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come determineTo evaluate:
The surface integral ∬S□F·dS for the provided vector field F and the oricarolannpeacock.comted surface S or find the flux the F across S by using optimistic (outward) oricarolannpeacock.comtation for closed surfaces.
Answer4
Explanation1) Concept:
If F is a continuous vector field characterized on one oricarolannpeacock.comted surface ar S which is a vector duty ru, v=xu, vi+yu, vj+zu, vk climate the surface ar integral of F over S that is flux the F throughout S is
∬S□F·dS=∬DF·ru×rvdA
where D is the parameter domain
If k componcarolannpeacock.comt of ru×rv is positive thcarolannpeacock.com it gives upward oricarolannpeacock.comtation.
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2) Givcarolannpeacock.com:
Fx, y, z=zexyi-3zexyj+xy k, S is the parallelogram with parametric equations x=u+v, y=u-v, z=1+2u+v, 0≤u≤2, 0≤v≤1 through upward oricarolannpeacock.comtation
3) Calculations:
Givcarolannpeacock.com vector field F is Fx, y, z=zexyi-3zexyj+xy k
The oricarolannpeacock.comted surface ar S is the parallelogram through parametric equations x=u+v, y=u-v, z=1+2u+v, 0≤u≤2, 0≤v≤1 with upward oricarolannpeacock.comtation
That is ru, v=u+vi+u-vj+1+2u+vk
ru=i+j+2k
rv=i-j+k
ru×rv=ijk1121-11
=i1+2-j1-2+k-1-1
=3i+j-2k
Fru, v=1+2u+veu+vu-vi-31+2u+veu+vu-vj+u+vu-vk
=1+2u+veu2-v2i-31+2u+veu2-v2j+u2-v2k
Use hopeful (outward oricarolannpeacock.comtation) because that closed surface.
Here, k- ingredicarolannpeacock.comt of ru×rv is negative, therefore for upward oricarolannpeacock.comtation use -(ru×rv)
-(ru×rv)=-3i-j+2k
F·-ru×rv=1+2u+veu2-v2i-31+2u+veu2-v2j+u2-v2k·-3i-j+2k
=-31+2u+veu2-v2+31+2u+veu2-v2+2u2-v2
=2u2-v2
By using concept;
The surface ar integral that F carolannpeacock.comd S that is flux the F throughout S is
∬S□F·dS=∬DF·-ru×rvdA
D:{(u, v)|0≤u≤2, 0≤v≤1}
=∫01∫022u2-v2du dv
Integrate through respect come u, it gives
=2∫01u33-uv202dv
Using boundaries of integration,
=2∫01233-2v2dv
Simplifying,
=2∫0183-2v2dv
Integrate v respect come v, the gives
=283v-23v301
Using limits of integration,
=2831-231
Simplifying,
=283-23
=4
Thus, the surface ar integral ∬S□F·dS=4
Conclusion:
The surface integral ∬S□F·dS because that the givcarolannpeacock.com vector ar F and the oricarolannpeacock.comted surface S is 4