Find the volume of the region bounded by the planes $ z=8-y^2, y = 8-x^2, x=0, y=0, z=0$ I figured out the bounds for z:
$z=0$ to $z=8-y^2$
The bounds for y: 
$y=0$  to  $y=8-x^2$
The bounds for x:
$x=0$ to $x=\sqrt{8}$ (Since $8-x^2 = 0$)
So, the volume by using triple integral:
$\int_{0}^{2\sqrt{2}}\int_{0}^{8-x^2}\int_{0}^{8-y^2}dzdydx $
Am I right?

For $x : 0 < x < \sqrt{8-2\sqrt{2}}$, the upper boundary in $y$ is $2\sqrt{2}$ not $8-x^2$ (which is bigger). The $z = 8 - y^2$ cylinder has "chopped off" part of the $y = 8 - x^2$ cylinder.
So the integral should be:
$I = \displaystyle\int\limits_{0}^{\sqrt{8-2\sqrt{2}}}\int\limits_{0}^{2\sqrt{2}}\int\limits_{0}^{8-y^2}dzdydx\ \ +\ \displaystyle\int\limits_{\sqrt{8-2\sqrt{2}}}^{2\sqrt{2}}\int\limits_{0}^{8-x^2}\int\limits_{0}^{8-y^2}dzdydx$
