15_5_9 - the cylinder lying between the nappes of the cone...

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9. One-quarter of the required area lies in the first octant. (See the figure.) In polar coordinates, the Cartesian equation x 2 + y 2 = 2 ay becomes r = 2 a sin θ . The arc length element on this curve is ds = s r 2 + ± dr d θ ² 2 d θ = 2 a d θ. Thus dS = p x 2 + y 2 ds = 2 ar d θ = 4 a 2 sin θ d θ on the cylinder. The area of that part of
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Unformatted text preview: the cylinder lying between the nappes of the cone is 4 Z π/ 2 4 a 2 sin θ d θ = 16 a 2 sq. units. . x y z d S x 2 + y 2 = 2 y ds z 2 = x 2 + y 2 Fig. 5-9...
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This note was uploaded on 03/08/2012 for the course MATH 120 taught by Professor Onurfen during the Spring '12 term at Middle East Technical University.

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