Check answer for volume by revolution

Can you please tell me if i worked the following out correctly?

$\displaystyle y=x^2$

$\displaystyle y=x+2$

1. Rotate this around the x-axisoutside radius = $\displaystyle x+2$

inside radius = $\displaystyle x^2$

$\displaystyle \pi\int_1^2 (x+2-x^2) = \frac{9\pi}{2}$

(thats suppose to be -1 to 2 but i cant get latex to work)

2. Rotate around $\displaystyle y = 4$outside radius = $\displaystyle 4-x^2$

inside radius = $\displaystyle 4-(x+2)$

$\displaystyle \pi\int_1^2 (4-x^2)-(4-(x+2)= \frac{9\pi}{2}$

(thats suppose to be -1 to 2 but i cant get latex to work)

3. Rotate around x = -2outside radius = $\displaystyle \sqrt{y}+2$

inside radius = $\displaystyle 2-y+2$

$\displaystyle \pi\int_0^4 (\sqrt{y}+2-(2-y+2)= \frac{16\pi}{3}$

Thanks in advance