Compute the volume of the solid of revolution generated by a region bounded by inequalities

Define a region to be the points $(x,y)$ such that

$0 \leq x \leq 2, \qquad \frac{1}{4} x^2 \leq y \leq 1.$

Sketch this region, and compute the volume of the solid of revolution under the following revolutions:

First, the sketch of the region is:

Rendered by QuickLaTeX.com

We compute the volume from revolving about the $x$ -axis as follows,
$\begin{align*} V &= \pi \int_0^2 \left( 1 - \left( \frac{x^2}{4} \right)^2 \right) \, dx \\ &= 2 \pi - \frac{\pi}{16} \int_0^2 x^4 \, dx \\ &= 2 \pi - \frac{2 \pi}{5} \\ &= \frac{8 \pi}{5}. \end{align*}$
First, we find an equation for $y = \frac{1}{4} x^2$ in terms of $x$ .
$y = \frac{1}{4} x^2 \quad \implies \quad |x| = 2 \sqrt{y}.$

Then, we compute the volume as follows from revolving about the $y$ -axis as follows,

$\begin{align*} V &= \pi \int_0^1 (2 \sqrt{y})^2 \, dy \\ &= \pi \int_0^1 4y \, dy \\ &= 2 \pi. \end{align*}$
We compute the volume as follows from revolving about the vertical line $x=2$ as follows,
$\begin{align*} V &= \pi \int_0^1 4 \, dy - \pi \int_0^1 (2 - 2 \sqrt{y})^2 \, dy \\ &= 4 \pi - \pi \int_0^1 (4 - 8 \sqrt{y} + 4 y) \, dy \\ &= 4 \pi - \pi \left( 4 - \frac{16}{3} + 2 \right) \\ &= 4 \pi - \frac{2 \pi}{3} \\ &= \frac{10 \pi}{3}. \end{align*}$
We compute the volume as follows from revolving about the horizontal line $y = 1$ as follows,
$\begin{align*} V &= \pi \int_0^2 \left( \frac{1}{4} x^2 - 1 \right)^2 \, dx \\ &= \pi \int_0^2 \left( \frac{x^4}{16} - \frac{x^2}{2} + 1 \right) \, dx \\ &= \pi \left( \frac{2}{5} - \frac{4}{3} + 2 \right) \\ &= \frac{16 \pi}{15}. \end{align*}$

Stumbling Robot