Surface Integrals of Vector Fields — Question 6

Section 19-4 · Calculus pages 3551–3552

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Question 6

Let EE be the tetrahedron in the first octant bounded by the coordinate planes and x+y+z=1x+y+z=1. Let S=∂ES=\partial E carry the outward orientation, and let 𝑭(x,y,z)=⟨x,y,z⟩.\mathbf F(x,y,z)=\langle x,y,z\rangle. Compute the total outward flux directly by summing the four faces.

Tasks

  1. Determine the flux through the three coordinate-plane faces.

  2. Parametrize the slanted face with its outward orientation.

  3. Sum the face contributions and verify the sign.

Question 6 – Solution

Strategy. The field has zero normal component on each coordinate face, leaving only the slanted triangular face.

Step 1: Coordinate faces On x=0x=0, the outward normal is −𝒊-\mathbf i and 𝑭⋅(−𝒊)=−x=0\mathbf F\cdot(-\mathbf i)=-x=0. Similarly, the flux densities on y=0y=0 and z=0z=0 vanish. These three fluxes are all zero.

Step 2: Slanted face Write z=1−x−yz=1-x-y over D={(x,y):x≥0,y≥0,x+y≤1}.D=\{(x,y):x\ge 0,\ y\ge 0,\ x+y\le 1\}. The outward side points away from the origin, and its vector element is 𝒏dS=⟨1,1,1⟩dxdy.\boxed{\mathbf n\,dS=\langle 1,1,1\rangle\,dx\,dy}.

Diagram shown in the original SVG worksheet.

On this face, 𝑭⋅⟨1,1,1⟩=x+y+z=1.\mathbf F\cdot\langle 1,1,1\rangle=x+y+z=1.

Step 3: Integrate and sum ∬∂E𝑭⋅𝒏dS=0+∬D1dA=12.\iint_{\partial E}\mathbf F\cdot\mathbf n\,dS =0+\iint_D1\,dA =\boxed{\frac 12}.

Verification The field points away from the origin and is tangent to the coordinate faces, so all nonzero flux must leave through the slanted face and must be positive.