Finding the bounds on a triple integral Problem: Find the volume enclosed by the cone $$x^2 + y^2 = z^2$$ and the plane $$2z - y -2 = 0$$
So I know that I need to do a triple integral over this region, and the integrand will be 1. 
My problem is with finding the bounds for the integral. I set the $z$s equal to each other and found the intersection is $x^2 + \frac{3}{4}(y-\frac{4}{9})^2 = \frac{13}{9}$ (although I may have made a mistake here.)
I believe that the bound for $z$ is from $\sqrt{x^2 + y^2}$ to $\frac{y+2}{2}$?
Now I am not sure what to do. How can I find the bounds for $x$ and $y$? Also, I think I should do a change of variable so that the plane lies flat across the cone instead of slanted, to make the upper bound for $z$ constant. Is this a good idea? How could I do it? I am really lost with this problem.

The intersection is indeed an ellipse, but you seem to have made some algebra mistakes:
\begin{align*}
x^2 + y^2 &= (\tfrac{y + 2}{2})^2 \\
x^2 + y^2 &= \tfrac{1}{4}y^2 + y + 1 \\
x^2 + \tfrac{3}{4}(y^2 - \tfrac{4}{3}y) &= 1 \\
x^2 + \tfrac{3}{4}(y^2 - \tfrac{4}{3}y + \tfrac{4}{9} - \tfrac{4}{9}) &= 1 \\
x^2 + \tfrac{3}{4}(y^2 - \tfrac{4}{3}y + \tfrac{4}{9}) - \tfrac{1}{3} &= 1 \\
x^2 + \tfrac{3}{4}(y - \tfrac{2}{3})^2 &= \tfrac{4}{3} \\
\end{align*}
Using the shadow method, notice that the above ellipse is the desired shadow that we want to integrate our $x$ and $y$ over. Solving for $x$ (for example) in the above ellipse yields:
$$
x = \pm \sqrt{\tfrac{4}{3} - \tfrac{3}{4}(y - \tfrac{2}{3})^2}
$$
A quick plot tells us that the min/max values of $y$ occur at the $y$-intercepts, where $y = \tfrac{2}{3} \pm \tfrac{4}{3}$.
so our triple integral is:
$$
V = \int_{-2/3}^2 \int_{-\sqrt{\frac{4}{3} - \frac{3}{4}(y - \frac{2}{3})^2}}^{\sqrt{\frac{4}{3} - \frac{3}{4}(y - \frac{2}{3})^2}} \int_\sqrt{x^2 + y^2}^{\frac{y + 2}{2}} \, dz \, dx \, dy
$$
To simplify this calculation consider using polar coordinates.
