Finding the volume of the tetrahedron with vertices $(0,0,0)$, $(2,0,0)$, $(0,2,0)$, $(0,0,2)$. I get $8$; answer is $4/3$. The following problem is from the 7th edition of the book "Calculus and Analytic Geometry Part II". It can be found in section 13.7. It is
problem number 5.

Find the volume of the tetrahedron whose vertices are the given points:
  $$ ( 0, 0, 0 ), ( 2, 0, 0 ), ( 0, 2, 0 ), ( 0, 0, 2 ) $$

Answer:
In this case, the tetrahedron is a parallelepiped object. If the bounds of such an object is given by the vectors $A$, $B$ and $C$ then
the area of the object is $A \cdot (B \times C)$. Let $V$ be the volume we are trying to find.
\begin{align*}
x^2 &= 6 - y^2 - z^2 \\[4pt]
A &= ( 2, 0, 0) - (0,0,0) = ( 2, 0, 0) \\
B &= ( 0, 2, 0) - (0,0,0) = ( 0, 2, 0) \\
C &= ( 0, 0, 2) - (0,0,0) = ( 0, 0, 2) \\[4pt]
V &= \begin{vmatrix}
a_1 & a_2 & a_3 \\
b_1 & b_2 & b_3 \\
c_1 & c_2 & c_3 \\
\end{vmatrix} =
\begin{vmatrix}
2 & 0 &0 \\
0 & 2 & 0 \\
0 & 0 & 2\\
\end{vmatrix} \\
&= 2 \begin{vmatrix}
2 & 0 \\
0 & 2\\
\end{vmatrix} = 2(4 - 0) \\
&= 8
\end{align*}
However, the book gets $\frac{4}{3}$.
 A: A tetrahedron is never a parallepiped. It is a pyramid with a triangular base. 
All six faces of a parallelopiped are parallelograms but a tetrahedron has only four faces and all are triangles. 
In short, what you are measuring is very unlike what you were asked to measure. 
A: Note that the given volume is a cone with the height 2 and a right isosceles  triangle of side 2 as the base. Thus, its volume can be calculated as
$$\frac13 Area_{base} \cdot Height = \frac13 (\frac12 \cdot 2\cdot 2)2=\frac43$$
A: your tetrahedron is also a pyramid. with the volume of $\frac{1}{3}\cdot \frac{2\cdot 2}{2}\cdot 2=\frac{4}{3}$
A: Referring to this, now you must divide it by $3!$ to get the final answer.
A: Yes, because what you are really doing is finding the volume of a volume of a parallelepiped. You can also approach the problem this way:
height= |a|
and the area is given by
|b × c|
So, then you have Volume = height * area
Volume= |a||b × c|=|a ∙( b × c)|
For b: (0,2,0) – (0,0,0) = (0,2,0)
For c: (0,0,2) – (0,0,0) = (0,0,2)
b ×c=[(2x2)-(0x0),(0x0)-(0x2),(0x0)-(2x0)]=(4,0,0)
For a: (2,0,0) – (0,0,0) = (2,0,0)
∴ Volume= a ∙( b × c)=(2x4)+(0x0)+(0x0)=8
The volume of tetrahedron is 1/6 that of the parallelepiped
∴ V_Tetrahedron=  (1/6) x 8=  4/3
