Divergence Theorem — Question 10

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Question 10

Let EE be a bounded solid with piecewise smooth closed boundary SS and outward unit normal 𝒏\mathbf n.

Tasks

  1. Use the Divergence Theorem with a suitable vector field to prove vol⁡(E)=13∬S⟨x,y,z⟩⋅𝒏dS\displaystyle \operatorname{vol}(E)=\frac 13\iint_S\langle x,y,z\rangle\cdot\mathbf n\,dS.

  2. Apply the identity to a right circular cone of base radius RR and height hh, with vertex at the origin and base in the plane z=hz=h.

  3. Explain why the lateral surface contributes zero.

Original worksheet page 1: question and worked solution for 6-6-010
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Question 10 – Solution

Strategy. Choose a field with divergence 11, then exploit the fact that the cone’s radial position vector is tangent to its lateral generators.

Step 1: Derive the identity Let 𝑮=13⟨x,y,z⟩.\mathbf G=\frac 13\langle x,y,z\rangle. Then ∇⋅𝑮=1\nabla\cdot\mathbf G=1, so vol⁡(E)=∭E1dV=∬S𝑮⋅𝒏dS=13∬S⟨x,y,z⟩⋅𝒏dS.\operatorname{vol}(E)=\iiint_E1\,dV =\iint_S\mathbf G\cdot\mathbf n\,dS =\boxed{\frac 13\iint_S\langle x,y,z\rangle\cdot\mathbf n\,dS}.

See the diagram in the original worksheet below.

Step 2: Lateral surface Every ray from the origin along the cone is tangent to the lateral surface. Hence ⟨x,y,z⟩\langle x,y,z\rangle is tangent there and ⟨x,y,z⟩⋅𝒏=0.\langle x,y,z\rangle\cdot\mathbf n=0.

Step 3: Base On the base, z=hz=h and 𝒏=𝒌\mathbf n=\mathbf k, so the dot product equals hh. Therefore vol⁡(E)=13∬basehdA=13h(πR2)=13πR2h.\boxed{\operatorname{vol}(E)=\frac 13\iint_{\mathrm{base}}h\,dA =\frac 13h(\pi R^2)=\frac 13\pi R^2h}.

Verification The formula has the expected “one-third base times height” form and correct cubic units.

Original worksheet page 2: question and worked solution for 6-6-010

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