Boundaries of a triple integral between $x^2+y^2=2z$ and $x+y=z$ I am calculating the integral $\iiint_V z\,dV$ between $x^2+y^2=2z$ and $x+y=z$, but I am not sure how to find boundaries for this case. I tried with cylindrical coordinates, but I was unable to solve the problem.
I would also like to know if there is any general procedure to solve this kind of problem.

Here I present a way to solve the integral. Maybe there's an easier way to integrate, because now you get some troubling terms in the last integral.
Since we have a function of $z$, we would like to integrate over $z$ first. The boundaries of $z$ evaluates to
$$\frac{x^2 + y^2}2 \le z \le x + y$$
since that's the only bounded volume. Therefore we get
$$\int_V z \ dV = \int_{\pi(V)} \left[\int_{\frac{x^2 + y^2}2}^{x + y} z \ dz \right] dA = \int_{\pi(V)} f(x, y) \ dA,$$
where $\pi(V)$ is the projection of $V$ on the $xy$-plane. The boundaries for $\pi(V)$ is where $\frac{x^2 + y^2}2 = x + y$. Solving for $x$ we get $x = 1 \pm \sqrt{1 + 2 y - y^2}$. Therefore $x$ in $\pi(V)$ is therefore contained by
$$1 - \sqrt{1 + 2 y - y^2} \le x \le 1 + \sqrt{1 + 2 y - y^2}.$$
and we get
$$\int_{\pi(V)} f(x, y) \ dA = \int_{Y} \left[\int_{1 - \sqrt{1 + 2 y - y^2}}^{1 + \sqrt{1 + 2 y - y^2}}  f(x, y) \ dx\right]dy$$
where $Y$ is the projection of $\pi(V)$ on the $y$-axis. Now the last step is to solve where $Y$ is located. Well, solve where $1 - \sqrt{1 + 2 y - y^2} = 1 + \sqrt{1 + 2 y - y^2}$. Here $y$ is bounded by
$$1 - \sqrt2 \le y \le 1 + \sqrt2,$$
so $Y = [1 - \sqrt2, 1 + \sqrt2]$.
