Volume of Irregular Tetrahedron Bounded By Axes

An irregular tetrahedron is bounded by the axes and the plane  
\[2x+y+z=4\]
. What is the volume of the tetrahedron?

We can take a volume element as a strip of width  
\[dx\]
, length  
\[y\]
  and height  
\[z\]
. The volume of the element is  
\[dV=ydxdz=(4-2x-z)dxdz\]
.
We can take the limits of  
\[x\]
  and  
\[y\]
  as  
\[, \; 2- \frac{z}{2}\]
  and  
\[0, \l 4\]
  respectively. The volume is
\[\begin{equation} \begin{aligned} V &= \int^4_0 \int^{2- \frac{z}{2}}_0 (4-2x-z)dxdz \\ &= \int^4_0 [ \int^{2- \frac{z}{2}}_0 (4-2x-z)dx ] dz \\ &= \int^4_0 [4x- x^2-xz]^{2- \frac{z}{2}}_0 dz \\ &= \int^4_0 4-2z+ \frac{z^2}{4} dz \\ &= [ 4z- z^2 + \frac{z^3}{6} ]^4_0 = \frac{32}{3}\end{aligned} \end{equation}\]

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