# How do you find the volume bounded by y=sqrtx and the lines y=0 and x=4 revolved about the y=-1?

Jun 23, 2016

$\frac{56 \pi}{3}$

#### Explanation:

the volume of a small element is given by

$\Delta V = \pi \left({\left(1 + \sqrt{x}\right)}^{2} - {\left(1\right)}^{2}\right) \Delta x$
$= \pi \left(1 + 2 \sqrt{x} + x - 1\right) \Delta x$
$= \pi \left(2 \sqrt{x} + x\right) \Delta x$

So $V = \pi \setminus {\int}_{0}^{4} \setminus 2 \sqrt{x} + x \setminus \mathrm{dx}$

$= \pi {\left[\frac{4}{3} {x}^{\frac{3}{2}} + {x}^{2} / 2\right]}_{0}^{4}$

$= \pi \left(\left[\frac{4}{3} \cdot 8 + 8\right] - 0\right)$

$= \frac{56 \pi}{3}$