Single Variable Calculus, Chapter 6, 6.3, Section 6.3, Problem 40

The region bounded by the curves $x = 1 - y^4, x = 0$ is rotated about the $x = 2$. Find the volume of the resulting solid by any method.

Let us use the water method together with horizontal strips to evaluate the volume more easy. Notice that the region rotated is about $x = 2$ has a cross section with outer radius $2$ and inner radius $2 - (1 - y^4)$. Thus, the cross sectional area is completed by subtracting the area of the outer circle from the inner circle. $A_{\text{washer}} = A_{\text{outer}} - A_{\text{inner}} = \pi(2)^2 - \pi (2 - (1 - y^4))^2$. Hence, the volume is..

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

The values of the upper and lower limits can be determined by the points of intersection of the curves.



$
\begin{equation}
\begin{aligned}

1 - y^4 =& 0
\\
y^4 =& 1
\\
y =& \pm 1


\end{aligned}
\end{equation}
$



Therefore, we have..


$
\begin{equation}
\begin{aligned}

V =& \int^1_{-1} \left[ \pi (2)^2 - \pi (2 - (1 - y^4))^2 \right] dy
\\
\\
V =& \pi \int^1_{-1} \left[4 - (1 + y^4)^2\right] dy
\\
\\
V =& \pi \int^1_{-1} \left[4 - 1 - 2y^4 - y^8\right] dy
\\
\\
V =& \pi \int^1_{-1} \left[3 - 2y^4 - y^8\right] dy
\\
\\
V =& \pi \left[ 3y - \frac{2y^5}{5} - \frac{y^9}{9} \right]^1_{-1}
\\
\\
V =& \frac{224 \pi}{45} \text{ cubic units}

\end{aligned}
\end{equation}
$