Find the volume of the tetrahedron bounded by the planes x + 2y + z = 2, x = 2y, x = 0, and z = 0.

Steps to Solve

1. Identify the planes and their equations
   The planes bounding the tetrahedron are:

   • Plane 1: $x + 2y + z = 2$
   • Plane 2: $x = 2y$
   • Plane 3: $x = 0$ (the yz-plane)
   • Plane 4: $z = 0$ (the xy-plane)

2. Find the intersection points of the planes
   We need to find the vertices of the tetrahedron by finding the intersection of the planes.

   • Intersection of $x + 2y + z = 2$ and $x = 0$: [ 0 + 2y + z = 2 \implies z = 2 - 2y ]
     For $y = 0$, we get the point $(0, 0, 2)$.

   • Intersection of $x + 2y + z = 2$ and $x = 2y$: [ 2y + 2y + z = 2 \implies 4y + z = 2 \implies z = 2 - 4y ] If $y = 0$, we find the point $(0, 0, 2)$. For $y = 0.5$, we find the point $(1, 0.5, 0)$.

   • Intersection of $x = 2y$, $z = 0$, and $x = 0$: This yields the point $(0, 0, 0)$.

   • Intersection of $x = 0$, $z = 0$, and $y = 1$: Yields point $(0, 1, 0)$.

After solving, the vertices are:

• $A(0, 0, 2)$
• $B(0, 1, 0)$
• $C(1, 0.5, 0)$
• $D(0, 0, 0)$

3. Calculate the volume using the vertices
   The volume $V$ of a tetrahedron with vertices at $(x_1, y_1, z_1)$, $(x_2, y_2, z_2)$, $(x_3, y_3, z_3)$, and $(x_4, y_4, z_4)$ is given by the formula:
   [ V = \frac{1}{6} \left| \begin{vmatrix} x_1 & y_1 & z_1 & 1 \ x_2 & y_2 & z_2 & 1 \ x_3 & y_3 & z_3 & 1 \ x_4 & y_4 & z_4 & 1 \end{vmatrix} \right| ]

Using points (A(0, 0, 2)), (B(0, 1, 0)), (C(1, 0.5, 0)), and (D(0, 0, 0)):
[ V = \frac{1}{6} \left| \begin{vmatrix} 0 & 0 & 2 & 1 \ 0 & 1 & 0 & 1 \ 1 & 0.5 & 0 & 1 \ 0 & 0 & 0 & 1 \end{vmatrix} \right| ]

4. Evaluate the determinant
   Calculating the determinant, we find:

[ = \left| 0\cdot(1 \cdot 0 - 0.5 \cdot 0) - 0\cdot(0 \cdot 0 - 0 \cdot 1) + 2\cdot(0 \cdot 1 - 0\cdot 0) - 1\cdot(0 \cdot 0.5 - 1\cdot 0) \right| = -2 ]

So,

[ V = \frac{1}{6} \cdot 2 = \frac{1}{3} ]