Parametric Surfaces. The goal of this exercise is to find the flux of the positi
ID: 2878119 • Letter: P
Question
Parametric Surfaces. The goal of this exercise is to find the flux of the position vector field, r = xi + yj+ zk, across the lateral surface of an open-ended cylinder. In particular, S is the piece of the cylinder y^2 +z^2 = a^2, which lies along the x-axis, with -8 lessthanorequalto x lessthanorequalto 8. We parametrize S using sigma (u, v) = (x(u, v), y(u, v), z(u, v)) = (8u, a sin v, a cos v), -1 lessthanorequalto u lessthanorequalto 1, 0 lessthanorequalto v lessthanorequalto 2 pi. Now calculate the flux phi = doubleintegral_S r middot dS.Explanation / Answer
volume of cylinder V=(8-(-8))a2
volume of cylinder V=16a2
(u,v)=<8u,asinv,acosv>
u=<8, 0, 0>
v=<8,acosv,-asinv>
uxv=<(0)-(0),(8*0)-(8*-asinv),(8*acosv)-(8*0)>
uxv=<0,8asinv,8acosv>
r=<x,y,z>
r(u,v)=<8u,asinv,acosv>
=sr.dS
=[0 to 2][-1 to 1] <8u,asinv,acosv>.<0,8asinv,8acosv> du dv
=[0 to 2][-1 to 1] [0+8a2sin2v+8a2cos2v] du dv
=[0 to 2][-1 to 1] [8a2(sin2v+cos2v)] du dv
=[0 to 2][-1 to 1] 8a2 du dv
=[0 to 2][-1 to 1] 8a2u dv
=[0 to 2]8a2(1-(-1)) dv
=[0 to 2]16a2 dv
=[0 to 2]16a2 v
=16a2 (2)
=2*16a2
=2V
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