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

Apr 14, 2017

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

#### Explanation:

$3 x + 2 y + 2 z = 6$

By solving for $z$,

$z = 3 - \frac{3}{2} x - y$

By setting $z = 0$ and solving for $y$,

$y = 3 - \frac{3}{2} x$

By setting $y = 0$ and solving for $x$,

$x = 2$

Now, the volume of the solid can be expressed as:

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

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

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

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

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

$= {\left[{\left(3 - \frac{3}{2} x\right)}^{3} / \left(\cancel{2} \cdot 3 \cdot \left(- \frac{3}{\cancel{2}}\right)\right)\right]}_{0}^{2} = {\left[- {\left(3 - \frac{3}{2} x\right)}^{3} / 9\right]}_{0}^{2}$

$= 0 - \left(- 3\right) = 3$

I hope that this was clear.