a question about the evaluation of triple integral, I am stuck! How to use the method of orthogonal transformation to figure out the triple
integral ?. I am stuck about it! The triple integral is:
$$
\iiint\cos\left(ax + by + cz\right)\,{\rm d}x\,{\rm d}y\,{\rm d}z
\qquad\mbox{and}\qquad x^{2} + y^{2} + z^{2} \leq 1
$$
My solution: I want to suppose
$\quad u = ax + by + cz\,,\quad v=y\quad$ and $\quad w=z$.
\begin{align}
&\mbox{Then}\quad\iiint\cos\left(ax + by +cz\right)\,{\rm d}x\,{\rm d}y\,{\rm d}z
=\iiint { 1\over{ a}}\cos\left(u\right)\,{\rm d}u\,{\rm d}v\,{\rm d}w
\\[3mm]&\mbox{and}\quad
\left({1\over a}\,u - {b\over a}\,v - {c\over z}\,w\right)^{2} + v^{2} + w^{2}
\leq 1
\end{align}
But I don't know how to continue. Is that right ?. Or can someone use other methods to solve the question ?. You don't need to use orthogonal transformation necessarily.

Let $\vec{u}$ be the vector $(a,b,c)$.
Let $\lambda = |\vec{u}| = \sqrt{a^2+b^2+c^2}$ and
$\displaystyle\;\hat{u} = \frac{\vec{u}}{|\vec{u}|}$ be the associated unit vector. 
Pick two more unit vectors $\hat{v}$, $\hat{w}$ such that
$\hat{u}, \hat{v}, \hat{w}$ forms an orthonormal basis. You then parametrize the points
in $\mathbb{R}^3$ as
$$\vec{r} = (x,y,z) = x\hat{x} + y\hat{y} + z\hat{z} = u\hat{u}+v\hat{v}+w\hat{w}$$
This is the sort of orthogonal transform you are supposed to use.  
You don't need to work out what are $\hat{v}$ and $\hat{w}$ exactly. What you
need to know is they exist and under this transform, both the unit sphere and the volume element are preserved. i.e.
$$|\vec{r}| \le 1 \quad\iff\quad x^2 + y^2 + z^2 \le 1 \quad\iff\quad u^2 + v^2 + w^2 \le 1$$
$$dx dy dz = du dv dw$$
Since $ax+by+cz = \lambda u$, your integral can be rewritten and evaluated as
$$\begin{align}
  \int_{|\vec{r}|\le 1}\cos(\lambda u) du dv dw
=& \pi \int_{-1}^1 (1-u^2)\cos(\lambda u)du
= \frac{\pi}{\lambda}\int_{-1}^1 (1-u^2)d \sin(\lambda u)\\
=& \frac{2\pi}{\lambda}\int_{-1}^1 \sin(\lambda u) u du
= -\frac{2\pi}{\lambda^2}\int_{-1}^1 u d\cos(\lambda u)\\
=& -\frac{2\pi}{\lambda^2}\left\{
\big[u\cos(\lambda u)\big]_{-1}^1 - \int_{-1}^1\cos(\lambda u) du
\right\}\\
=& \frac{4\pi}{\lambda^2}\left(\frac{\sin\lambda}{\lambda} - \cos\lambda\right)
\end{align}
$$
As a double check, consider what happens for small $\lambda$. We have
$$\frac{4\pi}{\lambda^2}\left(\frac{\sin\lambda}{\lambda} - \cos\lambda\right) \sim 
\frac{4\pi}{\lambda^2}\left(\left(1 - \frac{\lambda^2}{6}\right) - \left(1 - \frac{\lambda^2}{2}\right) + O(\lambda^4)\right)
= \frac{4\pi}{3} + O(\lambda^2)$$
In the limit $\lambda \to 0$, one recover the volume of the unit sphere $\displaystyle\;\frac{4\pi}{3}\;$ as expected.
