Volume using disk method $y = 2\sqrt{x} \quad y=x$ $y = 2\sqrt{x}$
$y=x$
about $x=-2$
I know that $y=x$ is on the outside and they meet at 4.
I need these in terms of y since I rotate about y.
$x = y$
$x = \frac{y^2}{4}$
$$\pi \int_0^4 (y - (-2))^2 - \left(\frac{y^2}{4} - (-2)\right)^2 dy$$
$$\pi \int_0^4 (y + 2)^2 - \left(\frac{y^2}{4} + 2\right)^2 dy$$
$$\pi \int_0^4 y^2 + 4y +4 - \left(\frac{y^4}{16} +4y+4\right) dy$$
$$\pi \int_0^4 y^2 - \frac{y^4}{16} dy$$
$$\pi \left(\frac{y^3}{3} - \frac{y^5}{16*5}\right)$$
Gives me $\dfrac{128\pi}{15}$
Where did I go wrong?

As pointed out in the other answer, your expansion of the subtracted term is off:
$$\pi \cdot \int_0^4 (y + 2)^2 -\left (\frac{y^2}{4} + 2\right)^2 dy$$
So far, so good. Your set up looks fantastic.
But note that expanding the subtracted square, we get:
$$\pi \cdot \int_0^4 y^2 + 4y +4 - \left(\frac{y^4}{16} +4\cdot\frac{y^2}{4} +4\right) dy = \pi \cdot \int_0^4 y^2 + 4y +4 - \frac{y^4}{16} - {y^2} -4 \,dy $$
$$\pi \cdot \int_0^4 4y - \frac{y^4}{16} dy$$
NOW integrating gives us $$\pi\left(\frac {4y^2}{2} - \frac{y^5}{5\cdot 16}\right)\Big|_0^4 = \pi\left(32 - \frac{64}{5}\right) = \left(\frac{96}{5}\right)\pi $$
