# If f (x)= x^2+ 1 , how do you find the volume of the solid generated by revolving the region under graph of f from x=-1 to x=1 about the x- axis?

Oct 21, 2017

$\frac{56 \pi}{15}$

11.73 (2. d.p.)

#### Explanation:

First we need to integrate ${\left({x}^{2} + 1\right)}^{2}$. $\int {\left({x}^{2} + 1\right)}^{2} \mathrm{dx}$

$\int {\left({x}^{2} + 1\right)}^{2} \mathrm{dx} = x + \left(\frac{1}{5}\right) \cdot {x}^{5} + \left(\frac{2}{3}\right) \cdot {x}^{3}$

So we have:

$\pi {\int}_{- 1}^{1} {\left({x}^{2} + 1\right)}^{2} \mathrm{dx}$

We have:

$\pi \left[\left(x + {x}^{5} / 5 + \frac{2 {x}^{3}}{3} + c\right) - \left(x + {x}^{5} / 5 + \frac{2 {x}^{3}}{3} + c\right)\right]$

Plugging in 1 and -1 for our upper and lower bounds:

$\pi \left[\left(\left(1\right) + {\left(1\right)}^{5} / 5 + \frac{2 {\left(1\right)}^{3}}{3} + c\right) - \left(\left(- 1\right) + {\left(- 1\right)}^{5} / 5 + \frac{2 {\left(- 1\right)}^{3}}{3} + c\right)\right]$

$\pi \left[\left(\frac{28}{15}\right) - \left(- \frac{28}{15}\right)\right] = \pi \left(\frac{56}{15}\right) \to$

$\frac{56 \pi}{15}$

11.73 (2. d.p.)