Finding the volume when the region between $y=x^2$ and $y=4-x^2$ is rotated about the $x$-axis.

So I want to find the volume when the region between $y=x^2$ and $y=4-x^2$ is rotated about the $x$-axis.

So I start by finding the roots where they meet, so I find :

$$\int_{ -\sqrt{2} }^{\sqrt{2}} \pi (4-x^2)^2 \, dx$$

But this got me the wrong answer $75.825$. Is that right or is the book wrong with $94.7$?

A typical cross-section is a washer, not a disk. So we must subtract the inner radius: $$V = \pi\int_{ -\sqrt{2} }^{\sqrt{2}} [(4-x^2)^2 - (x^2)^2] \, dx = \frac{64\sqrt 2}{3}\pi = 94.7815\ldots$$