Volume of A Solid I wonder if some one can clarify the confusion I have over the following problem:
Find the volume of the solid bounded by the parabolic cylinder $z = 1- y^2$, the planes $x = 0, z = 0$ and the plane $x+z = 1$. My struggle is to visualize the solid and find the set up of the triple integral for the volume. 
What is your approach?

Here is your region.

The yellow bit is what is left of the parabolic cylinder.  The green bit is the plane $x+z = 1$
When $z = 0,$ The intersection with $z=1-y^2$ leaves $y^2 = 1$ or $y=1, y=-1$
As for the integral
$\int_{-1}^1 \int_0^{1-y^2}\int_0^{1-z} \;dx \;dz \;dy$
alternatively you could say:
$\int_0^1 \int_{-\sqrt{1-z}}^{\sqrt{1-z}}\int_0^{1-z}\; dx \;dy \;dz$
then substiute u = 1-z
$\int_0^1 \int_{-\sqrt{u}}^{\sqrt{u}}\int_0^{u}\; dx \;dy \;du$
