# How do you find the volume of the solid obtained by rotating the region bounded by the curves y=x^2+1 and y=-x+3 rotated around the x-axis?

Jun 17, 2015

I got area is $23.4 \pi$. The "How" has no one sentence answer -- see below.

#### Explanation:

The curves: $y = {x}^{2} + 1$ and $y = - x + 3$ intersect at $- 2$ and at $1$.
On the interval $\left[- 2 , 1\right\}$, the value of $- x + 3$ is greater than that of ${x}^{x} + 1$. (The line is above the parabola.)

We'll use washers and find $\int \left(\pi {R}^{2} - \pi {r}^{2}\right) \mathrm{dx}$ (Where the larger radius is $R$ and the smaller $r$.

$V = {\int}_{-} {2}^{1} \left(\pi {\left(- x + 3\right)}^{2} - \pi {\left({x}^{2} + 1\right)}^{2}\right) \mathrm{dx}$

$= \pi {\int}_{-} {2}^{1} \left({\left(x - 3\right)}^{2} - {\left({x}^{2} + 1\right)}^{2}\right) \mathrm{dx}$

$= \pi {\int}_{-} {2}^{1} \left(\left({x}^{2} - 6 x + 9\right) - \left({x}^{4} + 2 {x}^{2} + 1\right)\right) \mathrm{dx}$

$= \pi {\int}_{-} {2}^{1} \left(- {x}^{4} - {x}^{2} - 6 x + 8\right) \mathrm{dx}$

$= \pi {\left[- {x}^{5} / 5 - {x}^{3} / 3 - 3 {x}^{2} + 8 x\right]}_{-} {2}^{1}$

$= \pi \left[\left(- \frac{1}{5} - \frac{1}{3} - 3 + 8\right) - \left(\frac{32}{5} + \frac{8}{3} - 12 - 16\right)\right]$

$= \pi \left[- \frac{33}{5} - \frac{9}{3} + 5 + 12 + 16\right]$

$= \pi \left[- \frac{66}{10} - 3 + 5 + 12 + 16\right]$

$= \pi \left[- 6.6 + 30\right]$

$= 23.4 \pi$