Trouble with finding volume of a solid for $y=2-2x^2, y=0$ Here is my problem:

$$y=2-2x^2, y=0$$

I first found points of intersection by solving $y=2-2x^2, y=0$ and found that my points are $-1, 1$

Then I found the area by doing the following:

$$A= \pi r^2$$
$$A=\pi(2-2x^2)^2$$
$$A=64\pi(x^4+1)$$

I then anti-differentiated:

$$64\pi\int_{-1}^{1}(x^4+1)dx$$
$$64\pi(\frac{x^5}{5}+x)|_{-1}^{1}$$

Which I found to be:

$$\frac{768\pi}{5}$$

I exceeded my maximum amount of submission for this problem, but I was wondering if this attempt at the answer is correct or if I made another mistake?
I assume you are rotating about the $x$-axis. Then the volume of the solid of revolution is
$$\int_{-1}^1\pi(2-2x^2)^2\,dx.$$
Expand. We want 
$$4\pi\int_{-1}^1 (1-2x^2+x^4)\,dx.$$
By symmetry, this is 
$$8\pi\int_0^1 (1-2x^2+x^4)\,dx.$$
An antiderivative is $x-\frac{2}{3}x^3+\frac{1}{5}x^5$.
When we do the calculation, we end up with $\frac{64\pi}{15}$.

Remark: There were errors of algebra in expanding $(2-2x^2)^2$. I do not know where the $64$ came from. The other error came from forgetting the "middle term" when you square $a+b$.
@T. Bongers

$$
\infty
\quad\mbox{since}\quad
\int_{-\infty}^{\infty}{\rm d}z \to \infty
$$

$\mathbf{\mbox{ADDENDUM}}$
$$
{\displaystyle{\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}}
 \atop
 y &lt; 2 - 2x^{2}\,,\ y&gt;0}\!\!\!\!
\displaystyle{{\rm d}x\,{\rm d}y\,{\rm d}z \to \infty\atop }
$$