Volume integral over a bounded region Class is over now and I am studying for my final and I have a problem with this question on our review sheet. If anyone can help I would appreciate it.
Question: Find the volume of the region in space bounded by $z=1-x^2-y^2$ and on the sides by the planes x=0, y=0 and x+y=1 and z=0.  
I first found the boundaries: $0\le z \le 1-x^2-y^2$ , $0\le y \le 1-x$, and $0\le x \le 1$.
Originally I just then computed the integral $\int\int\int dzdydx$ which gave me the answer of 1/3 however I feel as though I should be using polar coordinates for this question but when I tried that I could not seem to figure out anything that made sense. I might be screwing up the integral when I switch over to polar coordinates. Worst part is our teacher hasn't posted any solutions so I am really shooting in the dark. Again if anyone can help I would appreciate it.

You did the limits right in Cartesian coordinates, but we can, as you pointed, do the triple integral by using the Cylindrical coordinates: $$0\leq z\leq 1-x^2-y^2\longrightarrow 0\leq z\leq 1-r^2 $$ since $x^2+y^2=r^2$. Since $$x+y=1\longrightarrow r(\sin\theta+\cos\theta)=1\\\ z=0\longrightarrow1=x^2+y^2\longrightarrow r=1$$ so the ranges of changing $r$ would be as: $$r|_{0}^{\frac{1}{(\sin\theta+\cos\theta)}}$$ For $\theta$, we have $$\theta|_{0}^{\pi/2}$$

