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The Divergence Theorem · Tutorial 964 of 999

Polyhedral Solids and Plane-Bounded Regions

Learn to convert planar face inequalities into iterated bounds and use them to calculate outward flux through polyhedral solids.

Advanced 12 min read

What You'll Learn

  • Define convex polyhedral solids using finitely many planar half-space inequalities.
  • Derive tetrahedral bounds by projecting onto a coordinate plane.
  • Set up bounds for triangular prisms and other plane-bounded regions.
  • Apply the divergence theorem when the vector field satisfies its smoothness conditions.
  • Evaluate polynomial divergence integrals and check the resulting values.
  • Identify how independent coordinate bounds can include points outside a sloping face.

Why Polyhedral Solids Are Useful

A polyhedral solid is bounded by planar faces. This makes its geometry well suited to the divergence theorem: each face lies in a plane, while the volume integral can often be written with simple iterated bounds. The main task is to translate the face inequalities into bounds that describe exactly the solid.

Tetrahedra are especially useful examples because their sloping faces produce bounds that depend on the other coordinates. Triangular prisms and other plane-bounded regions show how the same procedure applies beyond tetrahedra.

Definition: In this tutorial, a convex polyhedral solid is a bounded region in three-dimensional space that is the intersection of finitely many closed half-spaces. Each half-space is specified by a linear inequality, and its boundary lies in a plane. The boundary of the solid consists of planar polygonal faces. A nonconvex region with planar faces is not, in general, a single intersection of half-spaces; it can instead be treated by decomposing it into convex pieces when that is appropriate.

The Divergence Theorem on a Polyhedral Boundary

For a vector field \(\mathbf{F}=(P,Q,R)\), its divergence is \(\nabla\cdot\mathbf{F}=\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}+\frac{\partial R}{\partial z}\). The divergence theorem converts outward flux across a closed boundary into a volume integral.

Theorem: Let \(E\) be a bounded solid whose boundary is a closed, piecewise smooth surface, oriented by its outward normal \(\mathbf{n}\). If \(\mathbf{F}\) has continuous first partial derivatives on an open set containing \(E\), then
$$ \iint_{\partial E}\mathbf{F}\cdot\mathbf{n}\,dS = \iiint_E \nabla\cdot\mathbf{F}\,dV. $$
The planar faces of a polyhedral solid are smooth individually. Its edges and vertices are where faces meet; they do not prevent the boundary from being piecewise smooth. The theorem applies provided the stated condition on \(\mathbf{F}\) also holds.

For a polyhedral solid, the theorem does not require a separate flux calculation on each face. Instead, find the divergence and describe the solid with bounds, then integrate over its volume.

Deriving Bounds for a Tetrahedron

Consider the tetrahedron in the first octant bounded by the coordinate planes and the plane \(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\), where \(a,b,c\) are positive. The solid is the portion satisfying \(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\leq 1\), together with \(x,y,z\geq 0\).

To find iterated bounds in the order \(dz\,dy\,dx\), first project the tetrahedron onto the \(xy\)-plane. Setting \(z=0\) in the plane inequality gives \(\frac{x}{a}+\frac{y}{b}\leq 1\). Thus \(x\) runs from \(0\) to \(a\), and for each \(x\), \(y\) runs from \(0\) to \(b(1-x/a)\). Finally, solve the plane inequality for \(z\).

$$ 0\leq x\leq a,\qquad 0\leq y\leq b\left(1-\frac{x}{a}\right),\qquad 0\leq z\leq c\left(1-\frac{x}{a}-\frac{y}{b}\right). $$

These bounds describe the whole tetrahedron and nothing outside it. For each permitted pair \((x,y)\), the final bounds include precisely the vertical segment from the coordinate plane \(z=0\) to the sloping face.

Why the bounds work: The three lower bounds come from the coordinate planes. The upper bound on \(y\) comes from the plane’s intersection with \(z=0\), and the upper bound on \(z\) is obtained by solving the plane inequality for \(z\). Together, these inequalities are exactly the four half-space conditions defining the tetrahedron.

Worked Example: Polynomial Divergence on an Intercept Tetrahedron

Worked Example: Polynomial Divergence on an Intercept Tetrahedron

Let \(E\) be the tetrahedron in the first octant bounded by \(\frac{x}{2}+\frac{y}{3}+\frac{z}{4}=1\). Find the outward flux of \(\mathbf{F}=(x^2/2,y^2/2,z^2/2)\) across its boundary.

The field is polynomial, so its first partial derivatives are continuous everywhere. Its divergence is

$$ \nabla\cdot\mathbf{F} = \frac{\partial}{\partial x}\left(\frac{x^2}{2}\right) + \frac{\partial}{\partial y}\left(\frac{y^2}{2}\right) + \frac{\partial}{\partial z}\left(\frac{z^2}{2}\right) =x+y+z. $$

Using the tetrahedral bounds with \(a=2\), \(b=3\), and \(c=4\), the divergence theorem gives

$$ \iint_{\partial E}\mathbf{F}\cdot\mathbf{n}\,dS = \int_0^2 \int_0^{3(1-x/2)} \int_0^{4(1-x/2-y/3)} (x+y+z)\,dz\,dy\,dx. $$

Scale the coordinates by \(x=2u\), \(y=3v\), and \(z=4w\). Then \(dV=24\,du\,dv\,dw\), and the tetrahedron becomes \(u\geq0\), \(v\geq0\), \(w\geq0\), \(u+v+w\leq1\). The integral is therefore

$$ 24\int_0^1\int_0^{1-u}\int_0^{1-u-v}(2u+3v+4w)\,dw\,dv\,du. $$

Evaluate the three coordinate contributions separately. For \(u\), integrating first with respect to \(w\) gives

$$ \int_0^1\int_0^{1-u}u(1-u-v)\,dv\,du = \frac12\int_0^1u(1-u)^2\,du = \frac12\left(\frac12-\frac23+\frac14\right) = \frac1{24}. $$

For \(v\), integrating first with respect to \(w\) and then \(v\) gives

$$ \int_0^1\int_0^{1-u}v(1-u-v)\,dv\,du = \int_0^1\frac{(1-u)^3}{6}\,du = \frac1{24}. $$

For \(w\), integration with respect to \(w\) gives \((1-u-v)^2/2\), so

$$ \int_0^1\int_0^{1-u}\int_0^{1-u-v}w\,dw\,dv\,du = \int_0^1\frac{(1-u)^3}{6}\,du = \frac1{24}. $$

Consequently, the flux is

$$ 24\left(2\cdot\frac1{24}+3\cdot\frac1{24}+4\cdot\frac1{24}\right) = 2+3+4 = 9. $$

As a check on the coordinate integrals, \(\int_0^1 u(1-u)^2/2\,du=1/24\), and \(\int_0^1(1-u)^3/6\,du=1/24\); their antiderivatives give the displayed endpoint values. The computed volume integral is therefore \(9\), which, by the divergence theorem, is the outward flux.

Triangular Prisms and Other Plane-Bounded Regions

A polyhedral solid need not be a tetrahedron. For instance, a triangular prism can be described by a triangular base in the \(xy\)-plane and constant lower and upper bounds on \(z\). The same principle applies: identify the projection, then use the face inequalities to give the remaining coordinate bounds.

Worked Example: Flux Through a Triangular Prism

Let \(E\) be the triangular prism defined by \(0\leq x\leq2\), \(0\leq y\leq2-x\), and \(0\leq z\leq1\). Find the outward flux of \(\mathbf{F}=(x^2,y^2/2,0)\) across its boundary.

The field has continuous first partial derivatives, and

$$ \nabla\cdot\mathbf{F} = \frac{\partial x^2}{\partial x} + \frac{\partial (y^2/2)}{\partial y} + \frac{\partial 0}{\partial z} = 2x+y. $$

The divergence theorem and the stated bounds give

$$ \iint_{\partial E}\mathbf{F}\cdot\mathbf{n}\,dS = \int_0^2\int_0^{2-x}\int_0^1(2x+y)\,dz\,dy\,dx. $$

The integrand does not depend on \(z\), and the \(z\)-interval has length \(1\). Integrating in \(y\) gives

$$ \int_0^{2-x}(2x+y)\,dy = 2x(2-x)+\frac{(2-x)^2}{2} = 4x-2x^2+2-2x+\frac{x^2}{2} = 2+2x-\frac32x^2. $$

Therefore,

$$ \int_0^2\left(2+2x-\frac32x^2\right)\,dx = \left[2x+x^2-\frac{x^3}{2}\right]_0^2 = 4+4-4 = 4. $$

The antiderivative differentiates to \(2+2x-\frac32x^2\), the integrand obtained after the \(y\)-integration. Thus the outward flux is \(4\).

Another Tetrahedral Polynomial Integral

Worked Example: A Quadratic Divergence on a Unit Tetrahedron

Let \(E\) be the unit tetrahedron \(x,y,z\geq0\), \(x+y+z\leq1\). Find the outward flux of \(\mathbf{F}=(x^3/3,0,0)\).

The field is smooth everywhere, and its divergence is \(\nabla\cdot\mathbf{F}=x^2\). The bounds, in the order \(dz\,dy\,dx\), are \(0\leq x\leq1\), \(0\leq y\leq1-x\), and \(0\leq z\leq1-x-y\). Hence

$$ \iint_{\partial E}\mathbf{F}\cdot\mathbf{n}\,dS = \int_0^1\int_0^{1-x}\int_0^{1-x-y}x^2\,dz\,dy\,dx. $$

Since \(x^2\) is constant with respect to \(z\), the \(z\)-integration gives \(x^2(1-x-y)\). Integrating with respect to \(y\) then gives

$$ \int_0^{1-x}x^2(1-x-y)\,dy = \frac{x^2(1-x)^2}{2}. $$

Thus the final integral is

$$ \frac12\int_0^1(x^2-2x^3+x^4)\,dx = \frac12\left(\frac13-\frac12+\frac15\right) = \frac1{60}. $$

The antiderivative for the final integrand is \(x^3/6-x^4/4+x^5/10\). Its derivative is \(x^2/2-x^3+x^4/2=\frac12x^2(1-x)^2\), and its endpoint difference is \(1/6-1/4+1/10=1/60\). The outward flux is therefore \(1/60\).

A Common Bounds Error

A sloping face usually makes one coordinate’s upper bound depend on another coordinate. Replacing that dependent bound with a constant can enlarge the region being integrated over.

Common mistake: For the intercept tetrahedron in the first example, taking \(0\leq x\leq2\), \(0\leq y\leq3\), and \(0\leq z\leq4\) describes the entire box, not the tetrahedron. For example, the point \((1,2,0)\) is in that box, but \(\frac12+\frac23>1\), so it lies beyond the tetrahedron’s sloping face. The correct projected bound is \(0\leq y\leq3(1-x/2)\), followed by \(0\leq z\leq4(1-x/2-y/3)\).

One reliable check is to substitute the proposed upper bounds into the plane equation. At the upper \(z\)-bound, \(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\); at \(z=0\), the allowed \(x,y\) values still satisfy \(\frac{x}{a}+\frac{y}{b}\leq1\). This checks that the iterated bounds meet the correct face and remain within its projection.

Why This Method Matters

Plane-bounded solids arise in flux problems because their faces are simple to describe, while direct surface calculations can require a separate integral on every face. The divergence theorem replaces those face calculations with one volume integral. Once the divergence is computed, the main geometric work is to express the solid accurately in an order of integration that matches its planar boundaries.

If a plane-bounded region is nonconvex, do not describe it as one intersection of half-spaces: such an intersection is convex. When a suitable decomposition into convex polyhedral pieces is available, the pieces can be handled separately; shared interior faces have opposite outward normals and cancel in the total flux. For a single convex polyhedron, the half-space description gives a direct route from faces to bounds.

Check Your Understanding

Use the definitions, bounds, and theorem in this tutorial to answer the following questions.

  1. What four inequalities describe the tetrahedron with intercepts \(a\), \(b\), and \(c\) on the positive coordinate axes?
  2. For the bounds \(dz\,dy\,dx\), why does the upper bound on \(y\) depend on \(x\)?
  3. What smoothness condition on \(\mathbf{F}\) is required to apply the divergence theorem?
  4. For \(0\leq x\leq1\), \(0\leq y\leq1-x\), and \(0\leq z\leq2\), write an iterated integral for \(\iiint_E (x+y)\,dV\).
  5. Why do independent bounds \(0\leq x\leq a\), \(0\leq y\leq b\), \(0\leq z\leq c\) generally describe a box rather than an intercept tetrahedron?

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