Triple Integrals over Tetrahedra and Triangular Prisms

From Planar Faces to Cartesian Bounds

A tetrahedron is bounded by four triangular faces. A triangular prism has two parallel triangular ends joined by three other faces. In both cases, the boundary faces are planes, so their equations can be used directly to describe the solid with Cartesian inequalities.

The key step is to choose an integration order and identify, in that order, the lower and upper limits for each variable. The outer variables describe a projection of the solid; the innermost variable runs between two faces. A correct set of bounds must describe the entire solid exactly once, without including points outside it.

For a solid \(E\) and an integrable function \(f\), a Cartesian iterated integral has the form \(\iiint_E f(x,y,z)\,dV\), with \(dV=dx\,dy\,dz\) or one of the other five orders. The limits in each order must describe the same region \(E\). Earlier material on Cartesian bounds and Fubini’s theorem supplies the general interpretation of these iterated integrals; here the emphasis is on reading the bounds from the planar faces of these particular solids.

A Standard Tetrahedron and Its Six Orders

Consider the tetrahedron in the first octant cut off by the plane \(x+y+z=1\). Its coordinate-plane faces impose \(x\geq 0\), \(y\geq 0\), and \(z\geq 0\), while its sloping face imposes \(x+y+z\leq 1\). To make one variable the innermost variable, solve the sloping-face inequality for that variable. For example, solving for \(z\) gives \(0\leq z\leq 1-x-y\). The remaining two variables must satisfy \(x+y\leq 1\).

The symmetry of \(x+y+z\leq 1\) means the same pattern works whichever variable is innermost. The six orders are:

For instance, the row for \(dy\,dz\,dx\) comes from fixing \(x\) first. The remaining inequality is \(y+z\leq 1-x\). Thus \(z\) ranges from \(0\) to \(1-x\), and after \(x\) and \(z\) are fixed, \(y\) ranges from \(0\) to \(1-x-z\).

Why the Bound Construction Works

For the standard tetrahedron, the region is defined by three nonnegative coordinates and one inequality, \(x+y+z\leq 1\). Fixing any two coordinates leaves an interval of allowed values for the third: it begins at zero and ends where the point meets the plane \(x+y+z=1\). The fixed coordinates themselves must be nonnegative and have sum at most one. Those conditions form a triangular projection in the corresponding coordinate plane.

This reasoning proves the listed bounds for every order. For example, if \(x\) is outermost and \(z\) is innermost, first choose \(x\) in \([0,1]\). For that fixed \(x\), the face and nonnegativity conditions require \(y+z\leq 1-x\). Therefore \(z\) lies between \(0\) and \(1-x-y\), while \(y\) lies between \(0\) and \(1-x\). The other five descriptions follow by choosing the outer and inner variables in turn; the original inequality is unchanged when the coordinate names are permuted.

Worked Example: Integrating Over the Standard Tetrahedron

Other Tetrahedra: Read the Sloping Face Carefully

A tetrahedron need not have the same intercept on each coordinate axis. Suppose its faces are the three coordinate planes and \(2x+3y+z=6\). The solid is in the first octant and satisfies \(2x+3y+z\leq 6\). If \(z\) is innermost, solving the face inequality for \(z\) gives \(0\leq z\leq 6-2x-3y\). The projection onto the \(xy\)-plane is determined by \(2x+3y\leq 6\), or \(0\leq y\leq (6-2x)/3\) with \(0\leq x\leq 3\).

Triangular Prisms: Separate the Base from the Height

For a vertical triangular prism, the base is a triangle in the \(xy\)-plane and \(z\) ranges between constant heights. For example, a base described by \(x\geq 0\), \(y\geq 0\), and \(x+y\leq 2\), together with \(0\leq z\leq 3\), gives a prism. Its vertical faces arise from the three edges of the base, while its top and bottom are the planes \(z=3\) and \(z=0\).

In an order with \(z\) innermost, the \(z\)-bounds are simply \(0\) and \(3\), and the two outer bounds describe the triangular base. Since these heights do not depend on \(x\) or \(y\), they can also be placed in the middle or outer position when the integration order is changed. The base bounds must still match the chosen order.

A Common Bounds Mistake

A frequent error is to choose the outer limits from the individual coordinate ranges but forget the restriction imposed by the sloping face. For the standard tetrahedron, it is true that \(0\leq x\leq 1\) and \(0\leq y\leq 1\) separately. It does not follow that \(y\) can range from \(0\) to \(1\) for every fixed \(x\).

The correction is to enforce \(x+y\leq 1\) in the projection. With \(x\) outermost, use \(0\leq x\leq 1\), then \(0\leq y\leq 1-x\), and finally \(0\leq z\leq 1-x-y\). At \(x=0.8\), this correctly restricts \(y\) to \(0\leq y\leq 0.2\), so the upper \(z\)-limit cannot be negative.

Choosing and Checking Bounds

Before evaluating an integral over one of these solids, check that each limit comes from an actual face or edge of the region. A useful procedure is to write the solid’s inequalities, select an order, solve the relevant plane equation for the inner variable, and then derive the projection limits from the remaining inequalities.

Planar faces make these regions especially suitable for Cartesian integration: their equations translate directly into linear bounds. The same geometric check is essential whether the integrand is \(1\), a coordinate function, or a more complicated expression. Once the bounds accurately describe the solid, an iterated integral evaluates the desired quantity over that region.