Divergence Theorem — Question 6

Section 19-6 · Calculus pages 3591–3592

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Question 6

Let EE be the first-octant tetrahedron x+y+z≤1x+y+z\le 1, and let S=∂ES=\partial E be outward oriented. For 𝑭(x,y,z)=⟨x2,y2,z2⟩,\mathbf F(x,y,z)=\langle x^2,y^2,z^2\rangle, compute the total outward flux.

Tasks

  1. Find the divergence.

  2. Evaluate the needed coordinate moments over the tetrahedron.

  3. Check the answer directly on the slanted face.

Question 6 – Solution

Strategy. Symmetry makes the three first moments equal; the coordinate-plane faces also provide a short direct check.

Step 1: Divergence ∇⋅𝑭=2x+2y+2z.\nabla\cdot\mathbf F=2x+2y+2z. The tetrahedron has volume 1/61/6 and centroid (1/4,1/4,1/4)(1/4,1/4,1/4), so ∭ExdV=∭EydV=∭EzdV=14⋅16=124.\iiint_E x\,dV=\iiint_E y\,dV=\iiint_E z\,dV =\frac 14\cdot\frac 16=\frac 1{24}.

Diagram shown in the original SVG worksheet.

Step 2: Apply the theorem ∬S𝑭⋅𝒏dS=2(124+124+124)=14.\boxed{\iint_S\mathbf F\cdot\mathbf n\,dS =2\left(\frac 1{24}+\frac 1{24}+\frac 1{24}\right)=\frac 14}.

Step 3: Direct check On each coordinate face the corresponding field component is zero, so those fluxes vanish. On z=1−x−yz=1-x-y, the outward vector element is ⟨1,1,1⟩dA\langle 1,1,1\rangle dA, giving ∬D(x2+y2+(1−x−y)2)dA=14.\iint_D\bigl(x^2+y^2+(1-x-y)^2\bigr)\,dA=\frac 14.

Verification The direct slanted-face integral agrees with the divergence result and is positive, as expected in the first octant.