Mathematics, 20.08.2021 03:40 andrew412603
This problem will illustrate the divergence theorem by computing the outward flux of the vector field F(x, y,z)=2xi+5yj+3zkF(x, y,z)=2xi+5yj+3zk across the boundary of the right rectangular prism: β3β€xβ€5,β5β€yβ€7,β4β€zβ€7β3β€xβ€5,β5β€yβ€7,β 4β€zβ€7 oriented outwards using a surface integral and a triple integral over the solid bounded by rectangular prism.
Note: The vectors in this field point outwards from the origin, so we would expect the flux across each face of the prism to be positive.
Part 1 - Using a Surface Integral
First we parameterize the six faces using 0β€sβ€10β€sβ€1 and 0β€tβ€10β€tβ€1:
The face with z = -4 : Ο1=(x1(s),y1(t),z1(s, t))Ο1=(x1(s),y1(t),z1(s, t))
x1(s)=x1(s)=
y1(t)=y1(t)=
z1(s, t)=β4z1(s, t)=β4
The face with z = 7 : Ο2=(x2(s),y2(t),z2(s, t))Ο2=(x2(s),y2(t),z2(s, t))
x2(s)=x2(s)=
y2(t)=y2(t)=
z2(s, t)=7z2(s, t)=7
The face with x = -3 : Ο3=x3(s, t),y3(s),z3(t))Ο3=x3(s, t),y3(s),z3(t))
x3(s, t)=β3x3(s, t)=β3
y3(s)=y3(s)=
z3(t)=z3(t)=
The face with x = 5 : Ο4=(x4(s, t),y4(s),z4(t))Ο4=(x4(s, t),y4(s),z4(t))
x4(s, t)=5x4(s, t)=5
y4(s)=y4(s)=
z4(t)=z4(t)=
The face with y = -5 : Ο5=(x5(s),y5(s, t),z5(t))Ο5=(x5(s),y5(s, t),z5(t))
x5(s)=x5(s)=
y5(s, t)=β5y5(s, t)=β5
z5(t)=z5(t)=
The face with y = 7 : Ο6=(x6(s),y6(s, t),z6(t))Ο6=(x6(s),y6(s, t),z6(t))
x6(s)=x6(s)=
y6(s, t)=7y6(s, t)=7
z6(t)=z6(t)=
Then (mind the orientation)
β«β«ΟFβ
ndSβ«β«ΟFβ
ndS =β«10β«10F(Ο1)β
(βΟ1βtΓβΟ1βs)dsdt=β«01β« 01F(Ο1)β
(βΟ1βtΓβΟ1βs)dsdt +β«10β«10F(Ο2)β
(βΟ2βsΓβΟ2βt)dsdt+β«01β« 01F(Ο2)β
(βΟ2βsΓβΟ2βt)dsdt +β«10β«10F(Ο3)β
(βΟ3βtΓβΟ3βs)dsdt+β«01β« 01F(Ο3)β
(βΟ3βtΓβΟ3βs)dsdt +β«10β«10F(Ο4)β
(βΟ4βsΓβΟ4βt)dsdt+β«01β« 01F(Ο4)β
(βΟ4βsΓβΟ4βt)dsdt +β«10β«10F(Ο5)β
(βΟ5βsΓβΟ5βt)dsdt+β«01β« 01F(Ο5)β
(βΟ5βsΓβΟ5βt)dsdt +β«10β«10F(Ο6)β
(βΟ6βtΓβΟ6βs)dsdt+β«01β« 01F(Ο6)β
(βΟ6βtΓβΟ6βs)dsdt
== + + + + +
==
Answers: 1
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