of Revolution Worksheet Next Time . . . . . .

Slide 15: Surface areas of Graphs Notation as in Stewart, p. 869ff. Suppose f(x, y) is a function of z . two variables with domain D. . S The graph of f over D is the surface in space given by S = { (x, y, z) | (x, y) ∈ D, z = f(x, y) } . y . . x D . . . . . . .

Slide 16: To find the surface area of S, use the parametrization r(x, y) = ⟨x, y, f(x, y)⟩ Then ⟨ ⟩ ∂f r x = 1, 0, ∂x ⟨ ⟩ ∂f r y = 0, 1, ∂y ⟨ ⟩ ∂f ∂f rx × ry = − , − , 1 ∂x ∂y So √ ∫∫ ( )2 ( )2 ∂f ∂f A= 1+ + dA ∂x ∂y D . . . . . .

Slide 17: Surfaces of Revolution A surface of revolution can be described by rotating the graph of y = f(x) over the interval [a, b] around the x-axis. . y . x . z . . . . . . .

Slide 18: Choose the parametrization r(x, θ) = ⟨u, f(u) cos θ, f(u) sin θ⟩ where a ≤ x ≤ b, 0 ≤ θ ≤ 2π . Then ⟨ ⟩ rx = 1, f′ (x) cos θ, f′ (x) sin θ rθ = ⟨0, −f(x) sin θ, f(x) cos θ⟩ ⟨ ⟩ rx × rθ = f′ (x)f(x), −f(x) cos θ, −f(x) sin θ √ |ru × rv | = f(x) 1 + f′ (x)2 So ∫ 2π ∫ b √ ∫ b √ ′ A= f(x) 1 + f (x)2 dx dθ = 2π f(x) 1 + f′ (x)2 dx 0 a a . . . . . .

Slide 19: Outline Last time: Integration over polar regions Integration in Polar Coordinates Easy Area Rectangles and Parallelograms in the plane Parallelograms in space Area of curved surfaces Divide Approximate Take the Limit Special cases Graphs Surfaces of Revolution Worksheet Next Time . . . . . .

Slide 20: Worksheet #1 Problem Find the surface area of the part of the plane z = 2 + 3x + 4y that lies above the rectangle [0, 5] × [1, 4]. . . . . . .

Slide 21: Worksheet #1 Problem Find the surface area of the part of the plane z = 2 + 3x + 4y that lies above the rectangle [0, 5] × [1, 4]. Solution This is a graph. So ∫ 5∫ 4√ √ A= 1 + 9 + 16 dy dx = 15 26 0 1 . . . . . .

Slide 22: Worksheet #2 2.0 1.5 Problem 1.0 Find the surface area of the 0.5 parameterized surface 0.0 2.0 x (u , v ) = u 2 1.5 y(u, v) = uv 1 1.0 z (u , v ) = v 2 2 0.5 where 0 ≤ u ≤ 1 and 0.0 0 ≤ v ≤ 2. 0.0 0.5 1.0 . . . . . .

Slide 23: Solution We have ru = ⟨2u, v, 0⟩ r v = ⟨0 , u , v ⟩ ⟨ ⟩ ru × rv = v2 , −2uv, 2u2 |ru × rv | = 2u2 + v2 So ∫ 1∫ 2 A= (2u2 + v2 ) dv du = 4 0 0 . . . . . .

Slide 24: Worksheet #3 4 Problem 2 Find the area of the part of the surface y = 4x + z2 that lies between the planes x = 0, x = 1, z = 0, and z = 1. 1.0 0 0.5 0.0 0.0 0.5 1.0 . . . . . .

Slide 25: Solution Use the parametrization ⟨ ⟩ r(u, v) = u, 4u + v2 , v where 0 ≤ u ≤ 1, 0 ≤ v ≤ 1. Then ru = ⟨1, 4, 0⟩ rv = ⟨0, 2v, 1⟩ ru × rv = ⟨4, −1, 2v⟩ So ∫ 1∫ 1√ ∫ 1√ A= 17 + 4v2 dv du = 17 + 4v2 dv 0 0 0 . . . . . .

Slide 26: About the integral ∫ 1√ √ To find 17 + 4v2 dv, use the substitution 2v = 17 tan θ. Then √0 √ √ 2 dv = 17 sec2 θ dθ and 17 + 4v2 = 17 sec θ. So ∫ ∫ √ 1√ 17 arctan(2/ 17) 17 + 4v2 dv = sec3 θ dθ 0 2 0 √ 17 θ=arctan(2/ 17) = [sec θ tan θ + ln | sec θ + tan θ|]θ=0 4[ ( ( )) ( ( )) 17 2 2 = sec arctan √ tan arctan √ 4 17 17 ( ( )) ( ( )) ] 2 2 + ln sec arctan √ + tan arctan √ 17 17 . . . . . .

Slide 27: ( ( )) √ 2 21 Now sec arctan √ = . So this simplifies (a little) to 17 17 [√ √ ] 17 21 2 21 2 √ √ + ln +√ 4 17 17 17 17 or+ √ ( √ ) 21 17 2 + 21 + ln √ 2 4 17 . . . . . .

Slide 28: Worksheet #4 Problem Find the area of the part of the sphere x2 + y2 + z2 = 4z that lies inside the paraboloid z = x2 + y2 . z . First find a better description of the surface. The two surfaces intersect when z + z2 = 4z =⇒ z = 0, 3 So we want the portion of the sphere where z ≥ 3. . x . . . . . . .

Slide 29: Solving the sphere equation gives √ z = 2 + 4 − x2 − y 2 So we want √ ∫∫ ( )2 ( )2 ∂z ∂z A= 1+ + dA D ∂x ∂y √ where D is the circle of radius 3 in the xy-plane. . . . . . .

Slide 30: 2 The integrand becomes √ , which makes it good for 4 − x2 − y2 integration in polar coordinates! ∫ ∫ √ ∫ √ 2π 3 3 2 r dr A= √ r dr dθ = 4π √ 0 0 4− x2 − y2 0 4 − r2 Regular u-substitution u = 4 − r2 , du = −2r dr gives A = 4π . . . . . .

Slide 31: Second parametrization Use a version of spherical coordinates to get r(u, v) = ⟨2 sin u cos v, 2 sin u sin v, 2 cos u + 2⟩ π where 0 ≤ u ≤ , 0 ≤ v ≤ 2π . Then 3 ru = ⟨2 cos u cos v, 2 cos u sin v, −2 sin u⟩ rv = ⟨−2 sin u sin v, 2 sin u cos v, 0⟩ ⟨ ⟩ ru × rv = 4 sin2 u cos v, 4 sin2 u sin v, 4 sin u cos 4 |ru × rv | = 4 sin u. So ∫ π/3 ∫ 2π u=π/3 A= 4 sin u dv du = 4 · 2π [− cos u]u=0 = 4π 0 0 . . . . . .

Slide 32: Outline Last time: Integration over polar regions Integration in Polar Coordinates Easy Area Rectangles and Parallelograms in the plane Parallelograms in space Area of curved surfaces Divide Approximate Take the Limit Special cases Graphs Surfaces of Revolution Worksheet Next Time . . . . . .

Slide 33: Next time: Triple Integrals . . . . . .