# How do you find the volume of the pyramid bounded by the plane 2x+3y+z=6 and the coordinate plane?

##### 2 Answers
Aug 8, 2016

$= 6$ cubic units

#### Explanation:

the normal vector is $\left(\begin{matrix}2 \\ 3 \\ 1\end{matrix}\right)$ which points out in the direction of octant 1, so the volume in question is under the plane and in octant 1

we can re-write the plane as $z \left(x , y\right) = 6 - 2 x - 3 y$

for $z = 0$ we have

• $z = 0 , x = 0 \implies y = 2$
• $z = 0 , y = 0 \implies x = 3$

and
- - $x = 0 , y = 0 \implies z = 6$

it's this:

the volume we need is

${\int}_{A} z \left(x , y\right) \mathrm{dA}$

$= {\int}_{x = 0}^{3} {\int}_{y = 0}^{2 - \frac{2}{3} x} 6 - 2 x - 3 y \setminus \mathrm{dy} \setminus \mathrm{dx}$

$= {\int}_{x = 0}^{3} {\left[6 y - 2 x y - \frac{3}{2} {y}^{2}\right]}_{y = 0}^{2 - \frac{2}{3} x} \setminus \mathrm{dx}$

$= {\int}_{x = 0}^{3} {\left[6 \left(2 - \frac{2}{3} x\right) - 2 x \left(2 - \frac{2}{3} x\right) - \frac{3}{2} {\left(2 - \frac{2}{3} x\right)}^{2}\right]}_{y = 0}^{2 - \frac{2}{3} x} \setminus \mathrm{dx}$

$= {\int}_{x = 0}^{3} 12 - 4 x - 4 x + \frac{4}{3} {x}^{2} - 6 - \frac{2}{3} {x}^{2} + 4 x \setminus \mathrm{dx}$

$= {\int}_{x = 0}^{3} 6 - 4 x + \frac{2}{3} {x}^{2} \setminus \mathrm{dx}$

$= {\left[6 x - 2 {x}^{2} + \frac{2}{9} {x}^{3}\right]}_{x = 0}^{3}$

$= 18 - 18 + \frac{54}{9}$

$= 6$

Aug 8, 2016

6

#### Explanation:

We are going to be performing a triple integral.
The cartesian coordinate system is the most applicable. The order of integration is not critical. We are going to go z first, y middle, x last.

$\underline{\text{Determining limits}}$

On the plane $z = 6 - 2 x - 3 y$ and on the coordinate plane $z = 0$ hence

$z : 0 \rightarrow 6 - 2 x - 3 y$

Along $z = 0$, $y$ goes from 0 to $3 y = 6 - 2 x$ hence

$y : 0 \rightarrow 2 - \frac{2}{3} x$

Along $y = 0 , z = 0$ hence

$x : 0 \rightarrow 3$

We are finding the volume so $f \left(x , y , z\right) = 1$. Integral becomes

${\int}_{0}^{3} {\int}_{0}^{2 - \frac{2}{3} x} {\int}_{0}^{6 - 2 x - 3 y} \mathrm{dz} \mathrm{dy} \mathrm{dx}$

$= {\int}_{0}^{3} {\int}_{0}^{2 - \frac{2}{3} x} {\left[z\right]}_{0}^{6 - 2 x - 3 y} \mathrm{dy} \mathrm{dx}$

$= {\int}_{0}^{3} {\int}_{0}^{2 - \frac{2}{3} x} \left(6 - 2 x - 3 y\right) \mathrm{dy} \mathrm{dx}$

$= {\int}_{0}^{3} {\left[6 y - 2 x y - \frac{3}{2} {y}^{2}\right]}_{0}^{2 - \frac{2}{3} x} \mathrm{dx}$

$= {\int}_{0}^{3} \left(6 \left(2 - \frac{2}{3} x\right) - 2 x \left(2 - \frac{2}{3} x\right) - \frac{3}{2} {\left(2 - \frac{2}{3} x\right)}^{2}\right) \mathrm{dx}$

$= {\int}_{0}^{3} \left(12 - 4 x - 4 x + \frac{4}{3} {x}^{2} - \frac{3}{2} \left(4 - \frac{8}{3} x + \frac{4}{9} {x}^{2}\right)\right) \mathrm{dx}$

$= {\int}_{0}^{3} \left(12 - 8 x + \frac{4}{3} {x}^{3} - 6 + 4 x - \frac{2}{3} {x}^{2}\right) \mathrm{dx}$

$= {\int}_{0}^{3} \left(6 - 4 x + \frac{2}{3} {x}^{2}\right) \mathrm{dx}$

$= {\left[6 x - 2 {x}^{2} + \frac{2}{9} {x}^{3}\right]}_{0}^{3}$

$= 6 \left(3\right) - 2 {\left(3\right)}^{2} + \frac{2}{9} {\left(3\right)}^{3}$

$= 6$