Solutions to Assignment-11 1. The aim of this question is to study surface integrals under change of parametrizations. Let S ⊂ R3 be a C1-regular surface with a non-degenerate parametrization ~r : D ! S. Like always we use coordinates (u; v) on D. Let ~γ(s; t) = ~r(u(s; t); v(s; t)) be another non-degenerate parametrization for S, where T (s; t) = (u(s; t); v(s; t) is a C1-diffoemorphism, from Ω to D. Given parametrizations ~r and ~γ, we have natural orienations for the surface S given by the unit normal vectors ~r × ~r ~γ × ~γ n^ = u v ; ν^ = s t j ~ru × ~rvj j ~γs × ~γtj We say that T is orientation preserving if @(u; v) > 0: @(s; t) (a) Show that @(u; v) ( ~γ × ~γ )(s; t) = ( ~r × ~r ); s t @(s; t) u v where the right hand side is evaluated at (u(s; t); v(s; t)). Solution: By chain rule, @u @v ~γ = ~r + ~r s @s u @s v @u @v ~γ = ~r + ~r ; t @t u @t v and so @u @v @u @v ( ~γ × ~γ ) = ~r + ~r × ~r + ~r s t @s u @s v @t u @t v @u @v @u @v = − ~r × ~r @s @t @t @s u v @(u; v) = ( ~r × ~r ): @(s; t) u v We used the distributive property to go from the first line to the second, and the fact that ~ru × ~ru = ~rv × ~rv = ~0 and ~rv × ~ru = − ~ru × ~rv. (b) The definition of surface integral in Lecture 23 is in terms of a given parametrization. Use part (a) to conclude that the definition is indeed independent of parametrizations. Solution: Change of variables (c) Show that if F~ is any vector field on S and T is orientation preserving, then the flux of F~ through S is independent of the parametrization if S is oriented using the standard unit normal for the parametrization. 1 Solution: Change of variables. 2. (a) Show that the surface area of the graph z = f(x; y) defined over a region D ⊂ R2 is given by s ZZ @f 2 @f 2 1 + + dA: D @x @y Solution: We can parametrize a graph by ~r(x; y) = hx; y; f(x; y)i: We have already shown in class that ^ ~rx × ~ry = −fx^i − fy^j + k; and so the area element is s @f 2 @f 2 dS = j ~r × ~r j dA = 1 + + dA; x y @x @y and the formula for the surface area is obtained by integrating. (b) Use this to compute the surface area of the part of the cone z2 = a2(x2 + y2) lying above the disc x2 + y2 ≤ R2 in the xy-plane. Express your answer in terms of R and the height h of the cone to recover the standard high-school formula. Solution: We apply the formula above with f(x; y) = apx2 + y2 and D = f(x; y) j x2 + y2 < R2g: Then it is easy to see that s @f 2 @f 2 p 1 + + dA = 1 + a2 dA; @x @y and so p ZZ A(S) = 1 + a2 dA D p = πR2 1 + a2: The height of the cone is H = aR, and so in terms of the height the surface area is p A(S) = πR R2 + H2: p The high-school formula is actually in terms of the slant length l, which is l = R2 + H2, and so in terms of l the surface area is πRl: 3.A surface of revolution is a surface obtained by rotating a curve lying in a plane about a line in that plane that is disjoint from the curve. (a) Let C be a simple curve in the xz-plane with parametrization (x(t); 0; z(t)), a ≤ t ≤ b such that x(t) ≥ 0 for all t. Let S be the surface obtained by rotating C about the z-axis. Write down a 2 parametrization for S, and show that the surface area is given by the formula Z b A(S) = 2π x(t)p[x0(t)]2 + [z0(t)]2 dt: a Hint. Take a look at the notes for Lecture 22. Solution: As in the lecture notes, the solid of revolution can be parametrized as ~r(t; θ) = hx(t) cos θ; x(t) sin θ; z(t)i: Then ^i ^j k^ 0 0 0 0 ^ 0 ^ 0 ^ ~rt × ~rθ = x (t) cos θ x (t) sin θ z (t) = −x(t)z (t) cos θi − x(t)z (t) sin θj + x(t)x (t)k:; −x(t) sin θ x(t) cos θ 0 and so the surface area element is given by p 0 2 0 2 dS = j ~rx × ~rθj = x(t) [x (t)] + [z (t)] dt dθ: Integrating out the dθ, Z b Z b A(S) = 2π jx(t)jp[x0(t)]2 + [z0(t)]2 dt = 2π x(t)p[x0(t)]2 + [z0(t)]2 dt; a a since x(t) ≥ 0. (b) Let b > a > 0 and f :[a; b] ! R be a real valued function with continuous derivative. Let S be the surface of revolution obtained by rotating the graph z = f(x) in the xz-plane about the z-axis. Write down a parametrization for the surface S, and show that the surface area is given by the formula Z b A(S) = 2π xp1 + [f 0(x)]2 dx: a Note. This is of course a standard formula taught in most single-variable calculus courses on integration, but it is only a special case of our general theory of surface integrals. Solution: Apply part(a) with x(t) = x and z(t) = f(x): (c)A torus (or doughnut) of inner radius a and outer radius b (0 < a < b) can be formed by rotating the circle (x − b)2 + z2 = a2, about the z-axis. Find it's surface area using part(b). Solution: We actually use part(a). The cricle can be parametrized by x = b + a cos t; z = a sin t: Then the surface area is Z 2π A(S) = 2πa (b + a cos t) dt 0 = 4π2ab: 3 (d) Now let C be a smooth simple curve lying in a plane Π with arc length L, and let l be any line in that plane that does not intersect C. Let S be the surface obtained by rotating C about the line l. The geometric centroid (xC ; 0; zC ) is defined by 1 Z 1 Z xC = x ds; zC = z ds: L C L C Prove Pappus' theorem that the surface area of S is given by A(S) = 2πLd; where d is the distance of the centroid (xC ; 0; zC ) from the line l. Hint. By rotation and translation you can assume that Π is the xz plane and the line l is the z-axis. Historical note. Pappus of Alexandria was a greek geometer who lived around 300AD. Solution: As in the hint, we assume that the plane is the xz-plane and l is the z-axis. Then d = xC . By the definition of xC and the fact that the arc length element is ds = p[x0(t)]2 + [z0(t)]2 dt, we see that Z Z b p 0 2 0 2 A(S) LxC = x ds = x(t) [x (t)] + [z (t)] dt = ; C a 2π by part(a). From this Pappus' theorem follows since d = xC . (e) Now use Pappus' theorem to solve part (c). Your answer should of course be the same as that in part (c). Solution: The centroid of the circle is the center which is at xc = b and zc = 0, and so d = b. L is the circumference of the circle, which is 2πa: So by Pappus' theorem A(S) = 2π(2πa)(b) = 4π2ab: 4.