# How would you find the volume of the tetrahedron T bounded by the coordinate planes and the plane 3x+4y+z=10?

Jan 2, 2017

Volume = 125/9 (${\text{units}}^{3}$)

#### Explanation:

The coordinate planes are given by $x = 0$, $y = 0$ and $z = 0$. The volume is that of a tetrahedron whose vertices are the intersections of three of the four planes given. The intersection of $x = 0$, $y = 0$ and $3 x + 4 y + z = 10$ is $\left(0 , 0 , 10\right)$, Similarly, the other three vertices are $\left(\frac{10}{3} , 0 , 0\right)$, $\left(0 , \frac{5}{2} , 0\right)$ and the origin $\left(0 , 0 , 0\right)$.

The given tetrahedron, $T$, is a solid that lies above the triangle $R$ in the $x y$-plane that has vertices $\left(0 , 0\right)$, $\left(\frac{10}{3} , 0\right)$ and $\left(0 , \frac{5}{2}\right)$.

The line joining $\left(0 , \frac{5}{2}\right)$ and $\left(\frac{10}{3} , 0\right)$ is given by:

$y - 0 = \left(\frac{\frac{5}{2}}{- \frac{10}{3}}\right) \left(x - \frac{10}{3}\right)$
$\therefore y = - \frac{3}{4} x + \frac{5}{2}$

And so the region $R$ is defined as:

 R = {(x, y) | 0 le x le 10/3, 0 le y le −3/4x +5/2 }

And the volume, $V$, of the tetrahedron is the double
integral of the function z=10 − 3x − 4y over $R$.

 :. V = int int_R (10 − 3x − 4y) \ dA
 " "= int_0^(10/3) int_(0)^(-3/4x+5/2) (10 − 3x − 4y) \ dy \ dx
$\text{ } = {\int}_{0}^{\frac{10}{3}} {\left[\left(10 - 3 x\right) y - 2 {y}^{2}\right]}_{y = 0}^{y = - \frac{3}{4} x + \frac{5}{2}} \setminus \mathrm{dx}$
$\text{ } = {\int}_{0}^{\frac{10}{3}} \left\{\left(10 - 3 x\right) \left(\frac{- 3 x}{4} + \frac{5}{2}\right) - 2 {\left(\frac{- 3 x}{4} + \frac{5}{2}\right)}^{2}\right\} - \left\{0\right\} \setminus \mathrm{dx}$
$\text{ } = {\int}_{0}^{\frac{10}{3}} \left(- \frac{30}{4} x + 25 + \frac{9}{4} {x}^{2} - \frac{15}{2} x\right) - 2 \left(\frac{9}{16} {x}^{2} - \frac{15}{4} x + \frac{25}{4}\right) \setminus \mathrm{dx}$
$\text{ } = {\int}_{0}^{\frac{10}{3}} \left(- \frac{15}{2} x + \frac{25}{2} + \frac{9}{8} {x}^{2}\right) \setminus \mathrm{dx}$
$\text{ } = {\left[- \frac{15}{4} {x}^{2} + \frac{25}{2} x + \frac{9}{24} {x}^{3}\right]}_{0}^{\frac{10}{3}}$
$\text{ } = \left\{- \frac{15}{4} \cdot \frac{100}{9} + \frac{25}{2} \cdot \frac{10}{3} + \frac{3}{8} \cdot \frac{1000}{27}\right\} - \left\{0\right\}$
$\text{ } = - \frac{125}{3} + \frac{125}{3} + \frac{125}{9}$
$\text{ } = \frac{125}{9}$