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?

The issue appears to be in your second expansion. $$\begin{align}\left(\frac{y^2}4-(-2)\right)^2 &= \left(\frac{y^2}4+2\right)^2\\ &= \left(\frac{y^2}4\right)^2+2\left(\frac{y^2}4\right)(2)+(2)^2\\ &= \frac{y^4}{16}+y^2+4.\end{align}$$
Can you get the rest of the way from there?
