Math 21A Parametrized Surfaces Fall 2016 1 Match the Following Surfaces to Their Parametric Equations

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

(d) ~r(u, v) = hu + 1, v − 2, 3 − u/2 − v/5i – VI √ (e) ~r(u, v) = hu, v, u2 + v2i –V √ √ (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(|x|+1))p4 |x1337 − x42 + y| + 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(|u|+1))p4 |u1337 − u42 + v| + 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) = |(eu + 1) cos v, (eu + 1) sin v, u|, 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

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. √ √ (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!