So let's say that we have I can only describe as a "truncated elliptic cone" (TEC) as on the picture:
Let's further say that this TEC is of height $h$, and let's say that it is centered at the origin, such that the origin is a midpoint of the TEC's height and the midpoint of the central cross-section. It goes without saying that we've picked our coordinate system such that the the $a$'s in the picture extend along the $x$-axis, the $b$'s extend along the $y$-axis, and the height along the $z$-axis. Let's further say that we have another point somewhere in space at $(x_0, y_0, z_0)$, and we wish to evaluate the integral
$$
\int \int \int_{\text{TEC}} ((x-x_0)^2+(y-y_0)^2+(z-z_0)^2) dx dy dz .
$$
If $a_1=b_1=r_1$ and $a_2=b_2=r_2$ (i.e. we're just looking at a truncated cone), the task becomes rather easy. Changing coordinates to cylindrical ones, the equation becomes
$$
\int^{h/2}_{-h/2} \int^{2 \pi}_{0} \int^{\frac{1}{h} (r_2-r_1 + \frac{1}{2}(r_2-r_1))}_0 ( r^2 + r_0^2 – 2 r r_0 \cos[\phi – \phi_0] + (z – z_0)^2) r dr d \phi dz
$$
which good old Lady Mathematica tells us is
$$
\frac{1}{60} h \pi [20 r_0^2 (r_1^2 + r_1 r_2 + r_2^2) +
h^2 (2 r_1^2 + r_1 r_2 + 2 r_2^2) +
6 (r_1^4 + r_1^3 r_2 + r_1^2 r_2^2 + r_1 r_2^3 + r_2^4) +
10 h (r_1 – r2) (r_1 + r_2) z_0 + 20 (r_1^2 + r_1 r_2 + r_2^2) z_0^2] .
$$
A long and cumbersome expression, perhaps, but impeccable for my purposes.
However, when I try to solve the general case using cylindrical coordinates, I end up with integrands that are so complicated that not even Lady Mathematica can integrate them.
As such, I am asking you if anyone of you perhaps can come up with some neat coordinate substitution or other trick that would allow me to evaluate this integral?
Looking forward to your responses!
Best Answer
Let the functions:
$$ a(z) := \underbrace{\frac{a_1+a_2}{2}}_{\alpha_1} - \underbrace{\frac{a_1-a_2}{2}}_{\alpha_2}\frac{2\,z}{h}, \quad \quad b(z) := \underbrace{\frac{b_1+b_2}{2}}_{\beta_1} - \underbrace{\frac{b_1-b_2}{2}}_{\beta_2}\frac{2\,z}{h} $$
the "generalized" truncated elliptic cone is:
$$ \Omega := \left\{(x,y,z) \in \mathbb{R}^3 : \frac{x^2}{a(z)^2} + \frac{y^2}{b(z)^2} \le 1, \, -\frac{h}{2} \le z \le \frac{h}{2}\right\} $$
where the "classic" truncated elliptic cone is obtained if $a_2/a_1 = b_2/b_1$, i.e. $\alpha_2/\alpha_1 = \beta_2/\beta_1$.
So, through a transformation of coordinates of the type:
$$ \begin{cases} \color{red}{x} = a(\color{blue}{z})\,\rho\cos\theta \\ \color{green}{y} = b(\color{blue}{z})\,\rho\sin\theta \\ \color{blue}{z} = \frac{h}{2}\,t \\ \end{cases} \quad \quad \text{with} \; (\rho,\theta,t) \in [0,1] \times [0,2\pi) \times \left[-1,1\right] $$
the following triple integral is easy to calculate:
$$ \begin{aligned} I & := \iiint\limits_{\Omega} \left((x-x_0)^2+(y-y_0)^2+(z-z_0)^2\right)\text{d}x\,\text{d}y\,\text{d}z \\ & = \int_{-1}^1 \text{d}t \int_0^{2\pi} \text{d}\theta \int_0^1 \left((\color{red}{x}-x_0)^2+(\color{green}{y}-y_0)^2+(\color{blue}{z}-z_0)^2\right)a(\color{blue}{z})\,b(\color{blue}{z})\,\frac{h}{2}\,\rho\,\text{d}\rho \\ & = \boxed{\frac{\pi\,h}{60}\left[15\alpha_1\beta_1\left(\alpha_1^2+\beta_1^2\right)+3\alpha_2\beta_2\left(\alpha_2^2+\beta_2^2\right)+5\left(\alpha_1\beta_2+\alpha_2\beta_1\right)\left(3\alpha_1\alpha_2+3\beta_1\beta_2\;\\+4\,h\,z_0\right)+\left(5\alpha_1\beta_1+3\alpha_2\beta_2\right)h^2+20\left(3\alpha_1\beta_1+\alpha_2\beta_2\right)\left(x_0^2+y_0^2+z_0^2\right)\right].} \\ \end{aligned} $$