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: −3≤x≤5,−5≤y≤7,−4≤z≤7−3≤x≤5,−5≤y≤7,− 4≤z≤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 0≤s≤10≤s≤1 and 0≤t≤10≤t≤1:
The face with z = -4 : σ1=(x1(s),y1(t),z1(s, t))σ1=(x1(s),y1(t),z1(s, t))
x1(s)=x1(s)=
y1(t)=y1(t)=
z1(s, t)=−4z1(s, t)=−4
The face with z = 7 : σ2=(x2(s),y2(t),z2(s, t))σ2=(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 : σ3=x3(s, t),y3(s),z3(t))σ3=x3(s, t),y3(s),z3(t))
x3(s, t)=−3x3(s, t)=−3
y3(s)=y3(s)=
z3(t)=z3(t)=
The face with x = 5 : σ4=(x4(s, t),y4(s),z4(t))σ4=(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 : σ5=(x5(s),y5(s, t),z5(t))σ5=(x5(s),y5(s, t),z5(t))
x5(s)=x5(s)=
y5(s, t)=−5y5(s, t)=−5
z5(t)=z5(t)=
The face with y = 7 : σ6=(x6(s),y6(s, t),z6(t))σ6=(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)
∫∫σF⋅ndS∫∫σF⋅ndS =∫10∫10F(σ1)⋅(∂σ1∂t×∂σ1∂s)dsdt=∫01∫ 01F(σ1)⋅(∂σ1∂t×∂σ1∂s)dsdt +∫10∫10F(σ2)⋅(∂σ2∂s×∂σ2∂t)dsdt+∫01∫ 01F(σ2)⋅(∂σ2∂s×∂σ2∂t)dsdt +∫10∫10F(σ3)⋅(∂σ3∂t×∂σ3∂s)dsdt+∫01∫ 01F(σ3)⋅(∂σ3∂t×∂σ3∂s)dsdt +∫10∫10F(σ4)⋅(∂σ4∂s×∂σ4∂t)dsdt+∫01∫ 01F(σ4)⋅(∂σ4∂s×∂σ4∂t)dsdt +∫10∫10F(σ5)⋅(∂σ5∂s×∂σ5∂t)dsdt+∫01∫ 01F(σ5)⋅(∂σ5∂s×∂σ5∂t)dsdt +∫10∫10F(σ6)⋅(∂σ6∂t×∂σ6∂s)dsdt+∫01∫ 01F(σ6)⋅(∂σ6∂t×∂σ6∂s)dsdt
== + + + + +
==