Close-form for triple integral $ \int_0^c \int_0^b \int_0^a \sqrt{x^2+y^2+z^2} dx dy dz$

Question

I am able to work out the double integral

$$\int_0^b \int_0^a \sqrt{x^2+y^2} dx dy $$ with brute-force (i.e. integrating $x$, then $y$) to arrive at the close-form result

$$\frac13ab\sqrt{a^2+b^2} +\frac16a^3\sinh^{-1}\frac ba +\frac16 b^3 \sinh^{-1}\frac ab$$

which has the expected parity between $a$ and $b$. However, it gets unwieldy to tackle the triple-integral extension $$\int_0^c \int_0^b \int_0^a \sqrt{x^2+y^2+z^2} dx dy dz$$ this way and I am unable to slug it out. Does anyone know the corresponding close-form expression for the triple version?

Answer 1

You can refer to J.M. Borwein's box integral

\begin{gather*} \int_0^c\int_0^b\int_0^a\dfrac{1}{\sqrt{x^2+y^2+z^2}}{\rm\,d}x{\rm\,d}y{\rm\,d}z\\ = \begin{array}{r} ab\ln\left(\frac{c}{\sqrt{a^2+b^2}}+\sqrt{1+\frac{c^2}{a^2+b^2}}\right)\\ +ac\ln\left(\frac{b}{\sqrt{a^2+c^2}}+\sqrt{1+\frac{b^2}{a^2+c^2}}\right) +bc\ln\left(\frac{a}{\sqrt{b^2+c^2}}+\sqrt{1+\frac{a^2}{b^2+c^2}}\right)\\ -\frac{a^2}{2}\arctan\left(\frac{bc}{a\sqrt{a^2+b^2+c^2}}\right)-\frac{b^2}{2}\arctan\left(\frac{ac}{b\sqrt{a^2+b^2+c^2}}\right)-\frac{c^2}{2}\arctan\left(\frac{ab}{c\sqrt{a^2+b^2+c^2}}\right) \end{array}\\ \\ \\ \int_0^c\int_0^b\int_0^a\sqrt{x^{\overset{\,}{2}}+y^2+z^2}{\rm\,d}x{\rm\,d}y{\rm\,d}z\\ = \begin{array}{r} \frac{abc}{4}\sqrt{a^2+b^2+c^2}+\frac{ab\left(a^2+b^2\right)}{6}\ln\left(\frac{c}{\sqrt{a^2+b^2}}+\sqrt{1+\frac{c^2}{a^2+b^2}}\right)\\ +\frac{ac\left(a^2+c^2\right)}{6}\ln\left(\frac{b}{\sqrt{a^2+c^2}}+\sqrt{1+\frac{b^2}{a^2+c^2}}\right) +\frac{bc\left(b^2+c^2\right)}{6}\ln\left(\frac{a}{\sqrt{b^2+c^2}}+\sqrt{1+\frac{a^2}{b^2+c^2}}\right)\\ -\frac{a^4}{12}\arctan\left(\frac{bc}{a\sqrt{a^2+b^2+c^2}}\right)-\frac{b^4}{12}\arctan\left(\frac{ac}{b\sqrt{a^2+b^2+c^2}}\right)-\frac{c^4}{12}\arctan\left(\frac{ab}{c\sqrt{a^2+b^2+c^2}}\right) \end{array}\\ \\ \\ \int_0^c\int_0^b\int_0^a\bigg(x^2+y^2+z^2\bigg)^{\frac{3}{2}}{\rm\,d}x{\rm\,d}y{\rm\,d}z\\ = \begin{array}{r} \frac{2abc\left(a^2+b^2+c^2\right)}{15}\sqrt{a^2+b^2+c^2}+\frac{ab\left(9a^4+10a^2b^2+9b^4\right)}{120}\ln\left(\frac{c}{\sqrt{a^2+b^2}}+\sqrt{1+\frac{c^2}{a^2+b^2}}\right)\\ +\frac{ac\left(9a^4+10a^2c^2+9c^4\right)}{120}\ln\left(\frac{b}{\sqrt{a^2+c^2}}+\sqrt{1+\frac{b^2}{a^2+c^2}}\right)\\ +\frac{bc\left(9b^4+10b^2c^2+9c^4\right)}{120}\ln\left(\frac{a}{\sqrt{b^2+c^2}}+\sqrt{1+\frac{a^2}{b^2+c^2}}\right)\\ -\frac{a^6}{30}\arctan\left(\frac{bc}{a\sqrt{a^2+b^2+c^2}}\right)-\frac{b^6}{30}\arctan\left(\frac{ac}{b\sqrt{a^2+b^2+c^2}}\right)-\frac{c^6}{30}\arctan\left(\frac{ab}{c\sqrt{a^2+b^2+c^2}}\right) \end{array} \end{gather*}

Answer 2

A start
Spherical coordinates. $(\rho,\theta,\phi)$. They are related to rectangular coordinates $(x,y,z)$ by: \begin{align} x &= \rho\sin\theta\cos\phi\\ y &= \rho\sin\theta\sin\phi\\ z &= \rho\cos\theta \end{align} and in reverse by \begin{align} \rho &= \sqrt{x^2+y^2+z^2}\\ \theta &=\arccos\frac{z}{\sqrt{x^2+y^2+z^2}}\\ \phi &= \arctan\frac{y}{x} \end{align} The box we want is $$ 0 \le x \le a,\\ 0 \le y \le b,\\ 0 \le z \le c. $$ In spherical coorcinates: $$ 0 \le \rho\sin\theta\cos\phi \le a,\\ 0 \le \rho\sin\theta\sin\phi \le b,\\ 0 \le \rho\cos\theta \le c. $$ If we think we will express our integrals using $\rho$ a function of $\theta,\phi$, do this as $$ 0 \le \rho \le a\csc\theta\sec\phi,\\ 0 \le \rho \le b\csc\theta\csc\phi,\\ 0 \le \rho \le c\sec\phi $$ Our triple integral will have three terms, depending on which of the three $a\csc\theta\sec\phi,b\csc\theta\csc\phi,c\sec\phi$ is smallest. That is, dependingon which of the three faces the ray from the origin with angles $\theta,\phi$ intersects.