Preview Activity 12.3.1.
A tetrahedron is a three-dimensional figure with four faces, each of which is a triangle. A picture of the tetrahedron \(T\) with vertices \((0,0,0)\text{,}\) \((1,0,0)\text{,}\) \((0,1,0)\text{,}\) and \((0,0,1)\) is shown at in Figure 12.3.1. If we place one vertex at the origin and let vectors \(\va\text{,}\) \(\vb\text{,}\) and \(\vc\) be determined by the edges of the tetrahedron that have one end at the origin, then a formula that tells us the volume \(V\) of the tetrahedron is
\begin{equation}
V = \frac{1}{6} \lvert \va \cdot (\vb \times \vc) \rvert.\tag{12.3.1}
\end{equation}
(a)
State the vectors \(\va\text{,}\) \(\vb\text{,}\) and \(\vc\) used for our tetrahedron \(T\text{.}\) Use these vectors in formula (12.3.1) to find the volume of the tetrahedron \(T\text{.}\)
(b)
Instead of memorizing or looking up the formula for the volume of a tetrahedron, we can use a double integral to calculate the volume of the tetrahedron \(T\text{.}\) To see how, notice that the top face of the tetrahedron \(T\) is the plane whose equation is
\begin{equation*}
z = 1-(x+y).
\end{equation*}
We will want to evaluate a pair of iterated integrals on on a non-rectangular region to calculate the volume of the region below this plane and in the first octant region. Reminder: the first octant is the part of \(\mathbb{R}^3\) that has positive \(x\)-, \(y\)-, and \(z\)-coordinates.
The volume of the tetrahedron will be given by the double integral
\begin{equation*}
\iint_R 1-(x+y) dA
\end{equation*}
where the region of integration will be the triangle at the base of our tetrahedron (in the \(xy\)-plane) as shown in Figure 12.3.2. We will need to understand how to express this double integral as an iterated integral using our cross sectional ideas from Section 12.2.
We will look at a cross sectional idea along a constant value of \(x=a\) which will give an iterated integral of the form
\begin{equation*}
\int_{x=?}^{x=?} \int_{y=?}^{y=?} 1-(x+y) \, dy \, dx.
\end{equation*}
If we look at \(R\) along the trace \(x=a\text{,}\) what will be the upper and lower bounds on the corresponding \(y\) coordinates?
\begin{equation*}
\text{If }x=a\text{, then } \underline{\hspace{1.0in}} \leq y \leq \underline{\hspace{1.0in}}
\end{equation*}
The lower bound will not depend on the value of \(a\) used but the upper bound will, so the expression for the upper bound should be written in terms of \(a\text{.}\) You should write a couple of sentences to describe how you came up with your upper and lower bounds.
(c)
Explain which part of the iterated integral
\begin{equation*}
\int_{x=0}^{x=1} \int_{y=0}^{y=1-x} 1-(x+y) \, dy \, dx
\end{equation*}
will correspond to the blue cross section in Figure 12.3.3.
(d)
Use the Fundamental Theorem of Calculus (twice!) to evaluate the iterated integral
\begin{equation*}
\int_{x=0}^{x=1} \int_{y=0}^{y=1-x} 1-(x+y) \, dy \, dx
\end{equation*}
and compare to your result from part (a). (As with iterated integrals over rectangular regions, start with the inner integral.)
