# How do you find the volume bounded by y=sqrt(x + 1), x = 0, x = 3, and y = 0 revolved about the x-axis?

We will use the disk method. Where $r = y$
The volume is $V = {\int}_{0}^{3} \pi {r}^{2} \mathrm{dx} = {\int}_{0}^{3} \pi {\left(\sqrt{x + 1}\right)}^{2} \mathrm{dx} =$
$\pi {\int}_{0}^{3} \left(x + 1\right) \mathrm{dx} = \pi {\left[{x}^{2} / 2 + x\right]}_{0}^{3} = \pi \left(\frac{9}{2} + 3\right) = \frac{15 \pi}{2}$