The problem asks to find the volume of the solid generated by revolving the region bounded by the curves $y = 3 - x$ and $y = x^2 + 1$ about the x-axis.

1. Problem Description

The problem asks to find the volume of the solid generated by revolving the region bounded by the curves $y = 3 - x$ and $y = x^2 + 1$ about the x-axis.

2. Solution Steps

First, find the points of intersection of the two curves by setting them equal to each other:

$3 - x = x^2 + 1$

$x^2 + x - 2 = 0$

$(x+2)(x-1) = 0$

$x = -2$ or $x = 1$

So the intersection points are at $x = -2$ and $x = 1$.

Now, we use the washer method to find the volume. The formula for the washer method is:

$V = \pi \int_a^b (R(x)^2 - r(x)^2) dx$

where $R(x)$ is the outer radius and $r(x)$ is the inner radius.

In this case, $R(x) = 3 - x$ and $r(x) = x^2 + 1$. The limits of integration are $a = -2$ and $b = 1$.

So the volume is:

$V = \pi \int_{-2}^1 ((3 - x)^2 - (x^2 + 1)^2) dx$

$V = \pi \int_{-2}^1 (9 - 6x + x^2 - (x^4 + 2x^2 + 1)) dx$

$V = \pi \int_{-2}^1 (9 - 6x + x^2 - x^4 - 2x^2 - 1) dx$

$V = \pi \int_{-2}^1 (-x^4 - x^2 - 6x + 8) dx$

$V = \pi [-\frac{x^5}{5} - \frac{x^3}{3} - 3x^2 + 8x]_{-2}^1$

$V = \pi [(-\frac{1}{5} - \frac{1}{3} - 3 + 8) - (-\frac{(-2)^5}{5} - \frac{(-2)^3}{3} - 3(-2)^2 + 8(-2))]$

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

$V = \pi [(-\frac{1}{5} - \frac{1}{3} + 5) - (\frac{32}{5} + \frac{8}{3} - 28)]$

$V = \pi [-\frac{1}{5} - \frac{1}{3} + 5 - \frac{32}{5} - \frac{8}{3} + 28]$

$V = \pi [-\frac{33}{5} - \frac{9}{3} + 33]$

$V = \pi [-\frac{33}{5} - 3 + 33]$

$V = \pi [-\frac{33}{5} + 30]$

$V = \pi [-\frac{33}{5} + \frac{150}{5}]$

$V = \pi [\frac{117}{5}]$

$V = \frac{117\pi}{5}$

3. Final Answer

The volume of the solid is $\frac{117\pi}{5}$.