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Calculus IIIChapter 6: Surface IntegralsSection 6-6

Divergence Theorem — Question 2

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Question 2

Let E=[0,1]×[0,2]×[0,3]E=[0,1]\times[0,2]\times[0,3] and let S=∂ES=\partial E have outward orientation. For 𝑭(x,y,z)=⟨x2,yz,z⟩,\mathbf F(x,y,z)=\langle x^2,yz,z\rangle, compute the total outward flux through the six faces.

Tasks

  1. Find the divergence.

  2. Evaluate the resulting triple integral by separating its terms.

  3. Explain why this is preferable to six direct surface integrals.

Original worksheet page 1: question and worked solution for 6-6-002
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Question 2 – Solution

Strategy. The box is closed and the field is smooth, so one volume integral replaces six oriented face calculations.

Step 1: Divergence ∇⋅𝑭=2x+z+1.\boxed{\nabla\cdot\mathbf F=2x+z+1}.

See the diagram in the original worksheet below.

Step 2: Integrate term by term The three contributions over EE are ∭E2xdV=(∫012xdx)(2)(3)=6,∭EzdV=(1)(2)∫03zdz=9,∭E1dV=(1)(2)(3)=6.\begin{align*} \iiint_E2x\,dV&=\left(\int_0^1 2x\,dx\right)(2)(3)=6,\\ \iiint_Ez\,dV&=(1)(2)\int_0^3z\,dz=9,\\ \iiint_E1\,dV&=(1)(2)(3)=6. \end{align*} Hence ∬S𝑭⋅𝒏dS=∭E(2x+z+1)dV=21.\boxed{\iint_S\mathbf F\cdot\mathbf n\,dS =\iiint_E(2x+z+1)\,dV=21}.

Step 3: Interpret A direct method would require a different outward normal and parameter domain on every face. The divergence automatically combines all six signed contributions.

Verification The divergence is positive throughout EE, so a positive net outward flux is consistent.

Original worksheet page 2: question and worked solution for 6-6-002

Original worksheet layout. Use Enlarge or open the PDF for a closer view.

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