## 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{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}