A manifold integral with multi-dimensional angle? I was given the following exercise in an advanced calculus course:

Let $f:\mathbb{R}\to\mathbb{R}$ be a continuous function and
  $a\in\mathbb{R}$. Show that  $$\int_{S^{n-1}}f\left(\left< x,a\right>\right)d\mathrm{vol}_{n-1}(x)=\omega_{n-1}\int^1_{-1}f\left(t\left|a\right|\right)\left(1-t^2\right)^\frac{n-3}{2}dt$$ where $\omega_{n-1}$ is the
  surface area of $S^{n-1}$, the $n-1$ dimensional sphere (embedded in $\mathbb{R}^n$).

I initially noticed two things. First, the right hand side seemed to suggest to me that the co-area formula might be useful here. Second, that $\left<x,\frac{a}{\left|a\right|}\right>=\cos\left(\theta\left(x\right)\right)$ when $\theta\left(x\right)$ is the angle between $a$ and $x$. 
Putting these two together, I thought I might be able to use the co-area formula on the function $\Phi(x)=\cos\left(\theta\left(x\right)\right)$ to work my way from RHS to LHS. I didn't manage it and couldn't come up with an alternative approach. I'd be thankful for any suggestion.
 A: Your idea is a good one.  You might have had trouble because the formula is incorrect; the $w_{n-1}$ should in fact be $w_{n-2}$.
First of all, both sides are rotationally invariant, so it suffices to assume that $a$ is parallel to $e_n$, i.e. $a=|a| e_n$. Second of all, we can assume $|a|=1$, since if we know the formula holds in this case, then we can deduce it holds for general $a$ by applying the formula to the function $g(x)=f(|a|x)$. Thus, $a=e_n$ without loss of generality. 
Now, apply the coarea formula to the function $F: S_{n-1}\to [-1,1],F(p)=<p,e_n>$. Before getting into the calculations, let's think geometrically about what's going on. The preimages of $F$ divide the sphere into n-2 dimensional slices at different heights. The value of $F$ is constant on each preimage (by definition). At any point on the sphere, we can either move within the preimage containing that point, or we can move along the line of longitude through that point, thus passing through other preimages. This gives $n-2+1=n-1=Dim(S_{n-1})$ independent directions, which is how we will construct our coordinate system. 
Now, let's compute $Jac(F)_p$ for any $p\in S_{n-1}$. As discussed above, we can choose an orthonormal system of local coordinates by first taking orthonormal local coordinates $z_1,\dots, z_{n-2}$ for the preimage $F^{-1}(F(p))$, and then letting $z_{n-1}$ correspond to the (downward) line of longitude. Then $\partial F/\partial z_i=0$ for $i\leq n-2$, because $F$ is constant on the preimage. 
To compute $\partial F/\partial z_{n-1}(p)$, we'll use the fact that $F$ is invariant with respect to rotations about $e_{n}$. Applying such a rotation to $p$, we may assume that $p=(\sqrt{1-q^2},0,..,0, q)$, where $q=F(p)$.  The unit vector at p pointing downward along the line of longitude is given by $v=(q,0,...,0,-\sqrt{1-q^2})$ (to get this, just take the orthogonal projection of $-e_n$ onto $p^{\perp}$ and renormalize the result). 
Now take any path $\gamma:(-\epsilon,\epsilon)\to S^{n-1}$ with $\gamma(0)=p, \gamma’(0)=v$. Then $\partial F/\partial z_{n-1}(p)={\frac d {dt}}_{t=0} F(\gamma(t))={\frac d {dt}}_{t=0} \gamma_n(t)=v_n=-\sqrt{1-q^2}=-\sqrt{1-F(p)^2}$. 
So, the Jacobian at $p$ $J(F)_{p, z_i}$ with respect to the coordinates $z_i$ is a row vector of length n-1, where the first n-2 entries are 0, and the last is $-\sqrt{1-F(p)^2}$. Since the tangent vectors ${{\partial}{\partial z_i}}$ are orthonormal, the transpose with respect to the Riemannian metric is equal to the regular matrix transpose. 
Finally, we're ready to apply the formula, which yields 
$$\int f(F(x))dV_{S_{n-1}}(x)=\int_{-1}^1 \int_{F^{-1}t} {\frac {f(F(p))}{\sqrt{det(J(F)J(F)^T)(p)}}}dV_{F^{-1}t}(p)dt$$
By the discussion above, for $p\in F^{-1}t$, we have $\sqrt{det(J(F)J(F)^T)(p)}=\sqrt{1-t^2}$. Furthermore, by definition of the inverse image, we have $f(F(p))=f(t)$ for any such $p$. Thus the integrand depends only on $t$, so we can pull it out of the inner integral: 
$$\int_{-1}^1 f(t)(1-t^2)^{-1/2}\int_{F^{-1}t}dV_{F^{-1}t}dt=\int_{-1}^1 f(t)(1-t^2)^{-1/2}Vol(F^{-1}t)dt$$
Now, $F^{-1}t$ is defined by the equations $x_n=t, x_1^2+...+x_{n-1}^2+t^2=1$, so it is an n-2 dimensional sphere of radius $\sqrt{1-t^2}$ embedded in the plane $x_n=t$. Accordingly, 
$$Vol(F^{-1}t)=(\sqrt{1-t^2})^{n-2}\omega_{n-2}$$
where $\omega_{n-2}$ is the volume of the unit n-2 sphere. This establishes the formula. 
A: The idea of using a particular change of variables involving a variable defined by the two nonzero vectors $x,a\in\mathbb{R}^n$ works, even if it is a little tricky. Precisely, on the plane defined by the given two vectors in $\mathbb{R}^n$, call $\theta$ the angle between $a$ and $x$ and put
$$
t=\Big\langle x,\frac{a}{\vert a \vert}\Big\rangle=\cos\theta(x)\,\Longleftrightarrow\, 
\begin{cases}
\theta(x)=\arccos(t)\\
\langle x,a\rangle=t\vert a\vert\\
\mathbf{t}=\frac{t}{\vert a\vert}a\in \mathbb{R}^n
\end{cases}
$$
The angle $\theta$ can be used as one of the angle polar coordinates on the unit hypersphere $S^{n-1}=S^{n-1}\!(\mathbf{0},1)$ , with associated differential form $d\theta$. A sketchy picture of the situation is shown in the image below:

This suggests the possibility of decomposing the surface measure (differential form) $d\mathrm{vol_{n-1}}$ on the hypersphere itself as
$$
d\mathrm{vol_{n-1}}(x)=d\theta d\theta_1\cdots d\theta_{n-2}=d\theta d{S^{n-2}} =d\theta d{S^{n-2}\!\big(\mathbf{t},(1-t^2)\big)},
$$
where $d{S^{n-2}}$ is the surface measure on the $2$-codimensional hypersphere ${S^{n-2}\!\big(\mathbf{t},(1-t^2)\big)}$ obtained from the intersection of a (affine) hyperplane in $\mathbb{R}^n$ passing through $x\in S^{n-1}$ and orthogonal to the direction of the vector $a$: this hypersphere is centered on $\mathbf{t}$ and has radius $R=(1-t^2)^\frac{1}{2}=\sin\theta(x)$. Remembering the definition of $t$, the coordinates change 
$$
\left(
\begin{array}{c}
\theta\\
\theta_1\\
\vdots\\
\theta_{n-2}
\end{array}
\right)\mapsto
\left(
\begin{array}{c}
t\\
\theta_1\\
\vdots\\
\theta_{n-2}
\end{array}
\right)
$$
is the one which gives the sought-for form of the integral. Indeed, by using the Jacobian of this transformation it is easy to see that
$$
d\mathrm{vol_{n-1}}(x)=(1-t^2)^\frac{1}{2}dt\, (1-t^2)^\frac{n-2}{2}d\theta_1\cdots d\theta_{n-2}=(1-t^2)^\frac{n-3}{2}dt\,d\mathrm{vol_{n-2}}.
$$
This implies
$$
\int_{S^{n-1}}f\left(\left< x,a\right>\right)d\mathrm{vol}_{n-1}(x)=\int_{S^{n-2}}\int^1_{-1} f\left(t\vert a\vert\right)\left(1-t^2\right)^\frac{n-3}{2}dt\,d\mathrm{vol_{n-2}},
$$
and thus the solution the exercise.


*

*The result is obtained by using a slightly simpler form of a standard change of variable used in the "classical" theory of singular integral operators, described by S. G. Mikhlin in his texts [1, ch. II, §5.6 pp. 43-44] and [2, ch. IX, §2.2 pp. 225].

*As Mike Hawk has already noted, there is a typo in the formula above, as it should be $\omega_{n-2}$, not $\omega_{n-1}$. This is a logical consequence of the fact that a (locally) injective (at least) continuous map does not change the dimensionality of the mapped set.


[1] Mikhlin, S.G. (1965), Multidimensional singular integrals and integral equations, International Series of Monographs in Pure and Applied Mathematics, 83, Oxford–London–Edinburgh–New York–Paris–Frankfurt: Pergamon Press, pp. XII+255, MR 0185399, Zbl 0129.07701.
[2] Mikhlin, Solomon G.; Prössdorf, Siegfried (1986), Singular Integral Operators, Berlin–Heidelberg–New York: Springer Verlag, p. 528, ISBN 3-540-15967-3, MR 0867687, Zbl 0612.47024.
