Single Variable Calculus, Chapter 6, 6.3, Section 6.3, Problem 38

The region bounded by the curves $y = -x^2 + 6x - 8, y = 0$ is rotated about the $x$-axis. Find the volume of the resulting solid by any method.

Let us use the disk method together with vertical strips to evaluate the volume more easy. If we use vertical strips, we'll get a cross section with radius $R(x) y_{\text{upper}} - y_{\text{lower}} R(x) = -x^2 + 6x - 8 - (0)$. Thus, the cross sectional area of the circle is $A(x) = \pi (R(x))^2 = \pi (-x^2 + 6x - 8)^2$. Thus, we have

$\displaystyle V = \int^b_a A(x) dx$

The value of the upper and lower limits can be completed by setting the points of intersection of the curves.

$-x^2 + 6x - 8 = 0$

by applying Quadratic Formula, we get

$x = 2$ and $x = 4$

Therefore, we have..


$\begin{equation} \begin{aligned} V =& \int^4_2 \pi (-x^2 + 6x - 8)^2 dx \\ \\ V =& \pi \int^4_2 \left(x^4 - 12x^3 + 52x^2 - 96x + 64\right) dx \\ \\ V =& \pi \left[ \frac{x^5}{5} - \frac{12x^4}{4} + \frac{52x^3}{3} - \frac{96x^2}{2} + 64x \right]^4_2 \\ \\ V =& \frac{16 \pi}{15} \text{ cubic units} \end{aligned} \end{equation}$