Equality of triple integrals over unit sphere $ \iiint_{\text{unit ball}} x e^{ax + by + cz} dV$ I have to calculate
$$ \iiint_{\text{unit ball}} x e^{ax + by + cz} \,dV,$$
where by "unit ball" I mean the region $x^2 + y^2 + z^2 \leq 1$.
I know how to calculate this (rotation matrix that takes $(a,b,c)$ to $(0,0,\sqrt{a^2 + b^2 + c^2})$ and then spherical coordinates). The answer gives
$$\frac{4\pi a}{r^5}((3 + r^2)\sinh(r) - 3r \cosh(r)), \,\,\,\,\,r = \sqrt{a^2 + b^2 + c^2}$$
Now this would suggest that we have the following equality:
$$ \iiint_{\text{unit ball}} \frac{x}{a} e^{ax + by + cz} \,dV = \iiint_{\text{unit ball}} \frac{y}{b} e^{ax + by + cz} \,dV = \iiint_{\text{unit ball}} \frac{z}{c} e^{ax + by + cz} \,dV$$
Is it true? How would one quickly prove it if it is? It would be relevant because if one could quickly spot and prove this, then by naming the value of the above integrals $I$, we would have
$$(a^2 + b^2 + c^2)I = \iiint_{\text{unit ball}} (ax+by+cz) e^{ax + by + cz} \,dV$$
This integral would be much easier to calculate than the first since it wouldn't require to calculate the rotation explicitly (just considering an arbitrary rotation that orients $(a,b,c)$ along the $z$-axis suffices.)
Any ideas?
 A: In the Cartesian coordinates $(u,v,w)$ with the unit vector $\hat w=\frac{1}{\sqrt{a^2+b^2+c^2}}(a,b,c)$,
$$\hat x= \cos\theta_u \hat u + \cos\theta_v \hat v + \cos\theta_w \hat w$$
where $\cos \theta_w =\frac{a}{\sqrt{a^2+b^2+c^2}}$. Then
\begin{align}
\iiint_{{r<1}} \frac xa  e^{ax + by + cz} \,dV
& =  \frac1a\iiint_{{r<1}}( \cos\theta_u u + \cos\theta_v v + \cos\theta_w w) e^{\sqrt{a^2+b^2+c^2}w }\,dV\\
&= \frac1a\iiint_{r<1} \cos\theta_w  w e^{\sqrt{a^2+b^2+c^2}w } \,dV\\
&= \frac{1}{\sqrt{a^2+b^2+c^2}} \iiint_{r<1} w e^{\sqrt{a^2+b^2+c^2}w } \,dV=I
\end{align}
where the integrations over $u$ and $v$ vanish due to symmetry of the unit ball. Likewise,
\begin{align}
\iiint_{r<1} \frac yb  e^{ax + by + cz} \,dV
= \iiint_{r<1} \frac zc e^{ax + by + cz} \,dV
= I
\end{align}
A: @Quanto has provided an elegant answer to your question, but probably the following ideas could also be useful.
Let's consider $I(r,R_0)=\iiint_{\text{ball radius Ro}} x e^{ax + by + cz}dV=\frac{\partial}{\partial{a}}\iiint_{\text{ball radius Ro}}e^{ax + by + cz}dV$.
We choose $\vec{r}=(a,b,c)$, $r=|\vec{r}|=\sqrt{a^2+b^2+c^2}$, then $ax+by+cz=(\vec{r},\vec{R})$, where $\vec{R}=(x,y,z)$ and $R=|\vec{R}|$.
In order to integrate over $d^3{\vec{R}}$ ($dV$) we have to choose a coordinate system (which is not defined apriory). So, we can orient it in such a way that the polar axis $Z$ has the direction of the vector $\vec{r}$. At this convenient choice we get $ax+by+cz=(\vec{r},\vec{R})=rR\cos\theta$ (polar angle).
Integration is straightforward:
$I(r,R_0)=\frac{\partial}{\partial{a}}\int_0^{2\pi}d\phi\int_0^{R_0}{R}^2{dR}\int_0^\pi{e}^{rR\cos\theta}\sin\theta{d}\theta=2\pi\frac{\partial}{\partial{a}}(\int_0^{R_0}{R}^2{dR}\int_{-1}^1{e}^{rRt}{d}t)=$$=2\pi\frac{\partial}{\partial{a}}(\frac{1}{r}\int_0^{Ro}(e^{rR}-e^{-rR})RdR)=2\pi\frac{\partial}{\partial{a}}(\frac{1}{r}\frac{\partial}{\partial{r}}\int_0^{Ro}(e^{rR}+e^{-rR})dR)=2\pi\frac{\partial}{\partial{a}}(\frac{1}{r}\frac{\partial}{\partial{r}}\frac{e^{rR_0}-e^{-rR_0}}{r})$
Given the fact that $\frac{\partial}{\partial{a}}f(r)=(\frac{\partial}{\partial{r}}f(r))\frac{\partial}{\partial{a}}r=(\frac{\partial}{\partial{r}}f(r))\frac{\partial}{\partial{a}}\sqrt{a^2+b^2+c^2}=(\frac{\partial}{\partial{r}}f(r))\frac{a}{r}$ we get
$$I(r,R_0)=\frac{2\pi{a}}{r}\frac{\partial}{\partial{r}}\left(\frac{1}{r}\frac{\partial}{\partial{r}}(\frac{e^{rR_0}-e^{-rR_0}}{r})\right)$$
Finally, $$I(r,R_0)=\frac{2\pi{a}}{r^5}\Biggl((r^2R_0^2+3)\sinh(rR_0)-3rR_0\cosh(rR_0)\Biggr)$$
At $R_0=1$ we get your answer.
