15_5_14 - z = 1 + y , then ZZ y dS = √ 3 ZZ E y dx dy =...

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14. Continuing the above solution, the cone z = p 2 ( x 2 + y 2 ) has area element dS = s 1 + ± z x ² 2 + ± z y ² 2 dx dy = s 1 + 4 ( x 2 + y 2 ) z 2 dx dy = 3 dx dy . If is the part of the cone lying below the plane
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Unformatted text preview: z = 1 + y , then ZZ y dS = √ 3 ZZ E y dx dy = √ 3 A y = √ 6 π....
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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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