This problem will illustrate the divergence theorem by computing the outward flux of the vector field F(x,y,z)=1xi+5yj+4zkF(x,y,z)=1xi+5yj+4zk across the boundary of the right rectangular prism: −5≤x≤6,−3≤y≤5,−5≤z≤6−5≤x≤6,−3≤y≤5,−5≤z≤6 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 = -5 :  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)=−5z1(s,t)=−5

The face with z = 6 :   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)=6z2(s,t)=6

The face with x = -5 :   σ3=x3(s,t),y3(s),z3(t))σ3=x3(s,t),y3(s),z3(t)) x3(s,t)=−5x3(s,t)=−5 y3(s)=y3(s)=  z3(t)=z3(t)= 

The face with x = 6 :   σ4=(x4(s,t),y4(s),z4(t))σ4=(x4(s,t),y4(s),z4(t)) x4(s,t)=6x4(s,t)=6 y4(s)=y4(s)=  z4(t)=z4(t)= 

The face with y = -3 :   σ5=(x5(s),y5(s,t),z5(t))σ5=(x5(s),y5(s,t),z5(t)) x5(s)=x5(s)=  y5(s,t)=−3y5(s,t)=−3 z5(t)=z5(t)= 

The face with y = 5 :   σ6=(x6(s),y6(s,t),z6(t))σ6=(x6(s),y6(s,t),z6(t)) x6(s)=x6(s)=  y6(s,t)=5y6(s,t)=5 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

==  + + + + +

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Part 2 - Using the Divergence Theorem

∫∫σF⋅ndS=∫∫∫GdivFdV=∫∫σF⋅ndS=∫∫∫GdivFdV=

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∫∫

 

∫∫

 

∫∫

    dxdydzdxdydz

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