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.
 A: 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.
A: I think if you change all variable to their negatives, $(x,y,z)\rightarrow (-x,-y,-z)$ the integral changes sign. Therefore $\int \int \int =0$
A: $\newcommand{\+}{^{\dagger}}
 \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
 \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,}
 \newcommand{\dd}{{\rm d}}
 \newcommand{\down}{\downarrow}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
 \newcommand{\fermi}{\,{\rm f}}
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\isdiv}{\,\left.\right\vert\,}
 \newcommand{\ket}[1]{\left\vert #1\right\rangle}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
 \newcommand{\sech}{\,{\rm sech}}
 \newcommand{\sgn}{\,{\rm sgn}}
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}
 \newcommand{\wt}[1]{\widetilde{#1}}$
$\ds{\iiint\cos\pars{ax + by + cz}\,\dd x\,\dd y\,\dd z:\ {\large ?}
     \qquad\qquad\mbox{and}\qquad\qquad x^{2} + y^{2} + z^{2} \leq 1}$

Lets $\ds{\vec{k} \equiv \pars{a,b,c}}$:
  \begin{align}&\color{#c00000}{\iiint\cos\pars{ax + by + cz}\,\dd x\,\dd y\,\dd z}
=\Re\iiint\expo{\ic \vec{k}\cdot\vec{r}}\,\dd^{3}\vec{r}
=\Re\int_{0}^{1}\color{#00f}{%
\int_{\Omega_{\vec{r}}}{\expo{\ic\vec{k}\cdot\vec{r}}}
\,{\dd\Omega_{\vec{r}} \over 4\pi}}
\,4\pi r^{2}\,\dd r
\end{align}

\begin{align}&\color{#00f}{%
\int_{\Omega_{\vec{r}}}{\expo{\ic\vec{k}\cdot\vec{r}}}
\,{\dd\Omega_{\vec{r}} \over 4\pi}}
={1 \over 4\pi}\int_{0}^{2\pi}\int_{0}^{\pi}
\expo{\ic kr\cos\pars{\theta}}\sin\pars{\theta}\,\dd\theta\,\dd\phi
={\sin\pars{kr} \over kr}
\end{align}

Then,
  \begin{align}&\color{#c00000}{\iiint\cos\pars{ax + by + cz}\,\dd x\,\dd y\,\dd z}
=\int_{0}^{1}{\sin\pars{kr} \over kr}\,4\pi r^{2}\,\dd r
={4\pi \over k^{3}}\
\overbrace{\int_{0}^{k}t\sin\pars{t}\,\dd t}
^{\ds{=\ \sin\pars{k} - k\cos\pars{k}}}
\end{align}

\begin{align}&\color{#66f}{\large
\iiint_{x^{2}\ +\ y^{2}\ +\ z^{2}\ \leq\ 1}
\cos\pars{ax + by + cz}\,\dd x\,\dd y\,\dd z}
\\[3mm]&=\color{#66f}{\large 4\pi\,
{\sin\pars{\root{a^{2} + b^{2} + c^{2}}}
- \root{a^{2} + b^{2} + c^{2}}\cos\pars{\root{a^{2} + b^{2} + c^{2}}}
\over \pars{a^{2} + b^{2} + c^{2}}^{3/2}}}
\end{align}
