15_5_13 - inside the cone, then the area element on is dS =...

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13. The intersection of the plane z = 1 + y and the cone z = p 2 ( x 2 + y 2 ) has projection onto the xy -plane the elliptic disk E bounded by ( 1 + y ) 2 = 2 ( x 2 + y 2 ) 1 + 2 y + y 2 = 2 x 2 + 2 y 2 2 x 2 + y 2 - 2 y + 1 = 2 x 2 + ( y - 1 ) 2 2 = 1 . Note that E has area A = π( 1 )( 2 ) and centroid ( 0 , 1 ) . If is the part of the plane lying
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Unformatted text preview: inside the cone, then the area element on is dS = s 1 + ± ∂ z ∂ y ² 2 dx dy = √ 2 dx dy . Thus ZZ y dS = √ 2 ZZ E y dx dy = √ 2 A y = 2 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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