Single Variable Calculus, Chapter 6, 6.3, Section 6.3, Problem 34

Use the graph of $y = x^3 - x + 1, y = -x^4 + 4x - 1 $ to estimate the $x$-coordinates of the points of intersection of the curves. Then, estimate the volume of the solid obtained by rotating about the $y$-axis the region enclosed by the curves.







Based from the graph, the $x$-coordinates of the points of intersection are $x \approx 0.4$ and $x \approx 1.25$. If we use a vertical strips, we can see that there are strips that have a distance of $x$ to the $y$-axis. If we revolve this distance about $y$-axis, you'll have a circumference of $C = 2 \pi x$. Also, the height of the strips resembles the height of the cylinder as $H y_{\text{upper}} - y_{\text{lower}} = -x^4 + 4x - 1 - (x^3 - x + 1)$.

Thus, we have


$
\begin{equation}
\begin{aligned}

V =& \int^b_a C(x) H(x) dx
\\
\\
V =& \int^{1.25}_{0.4} (2 \pi x) \left[-x^4 + 4x - 1 - (x^3 - x + 1)\right] dx
\\
\\
V =& \int^{1.25}_{0.4} (2 \pi x) \left[-x^4 - x^3 + 5x - 2\right] dx
\\
\\
V =& 2 \pi \int^{1.25}_{0.4} \left[-x^5 - x^4 + 5x^2 - 2x\right] dx
\\
\\
V =& 2 \pi \left[ \frac{-x^6}{6} - \frac{x^5}{5} + \frac{5x^3}{3} - \frac{2x^2}{2} \right]^{1.25}_{0.4}
\\
\\
V =& 3.1582 \text{ cubic units}


\end{aligned}
\end{equation}
$