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o(u, v) = (sin 2u cos v, sin 2u sin v, sin² u sin 2v) over all (u, v). To make o injective, and to simplify computations, in this problem we only consider the domain π U = {(u, v): 0 < u < 7,5 –

o(u, v) = (sin 2u cos v, sin 2u sin v, sin² u sin 2v) over all (u, v). To make o injective, and to simplify computations, in this problem we only consider the domain π U = {(u, v): 0 < u < 7,5 –

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter11: Topics From Analytic Geometry
Section11.4: Plane Curves And Parametric Equations
Problem 20E
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Question 3. The Roman surface was studied by Steiner in 1844. It is defined as the image of the function o(u, v) = (sin 2u cos v, sin 2u sin u, sin² u sin 2v) over all (u, v). To make o injective, and to simplify computations, in this problem we only consider the domain π U = {(u, v): 0 < u < 7/7, 4 π </}. 2 (However, you may not assume o is injective.) Show that olu is a regular surface patch, and that its image is a smooth surface. <U< (The Roman surface has self-intersections, and is an immersion of the real projective plane in R³; see the left image. The right image is our restriction.)
Transcribed Image Text:Question 3. The Roman surface was studied by Steiner in 1844. It is defined as the image of the function o(u, v) = (sin 2u cos v, sin 2u sin u, sin² u sin 2v) over all (u, v). To make o injective, and to simplify computations, in this problem we only consider the domain π U = {(u, v): 0 < u < 7/7, 4 π </}. 2 (However, you may not assume o is injective.) Show that olu is a regular surface patch, and that its image is a smooth surface. <U< (The Roman surface has self-intersections, and is an immersion of the real projective plane in R³; see the left image. The right image is our restriction.)
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