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.
 A: In cylindrical coordinates, those conditions become $r^2=2z$ and $r(\cos\theta+\sin\theta)=z$. But then $\frac12r^2=r(\cos\theta+\sin\theta)$ and therefore $r=2(\cos\theta+\sin\theta)$. So, $\cos\theta+\sin\theta\geqslant0$, and therefore $-\frac\pi4\leqslant\theta\leqslant\frac{3\pi}4$. For each such $\theta$, the possible values of $r$ go from $0$ to $2\sqrt{1+\sin(2\theta)}$ (taken at $\bigl(2\cos^2(\theta)+\sin(2\theta),2\sin^2(\theta)+\sin(2\theta),2+2\sin(2\theta)\bigr)$). So, since we are interested in the region above the paraboloid $z=\frac12(x^2+y^2)$ and below the plane $x+y=z$, take$$\int_{-\pi/4}^{3\pi/4}\int_0^{2\sqrt{1+\sin(2\theta)}}\int_{r^2/2}^{r(\cos\theta+\sin\theta)}rz\,\mathrm dz\,\mathrm dr\,\mathrm d\theta.\tag1$$Then\begin{align}(1)&=\int_{-\pi/4}^{3\pi/4}\int_0^{2\sqrt{1+\sin(2\theta)}}-\frac{r^5}8+\frac{r^3}2\sin(2\theta)+\frac{r^3}2\,\mathrm dr\,\mathrm d\theta\\&=\int_{-\pi/4}^{3\pi/4}\frac23\bigl(\sin(\theta)+\cos(\theta)\bigr)^6\,\mathrm d\theta\\&=\frac{5\pi}3.\end{align}
A: 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 $Y = [1 - \sqrt2, 1 + \sqrt2]$.
A: Caution: I have very little experience of solving this kind of problem, so the way I have evaluated the integral is probably not to be recommended! However, it only involves integration of polynomials of low degree, and in any case, the question was not how to evaluate the integral, but how to describe the region over which one must integrate, so perhaps these remarks will be of some use, even though I cannot suggest a general method for approaching such problems.
The paraboloid $x^2 + y^2 = 2z$ is the common boundary of two regions of $\mathbb{R}^3,$ and the plane $x + y = z$ is the common boundary of another two regions of $\mathbb{R}^3.$ The easiest way I find to visualise the way in which these four regions intersect, partitioning $\mathbb{R}^3$ into four regions, is to consider the coordinate system in the $(x, y)$ plane with basis vectors $\left(\frac1{\sqrt2}, \frac1{\sqrt2}\right)$ and $\left(-\frac1{\sqrt2}, \frac1{\sqrt2}\right)$:
$$
(u, v) = \left(\frac{x + y}{\sqrt2}, \frac{y - x}{\sqrt2} \right).
$$
In the plane $v = 0,$ the paraboloid $x^2 + y^2 = 2z$ projects onto the parabola $z = \frac12u^2,$ and the plane $x + y = z$ projects onto the line $z = \sqrt2u.$ It is clear from this projection that the only one of the four regions of intersection that is bounded is:
$$
\left\{ (x, y, z) \colon \frac{x^2 + y^2}2 \leqslant z \leqslant x + y \right\}.
$$
In terms of $(u, v)$ rather than $(x, y),$ this is:
$$
\left\{ (u, v, z) \colon \frac{u^2 + v^2}2 \leqslant z \leqslant \sqrt2u \right\}.
$$
The region of integration projects onto a circle of radius $\sqrt2$ in the $(u, v)$ plane:
$$
\{ (u, v) \colon u^2 - 2\sqrt2u + v^2 \leqslant 0 \} =
\{ (u, v) \colon (u - \sqrt2)^2 + v^2 \leqslant 2 \}.
$$
Let us therefore transform coordinates again, by translating $(u, v)$ to $(t, v)$ where $t = u - \sqrt2.$
After some simplification, the region of integration becomes:
$$
\left\{ (t, v, z) \colon 1 + \sqrt2t + \frac{t^2 + v^2}2 \leqslant z \leqslant 2 + \sqrt2t \right\}.
$$
One final transformation: $(t, v) = (r\cos\theta, r\sin\theta),$ where $0 \leqslant r \leqslant \sqrt2$ and $-\pi < \theta \leqslant \pi.$ (I hope it doesn't matter that I haven't expressed this very precisely.) The Jacobian is (if I remember the notation for this sort of thing correctly! - please bear with me, I've got about half a century's worth of rust to get through):
$$
\frac{\partial(t, v)}{\partial(r, \theta)} =
\begin{pmatrix}
\cos\theta & -r\sin\theta \\ \sin\theta & r\cos\theta
\end{pmatrix},
$$
with determinant $r,$ so we are finally interested in:
$$
I = \iiint_Ezr\,dE,
$$
where:
$$
E = \left\{ (r, \theta, z) \colon 1 + \sqrt2r\cos\theta + \frac{r^2}2 \leqslant z \leqslant 2 + \sqrt2r\cos\theta \right\}.
$$
Evaluating:
\begin{align*}
I & = \int_0^\sqrt2\int_{-\pi}^\pi\int_{1 + \sqrt2r\cos\theta + \frac{r^2}2}^{2 + \sqrt2r\cos\theta}zr\,dz\,d\theta\,dr \\
& = \int_0^\sqrt2r\int_{-\pi}^\pi\left[\frac{z^2}2\right]_{1 + \sqrt2r\cos\theta + \frac{r^2}2}^{2 + \sqrt2r\cos\theta}\,d\theta\,dr \\
& = \int_0^\sqrt2r\int_{-\pi}^\pi\frac12\left(1 - \frac{r^2}2\right)\left(3 + 2\sqrt2r\cos\theta + \frac{r^2}2\right)\,d\theta\,dr \\
& = \frac12\int_0^\sqrt2r\left(1 - \frac{r^2}2\right)\left[3\theta + 2\sqrt2r\sin\theta + \frac{r^2\theta}2\right]_{-\pi}^{\pi}\,dr \\
& = \pi\int_0^\sqrt2r\left(1 - \frac{r^2}2\right)\left(3 + \frac{r^2}2\right)\,dr \\
& = \pi\int_0^\sqrt2\left(3r - r^3 - \frac{r^5}4\right)\,dr \\
& = \pi\left[\frac{3r^2}2 - \frac{r^4}4 - \frac{r^6}{24}\right]_0^{\sqrt2} \\
& = \pi\left(3 - 1 - \frac13\right) \\
& = \frac{5\pi}3.
\end{align*}
