Single Variable Calculus, Chapter 6, 6.3, Section 6.3, Problem 28

Use the Midpoint Rule with $n = 5$ to estimate the volume of the region shown in the figure below, if the region is rotated about the $y$-axis.

*Refer to the figure in the book.*

By using Midpoint Rule with $n = 5$, we can determine the thickness of the approximately rectangles as...

$\displaystyle \Delta x = \frac{12 - 2}{5} = 2$ thus, the midpoints are $\displaystyle \left( \frac{2 + 4}{2} \right) = 3, \left( \frac{4 + 6}{2} \right) = 5, \left( \frac{6 + 8}{2} \right) = 7, \left( \frac{8 + 10}{2} \right) = 9, \left( \frac{10 + 12}{2} \right) = 11$

Notice that the distance of these approximating rectangles at $y$-axis is $x$. If you revolve, this length to about the $y$-axis, you'll have a circumference of $2 \pi x$ and the total volume is equal to the product of the circumference and the total area of the curve. So..


$
\begin{equation}
\begin{aligned}

V =& \int^{12}_2 2 \pi x f(x) \Delta x
\\
\\
\approx & 2 \pi [3 f(3) + 5f(5) + 7f(7) + 9 f(9) + 11f(11)] \cdot 2
\\
\\
\approx & 2 \pi [3(2) + 5 (4) + 7(4) + 9(2) + 11(1)] \cdot 2
\\
\\
\approx & 4 \pi [83]
\\
\\
\approx & 332 \pi \text{ cubic units}

\end{aligned}
\end{equation}
$