Find the volume of the solid generated by revolving the region bounded by the parabola $\frac{-x^2}{4}$ and the line $y=-1$ about the following lines.
$y = -1, y = -2, y = 1$
I am struggling to understand the setup when the line to revolve around is not directly on the x or y axis. I am not looking for anyone to solve the integral, I just need help with setup for $\left[R\left(x\right)\right]$ and $\left[r\left(x\right)\right]$
$V = \int_{-2}^2 \pi\left(\left[R\left(x\right)\right]^2-\left[r\left(x\right)\right]^2\right)dx$
Set up for $y = -1$
Since the first revolution is being revolved around the same axis as the lower boundary. I'm no sure if I can just use Disk Method only, but I know I need to account for the -1 offset.
I would assume the radii is $\left[R\left(x\right)\right] = -1 - \frac{-x^2}{4}$
Set up for $y = -2$
Set up for $y = 1$
Thanks in advance for any support you can give
In the following figure, the coordinate axes have deliberately been omitted so that one can concentrate on the essentials.
The upper green region is rotated about the horizontal axis of revolution $$ y=a $$ which does not pass through the figure. Note that $a$ could be positive or negative.
The figure is bounded between the graphs of
\begin{eqnarray} y&=&f(x)\text{ and}\\ y&=&g(x) \end{eqnarray} with $y=f(x)$ the farthest from the axis of revolution and $y=g(x)$ the closest.
Then the outer radius will always be
$$ R=\vert f(x)-a \vert $$
and the inner radius will always be
$$ r=\vert g(x)-a \vert $$
and the volume of the solid of revolution will be
$$ \pi\int (R^2-r^2)\,dx $$
between appropriate limits of integration.