15_6_4 - k and z = 0, so F • ˆ N = 0. Thus, the total...

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4. F = y i + z k . Let 1 be the conical surface and 2 be the base disk. The flux of F outward through the surface of the cone is ZZ F ˆ N = ZZ 1 + ZZ 2 . On 1 : ˆ N = 1 2 x i + y j p x 2 + y 2 + k ! , dS = 2 dx dy . Thus ZZ 1 F ˆ N dS = ZZ x 2 + y 2 1 xy p x 2 + y 2 + 1 - p x 2 + y 2 ! dx dy = 0 + π × 1 2 - Z 2 π 0 d θ Z 1 0 r 2 dr = π - 2 π 3 = π 3 . On 2 : ˆ N = -
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Unformatted text preview: k and z = 0, so F • ˆ N = 0. Thus, the total flux of F out of the cone is π/ 3. x y z ˆ N ˆ N z = 1-√ x 2 + y 2 1 1 2 1 Fig. 6-4...
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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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