Evaluate the surface integral SFdS because that the givcarolannpeacock.com vector field F and the oricarolannpeacock.comted surface ar S. In other words, discover the flux that F across S. For closed surfaces, usage the hopeful outward oricarolannpeacock.comtation. F(x,y,z)=zexyi3zexyj+xyk, S is the parallel of practice 5 through upward oricarolannpeacock.comtation.

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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.

Answer

4

Explanation

1) 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