Triple integral - problem with cylindric coordinates for volume between cylinder, paraboloid and plane I'm having problems with this exercise.
The volume is bound by:
$$x^2 + y^2 = 2ax$$
$$x^2 + y^2 = 2az$$
$$z=0$$
So, it seemed natural to me to introduce cylindric coordinates.
$x=r\cos\phi$
$y=r\sin\phi$
$z=z$
The first equation then transforms to $$r=2a\cos\phi$$
And the second transforms to $$r^2=2az$$
Now I am completely stumped on what to do, because I have two equations and three variables.I have only deduced the range for the angle, which is $[-\frac{\pi}{2}, \frac{\pi}{2}]$, and it is because $\cos\phi$ has to be $\ge 0$
But where do I go from here? If I try to get the boundaries for $z$, I end up with:
$$z \in [0, \frac{r^2}{2a}]$$, but I can't express boundaries for $r$ because it depends on $\phi$!
What should I do?
 A: You have already identified the boundary of $r$. It is,
$0 \leq r \leq 2a\cos\phi$ and you rightly said that $-\frac{\pi}{2} \leq \phi \leq \frac{\pi}{2}$.
So integral to found volume of the region is,
$\displaystyle \int_{-\pi/2}^{\pi/2} \int_0^{2a \cos\phi} \int_0^{r^2/(2a)} r \ dz \ dr \ d\phi$
A: The second equation, $x^2+y^2=2az$ defines a circle in the $x,y$-plane with variable radius along $z$, which cuts
a cone with axis along $+z$. The first equation, $x^2+y^2=2ax$ can be brought to a similar format
by completing the square,
$$
x^2-2ax+y^2=0;
$$
$$
(x-a)^2-a^2+y^2=0;
$$
$$
(x-a)^2+y^2=a^2;
$$
which is not related to $z$, so this is a cylinder centered at $(x,y)=(a,0)$ of radius $a$ (i.e., touching the origin
of coordinates).
The sign of $a$ flips some coordinates, so consider $a>0$ for now. Because it only makes sense to define a finite volume
here, we don't compute the volume which is inside the cone and inside the cylinder, but the volume outside
the cone and inside the cylinder, and furthermore assume the portion $z\ge 0$.
For constant $z$ we look at the intersection of 2 circles, the first circle of the first equation centered at $(x,y)=(a,0)$,
the second circle of the second equation centered at $(x,y)=(0,0)$. The symmetry about flipping $y\leftrightarrow -y$
introduces a factor of 2.
$z$ becomes exhausted once the two circles have only one point in common, i.e., when the radius of the second circle
has grown to $2a$, the diameter of the first circle; $r^2=2az$ with $r=2a$ leads to $0\le z\le 2a$. The Jacobian
is $r$ as usual.
The limits of $r$ are set by staying outside second circle $r=\sqrt{2az}$ and inside the first circle $r=2a\cos\phi$. The upper limit of $\phi$ is set by the intersection of the two circles at some common $r$,
$2a\cos\phi=\sqrt{2az}$:
$$
V=\int_0^{2a} dz \int_{2a\cos\phi> \sqrt{2az}}d\phi \int_{\sqrt{2az}}^{2a\cos\phi} rdr
$$
$$
=2\int_0^{2a} dz \int_{\phi=0}^{\arccos\sqrt{z/(2a)}}d\phi \int_{\sqrt{2az}}^{2a\cos\phi} rdr
$$
with $\int rdr=r^2/2$,
$$
V=\int_0^{2a} dz \int_0^{\arccos\sqrt{z/(2a)}}d\phi [4a^2\cos^2\phi-2az].
$$
With $\int\cos^2\phi d\phi = (\phi+\cos\phi\sin\phi)/2$ and $\int d\phi=\phi$:
$$
V=\int_0^{2a} dz [4a^2\frac12(\phi+\cos\phi\sin\phi)-2az \phi]\mid_{\phi=0}^{\arccos\sqrt{z/(2a)}}
$$
$$
=\int_0^{2a} dz [2a^2(\phi+\cos\phi\sin\phi)-2az \phi]\mid_{\phi=0}^{\arccos\sqrt{z/(2a)}}
$$
$$
=2a\int_0^{2a} dz [a(\phi+\cos\phi\sin\phi)-z \phi]\mid_{\phi=0}^{\arccos\sqrt{z/(2a)}}
$$
$$
=2a\int_0^{2a} dz [(a-z)\phi+a\cos\phi\sin\phi)]\mid_{\phi=0}^{\arccos\sqrt{z/(2a)}}
$$
$$
=2a\int_0^{2a} dz [(a-z)\arccos\sqrt{z/(2a)}+a\sqrt{z/(2a)}\sqrt{1-z/(2a)}]
$$
with $u=z/(2a)$
$$
V=(2a)^2\int_0^1 du [(a-2au)\arccos\sqrt u +a\sqrt u\sqrt{1-u}]
$$
$$
V=4a^3\int_0^1 du [(1-2u)\arccos\sqrt u +\sqrt u\sqrt{1-u}]
$$
Here
$$
\int \arccos\sqrt udu = u\arccos\sqrt{u} -\sqrt{u}\sqrt{1-u}/2+\frac12\arcsin\sqrt u,
$$
$$
\int 2u\arccos\sqrt udu = u^2\arccos\sqrt{u} -(2u-3)\sqrt u\sqrt{1-u}/8+\frac38\arcsin\sqrt u,
$$
$$
\int \sqrt u\sqrt{1-u}du = \frac14 (1-2(1-u))\sqrt{u}\sqrt{1-u}+\frac18 \arcsin(2u-1),
$$
$$
V=4a^3\frac{3\pi}{16} = \frac34\pi a^3.
$$
