Math 21a Parametrized surfaces Fall 2016 1 Match the following surfaces to their parametric equations. I II III IV V VI (a) ~r(u; v) = hsin u cos v; cos u; sin u sin vi { III (x2 + y2 + z2 = 1) (b) ~r(u; v) = hu; sin u cos v; sin u sin vi { IV (c) ~r(u; v) = hv cos u; 2v sin u; 2v2i { II (4x2 + y2 = 2z) (d) ~r(u; v) = hu + 1; v − 2; 3 − u=2 − v=5i { VI p (e) ~r(u; v) = hu; v; u2 + v2i {V p p (f) ~r(u; v) = h 1 + u2 cos v; 1 + u2 sin v; ui {I(x2 + y2 − z2 = 1) 2 Give a parametric equation for the plane containing the points (2; 2; 2), (−4; 0; 4) and (1; 3; 5). −! −! Solution. We have PQ = h−6; −2; 2i and PR = h−1; 1; 3i, so a parametric equation for the plane is ~r(u; v) = (2; 2; 2) + uh−6; −2; 2i + vh−1; 1; 3i = h2 − 6u − v; 2 − 2u + v; 2 + 2u + 3vi: 3 Give a parametric equation for a sphere of radius 6 centered at (3; −1; −2). (Hint: use spherical coordinates, then modify the equation.) Solution. We use spherical coordinates to write down equations for a sphere centered at the origin, then translate it. ~r(u; v) = h3 + 6 sin u cos v; −1 + 6 sin u sin v; −2 + 6 cos ui; for 0 ≤ u ≤ πand 0 ≤ v < 2π. 4 Let f(x; y) = yesin(log log(jxj+1))p4 jx1337 − x42 + yj + 9001y. Find a parametric equation for the graph of f, i.e., the surface z = f(x; y). Solution. This is a graph: ~r(u; v) = hu; v; f(u; v)i = hu; v; vesin(log log(juj+1))p4 ju1337 − u42 + vj + 9001vi: 5 Let x = ez + 1 be a curve in the xz-plane. Find a parametric equation for the surface obtained by revolving this parabola around the z-axis. Solution. This is a surface of revolution. ~r(u; v) = j(eu + 1) cos v; (eu + 1) sin v; uj; 0 ≤ v < 2π: How to think about this? At height z = u, we want a circle of radius eu + 1 in the xy-slice. 6 Find parametrizations for each of the following surfaces, given by an equation in cylin- drical or spherical coordinates. Then match each equation to its plot. (a) r = 3 + 2 sin(z) cos(5θ) (b) ρ = 6 + sin(7φ) sin(5θ) (c) r = 4 + 2 sin(z + θ) I II III Solution. (a) This is II. A parametrization is h(3 + 2 sin z cos 5θ) cos θ; (3 + 2 sin z cos 5θ) sin θ; zi for 0 ≤ θ < 2π. (b) This is I. A parametrization is h(6 + sin φ sin 5θ) sin φ cos θ; (6 + sin 7φ sin 5θ) sin φ sin θ; (6 + sin 7φ sin 5θ) cos φi for 0 ≤ φ ≤ π. (c) This is III. A parametrization is h(4 + 2 sin(z + θ)) cos θ; (4 + 2 sin(z + θ)) sin θ; zi. 7 Here are some more surfaces for you to parametrize as practice. (a)3 x + 2y + z = 6 (b) y = 2z2 − 3x2 (c) z = x2 + y2 + 1; what if we only want to parametrize the part of this elliptic paraboloid under the plane z = 10? x2 y2 2 (d) 4 + 9 + z = 1 (e) The graph of the curve y = 1 + sin x revolved around the x-axis (f) The graph of the curve y = z2 + z revolved around the z-axis Solution. (a) Two ways: as a graph, hu; v; 6−3u−2vi; as a plane, e.g., (0; 0; 6)+uh1; 0; −3i+vh0; 1; −2i. (b) As a graph, hu; 2v2 − 3u2; vi. p p (c) As a graph, hu; v; u2 + v2 + 1i; as a surface of revolution, h u − 1 cos v; u − 1 sin v; ui. (d) As a \stretched" sphere (ellipsoid), h2 sin u cos v; 3 sin u sin v; cos vi for 0 ≤ u < 2π and 0 ≤ v < π. (Can you do something similar to a surface of revolution?) (e) As a surface of revolution, hu; (1 + sin u) cos v; (1 + sin u) sin vi for 0 ≤ v < 2π. (f) As a surface of revolution, h(v2 + v) cos u; (v2 + v) sin u; vi for 0 ≤ u < 2π. Notice that v2 + v is not always nonnegative!.