Question

Use a triple integral to compute the volume of the following regions. The larger of two solids formed when the parallelepiped (slanted box) with vertices (0,0,0),(2,0,0),(0,2,0),(2,2,0),(0,1,1) \((2,1,1),(0,3,1),\) and (2,3,1) is sliced by the plane \(y=2\)

Short answer

Answer: The volume of the larger solid is the same as the entire parallelepiped because the plane intersects one of its faces, and the object is not divided into two different parts.

Step-by-step solution

Obtain the equation of the slicing plane

The plane that slices the parallelepiped is given by the equation \(y=2\). Hence, we will use this plane equation to determine the boundaries of our solid in the \(y\)-axis.

Determine the parametrization of the solid

The given vertices of the parallelepiped indicate that it is formed by the vectors \(\vec{a} = \begin{pmatrix} 2 \\ 0 \\ 0 \end{pmatrix}\), \(\vec{b} = \begin{pmatrix} 0 \\ 2 \\ 0 \end{pmatrix}\), and \(\vec{c} = \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}\). We can describe a point inside the parallelepiped as a linear combination of these vectors. A point \(P\) inside the parallelepiped can be written as $$ P = \begin{pmatrix} x \\ y \\ z \end{pmatrix} = t\,\vec{a} + u\,\vec{b} + v\,\vec{c}, $$ where \(0 \leq t,u,v \leq 1\). Now that we have the parametrization, we will determine the range of \(t\), \(u\), and \(v\) to set up the triple integral.

Set up the triple integral

Since \(y=2\) slices the parallelepiped in such a way that it does not intersect two of the edges emanating from \((0,0,0)\), we can express the vertices that correspond to \(y=2\) as those with \(u=1\). The remaining vertices are described by \(0 \le t,v \le 1\). Thus, our limits of integration will be: $$ 0 \le t \le 1, \quad 1 \le u \le 1, \quad 0 \le v \le 1. $$ We need to find the Jacobian determinant \(J\) of the transformation from \((t,u,v)\) to \((x,y,z)\). We compute the Jacobian matrix and its determinant: $$ \mathcal{J} = \begin{pmatrix} 2 & 0 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 1 \end{pmatrix},\quad J = \begin{vmatrix} 2 & 0 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 1 \end{vmatrix} = 2(2) = 4. $$ Now, we can set up the triple integral to compute the volume of the larger solid: $$ V = \iiint_{V} J \, dt \, du \, dv = 4 \int_{0}^{1} \int_{1}^{1} \int_{0}^{1} dt \, du \, dv. $$

Evaluate the triple integral

We will now evaluate the triple integral using the limits of integration found in Step 3: $$ V = 4 \int_{0}^{1} \int_{1}^{1} \int_{0}^{1} dt \, du \, dv = 4 \int_{0}^{1} \int_{1}^{1} \left[ t \right]_{0}^{1} du \, dv = 4 \int_{0}^{1} \int_{1}^{1} dv, $$ since \(\int_{1}^{1} du = 0\), the volume becomes: $$ V = 0. $$ The volume of the larger solid is zero because the plane \(y=2\) coincides with one of the faces of the parallelepiped, and the object is not sliced in two parts as the question suggests. Therefore, the volume of the "larger" solid becomes the entire parallelepiped.