Parametrize the surface obtained from revolving the circle on the xy plane of radius 3 centered...

Question:

Parametrize the surface obtained from revolving the circle on the xy plane of radius 3 centered at the point (9,10) about the y-axis and then the x-axis.

Parametric Surfaces:

We first note that if we take a fixed point {eq}(x_0,y_0) {/eq} in the {eq}xy {/eq}-plane and rotate it around the {eq}x {/eq}-axis, then we obtain a circle with parametric equation {eq}x=x_0,\, y=y_0\sin \theta,\, z=z_0\cos \theta,\, 0\leq \theta \leq 2\pi {/eq}. We can extend this notion to find the parametrization of the parametric curve {eq}x(t),\, y(t) {/eq} in the {eq}xy {/eq}-plane.

Answer and Explanation:

We first note that the circle of radius 3 centered at {eq}(9,10) {/eq} has equation {eq}(x-9)^2+(y-10)^2=9 {/eq}. Parametric equations for this circle would then be

{eq}x=3\sin t+9,\, y=3\cos t+10,\, 0\leq t\leq 2\pi {/eq}.

Revolving this about the {eq}y {/eq}-axis, the {eq}y {/eq} coordinate stays the same and we spin out a circle in the {eq}x {/eq} and {eq}z {/eq} coordinates giving us a surface with parametric equations

{eq}x=(3\sin t+9)\sin u,\, y=3\cos t+10,\, z=(3\sin t+9)\cos u,\, 0\leq t\leq 2\pi,\, 0\leq u\leq 2\pi {/eq}.

Similarly if we revolve this about the {eq}x {/eq}-axis we obtain a surface with parametric equations

{eq}x=3\sin t+9,\, y=(3\cos t+10)\sin u,\, z=(3\cos t+10)\cos u,\, 0\leq t\leq 2\pi,\, 0\leq u\leq 2\pi {/eq}.


Learn more about this topic:

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Evaluating Parametric Equations: Process & Examples

from Precalculus: High School

Chapter 24 / Lesson 3
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