Finding the volume of an oblique cylinder using triple integrals I want to find the volume of an oblique cylinder using triple integrals. The triple integral part is important because I plan to use the limits of integration for computing the moment of inertia tensor. So only answers that find the volume using a triple integral in either Cartesian, spherical, or cylindrical coordinates will be useful to me.
How can I do this?
 A: I am answering only question in title. Let me suggest change of variables by formulas:
$$
\begin{cases}
x-C\cdot z=x_1 \\
y=y_1 \\
z=z_1
\end{cases} \Leftrightarrow \begin{cases}
x=x_1+C\cdot z_1 \\
y=y_1 \\
z=z_1
\end{cases}
$$
Where $C=\tan \alpha$. Jacobian $=1$.  Equations
\begin{align}
(x-z\tan(\alpha))^2 + y^2 \leq R^2 \\
|z| \leq \frac{h}{2} 
\end{align}
become
\begin{align}
x_1^2 + y_1^2 \leq R^2 \\
|z_1| \leq \frac{h}{2} 
\end{align}
So volume is
$$4\int\limits_{0}^{R}\int\limits_{0}^{\sqrt{R^2-x_1^2}}\int\limits_{-\frac{h}{2}}^{\frac{h}{2}}dx_1dy_1dz_1$$
If you insist, of course, here can be used now cylindrical coordinates. I am aware, that it is turning back oblique cylinder to usual one, but your title is ordered only calculating volume using 3-ple integrals.
A: An oblique cylinder looks like this:

where $\alpha$ is the angle from the vertical
The equation for a solid oblique cylinder sheared in the $xz$ plane is
$$\begin{align}
(x-z\tan(\alpha))^2 + y^2 \leq R^2 \\
|z| \leq \frac{h}{2} 
\end{align}
$$
In this answer, I show my work for how I got this:
https://math.stackexchange.com/a/3774822/197705
I can find the volume for a regular cylinder as follows using cylindrical coordinates:
$$\int_{-\frac{h}{2}}^{\frac{h}{2}}\int_{0}^{2\pi}\int_{0}^{R} r dr d\theta dz$$
For an oblique cylinder,  I reasoned since the oblique cylinder still runs from $-\frac{h}{2}$ to $\frac{h}{2}$, I assume then $z \in [-\frac{h}{2},\frac{h}{2}]$
Similarly, on any given cross-section of the cylinder, I will still want to consider all of the cross section, so $\theta \in [0,2\pi]$.
Thus, it seems like $r$ would be the tricky one. In the original cylinder, I had
$$\begin{align}
x^2 + y^2 &\leq R^2\\
(r\cos(\theta)^2+(r\sin(\theta))^2 &\leq R^2\\
r^2&\leq R^2\\
\end{align}
$$
So it was logical to determine that $r\in [0,R]$.
However, for an oblique cylinder
$$\begin{align}
(x-z\tan(\alpha))^2 + y^2 &\leq R^2\\
(r\cos(\theta)-z\tan(\alpha))^2+(r\sin(\theta))^2 &\leq R^2\\
r^2 \cos^2(\theta)-2rz\cos(\theta)\tan(\alpha)+z^2\tan^2(\alpha)+r^2\sin^2(\theta)\leq R^2\\
r^2 -2rz\cos(\theta)\tan(\alpha)+z^2\tan^2(\alpha)\leq R^2
\end{align}
$$
I tried then setting up a quadratic
$$r^2 + (-2z\cos(\theta)\tan(\alpha))r+(z^2\tan^2(\alpha)-R^2)=0$$
and then treating
$$\begin{align}
A&=1\\
B&=-2z\cos(\theta)\tan(\alpha)\\
C&=z^2\tan^2(\alpha)-R^2
\end{align}
$$
I then solved using the quadratic formula
$$r = \frac{-B\pm \sqrt{B^2-4AC}}{2A}$$
which has solutions
$$z\cos(\theta)\tan(\alpha)+\sqrt{R^2-z^2\sin^2(\theta)\tan^2(\theta)}$$
$$z\cos(\theta)\tan(\alpha)-\sqrt{R^2-z^2\sin^2(\theta)\tan^2(\theta)}$$
If I assume $z=3$, $\theta = \frac{\pi}{4}$,$\alpha = \frac{\pi}{45} = 4^{\circ}$, $R=2$, then I get negative values with the second answer, which can't be since the radius can't be negative. So the first answer must be correct then.
Then my limits of integration are
$$\int_{-\frac{h}{2}}^{\frac{h}{2}}\int_{0}^{2\pi}\int_{0}^{q} r dr d\theta dz$$
where
$$q = z\cos(\theta)\tan(\alpha)+\sqrt{R^2-z^2\sin^2(\theta)\tan^2(\theta)}$$
If you integrate this, the volume equals the volume of a regular cylinder, which is what we would expect since their volumes are the same.
