# How do you find the volume of the solid generated by revolving the region bounded by the graphs y=3(2-x), y=0, x=0, about the y axis?

Dec 30, 2016

Volume $= 8 \pi$

#### Explanation:

When $y \left(2 - x\right)$ is rotated about the $y$-axis it will generate a cone of base radius $2$ and height $6$, hence the volume is:

$\setminus \setminus \setminus \setminus \setminus V = \frac{1}{3} \pi {r}^{2} h$
$\therefore V = \frac{1}{3} \pi \cdot {2}^{2} \cdot 6$
$\therefore V = 8 \pi$

If you want a calculus solution, then the Volume of revolution about the $y$-axis is given by:

$V = {\int}_{a}^{b} \pi {x}^{2} \mathrm{dy}$

Now:

$y = 3 \left(2 - x\right) \implies y = 6 - 3 x$
$\therefore x = \frac{1}{3} \left(6 - y\right)$

And so:

$V = {\int}_{0}^{6} \pi {\left(\frac{1}{3} \left(6 - y\right)\right)}^{2} \setminus \mathrm{dy}$
$\setminus \setminus \setminus = \frac{1}{9} \pi {\int}_{0}^{6} {\left(6 - y\right)}^{2} \setminus \mathrm{dy}$
$\setminus \setminus \setminus = \frac{1}{9} \pi {\int}_{0}^{6} \left(36 - 12 y + {y}^{2}\right) \setminus \mathrm{dy}$
$\setminus \setminus \setminus = \frac{1}{9} \pi {\left[36 y - 6 {y}^{2} + \frac{1}{3} {y}^{3}\right]}_{0}^{6}$
$\setminus \setminus \setminus = \frac{1}{9} \pi \left\{\left(36 \cdot 6 - 6 \cdot 36 + \frac{216}{3}\right) - \left(0\right)\right\}$
$\setminus \setminus \setminus = \frac{1}{9} \pi \left(72\right)$
$\setminus \setminus \setminus = 8 \pi$