Single Variable Calculus, Chapter 6, 6.3, Section 6.3, Problem 42

The region bounded by the curves $x = (y - 3)^2, x = 4$ is rotated about the $y = 1$. Find the volume of the resulting solid by any method.

Let us use the shell method together with horizontal strips to evaluate the volume more easy. Notice that these strips have a distance of $y - 1$ from the line $y = 1$. If you rotate this length about $y = 1$, you'll get a circumference of $C = 2 \pi (y - 1)$. Also, the height of the strips resembles the height of the cylinder as $H = x_{\text{right}} - x_{\text{left}} = 4 - (y - 3)^2$. Thus, the volume is..

$\displaystyle V = \int^b_a C(y) H(y) dy$

The values of the upper and lower limits can be determined by simply getting their points of intersection.



$
\begin{equation}
\begin{aligned}

& (y - 3)^2 = 4
\\
& y - 3 = \pm 2
\\
& y = \pm 2 + 3
\\
& \text{We have,}
\\
& y = 1 \text{ and } y = 5


\end{aligned}
\end{equation}
$



Therefore, we have..


$
\begin{equation}
\begin{aligned}

V =& \int^5_1 2 \pi (y - 1) (4 - (y - 3)^2) dy
\\
\\
V =& 2 \pi \int^5_1 (y - 1)(4 - y^2 + 6y - 9) dy
\\
\\
V =& 2 \pi \int^5_1 (y - 1)(-y^2 + 6y - 5) dy
\\
\\
V =& 2 \pi \int^5_1 [-y^3 + 6y^2 - 5y + y^2 - 6y + 5] dy
\\
\\
V =& 2 \pi \int^5_1 [-y^3 + 7y^2 - 11y + 5] dy
\\
\\
V =& 2 \pi \left[ \frac{-y^4}{4} + \frac{7y^3}{3} - \frac{11y^3}{2} + 5y \right]^5_1
\\
\\
V =& \frac{128 \pi}{3} \text{ cubic units}

\end{aligned}
\end{equation}
$