So I was given the following prompt with the graph
:
"Let $R$ be the region bounded above by the graph of $y=1-x^2$ and below by the graph of $y=x^2-1$, for $-1\le x\le1$, as shaded in the figure above. What is the volume of the solid generated when the region $R$ is revolved about the horizontal line $y=3$?"
I understand that the bounds of the integral that would solve for the volume would be from $-1$ to $1$, but I'm a bit confused about what the equation for the volume might look like here. I was thinking it might look something like the following: $\int((1-x^2+3)^2-(x^2-1+3)^2)$, but I'm not too sure. Any help would be appreciated!
Best Answer
Instead of rotating around the line $y=3$, we translate everything $3$ units down and solve $$\pi\int_{-1}^1 \left[\left(x^2-4\right)^2-\left(-x^2-2\right)^2\right]\,dx=16\pi$$
alternative Pappus' theorem
Compute di area $A$ between the "blue" parabolas, $A=\frac83$ and multiply by the circumference $C=6\pi$ described by the centroid $O$ of the region $A$
$$V=(6\pi)\cdot \frac83=16\pi$$
$$...$$