I'm not sure if I'm doing this correctly, it's been a while since I've done any calculus.
I drew the graph, and since I'm not considering the part above $y=-1$ I broke it up into two integrals, one from $x=0$ to $x=1$ with $r=1$ and one from $x=1$ to $x=\sqrt2$ with $r=x^2-2$.
So I have:
$V=\int_0^1 2\pi(1)^2dx + \int_1^{\sqrt2}2\pi(x^2-2)^2dx$
Which I solved to get $V=7.2277$.
Am I setting up my integrals properly?
$\endgroup$2 Answers
$\begingroup$The definition for the volume of a solid of revolution using the circular-ring method given $f(x)$ is the outer-radius and $g(x)$ is the inner-radius for all $x$ in $[a,b]$ is
$$ V = \pi \int_a^b ([f(x)]^2 - [g(x)]^2) \, \textrm{d}x \\ $$
For $[0,1]$, you have a correct integral for what is essentially a right circular cylinder; $\pi r^2 h$ could've worked too.
For $[1,\sqrt{2}]$, you have an incorrect integral for a couple reasons. First, you need to adjust $x^2 - 2$ for the offset of the axis of revolution. Second, you're assuming that $x^2 - 2$ is the outer-radius when it's actually the inner-radius; the outer-radius is $1$. So, you should change the second part of your equation to
$$ 2 \pi \int_1^{\sqrt 2} (1^2 - (1 + x^2 - 2)^2) \, \textrm{d}x \\ $$
You should get a total of $V = 8.0901$.
$\endgroup$ 3 $\begingroup$Splitting the integral is a good idea, but you have to take in account the axis of revolution.
It may be easier to see if you graph the area that you want revolve. What you get is something like the following:
In purple is the revolution axis.
On the one hand, there is no reason to just consider $x \ge 0$, so if you do that you have to multiply by $2$ at the end.
On the other hand, since the revolution axis is $y=-1$ and not $y=0$, in the region between $x=1$ and $x=\sqrt2$ you have 2 radius to consider. So you have to calculate the volume with the outer and subtract the empty volume with the inner radius. Please tell me if you need further explanation.
$\endgroup$
