# How do you find the volume of the solid obtained by rotating the region bound by the curve and y=x^2+1 and x-axis in the interval (2,3)?

Mar 22, 2017

#### Explanation:

The volume of the solid obtained by rotating the region bound by the curve $y = R \left(x\right)$ and $x$-axis in the interval $\left(c , d\right)$ is

$V = \pi {\int}_{c}^{d} {\left[R \left(x\right)\right]}^{2} \mathrm{dx}$

Here, we have $R \left(x\right) = {x}^{2} + 1$, $c = 2$ and $d = 3$ and volume is

$\pi {\int}_{\textcolor{red}{2}}^{\textcolor{red}{3}} {\left[{x}^{2} + 1\right]}^{2} \mathrm{dx}$
(once we have put $R \left(x\right)$, put $c$ amd $d$ as well)

= $\pi {\int}_{\textcolor{red}{2}}^{\textcolor{red}{3}} \left[{x}^{4} + 2 {x}^{2} + 1\right] \mathrm{dx}$

= color(red)(pi[x^5/5+(2x^3)/3+x+C]_2^3
(once integral has been done no need to write it)

= color(red)(pi[(3^5/5+(2xx3^3)/3+3+C)-(2^5/5+(2xx2^3)/3+2+C)]

= color(red)(pi[(243/5+54/3+3+C)-(32/5+16/3+2+C)]

= color(red)(pi[(243-32)/5+(54-16)/3+(3-2)]

= color(red)(pi[211/5+38/3+1]

= color(red)(pi[(211xx3+38xx5+15)/15]

= color(red)(pi[(633+190+15)/15]

= color(red)(pi[(838)/15]=55.86pi