Volume of Lie groups and homogeneous manifolds with respect to Riemannian volume forms (Motivations / background below, but the question is self contained.)
Question
Suppose $G$ is a compact Lie group and $K$ a closed subgroup. Let $X=G/K$ the corresponding homogeneous manifold (to be concrete: we quotient by multiplication on the right by $K$, namely elements of $X$ are cosets $[g]=gK$).
Suppose also that $G$ is equipped with a bi-invariant Riemannian metric $\gamma$. Let $\Omega_G$ be the associated volume form. (In coordinates $\gamma=\gamma_{ij}(x)dx^idx^j\Rightarrow\Omega_G=\sqrt{\det\gamma_{ij}(x)}dx^1\wedge...\wedge dx^n$.)
The restricted Riemann metric $\gamma|_K$ induces a volume form $\Omega_K$ on $K$.
Moreover $\gamma$ induces an invariant Riemannian metric on $X$; the way I see this is because $T_{[g]}X \simeq T_gG / T_g(g.K)\simeq(T_g(g.K))^{\perp_\gamma}$ (where $g.K$ is the orbit of $g$ under right multiplication by $K$ and the superscript $\perp_\gamma$ denotes the orthogonal subspace with respect to $\gamma$) so that we can define the metric on $T_{[g]}X$ by restricting $\gamma$ to $(T_g(g.K))^{\perp_\gamma}$.
What I would like to prove (or disprove?) is that
$$
(\star)\qquad\qquad\int_G\Omega_G=\int_X\Omega_X\int_K\Omega_K,
$$
or even some version for integration of functions on $G$ if possible. (Maybe I need some version of the "coarea formula"?)
Motivations / Background
Let $\mathcal U_n$ be the real Lie group of unitary matrices of size $n$ and let $\mathcal H_n$ be the real vector space of Hermitian matrices of size $n$.
We can define the bi-invariant Riemannian metric $\gamma_U^{(n)}(M,N)=tr(MN^\dagger)$ on $\mathcal U_n\ni U$ where $M,N\in T_U\mathcal U_n=\sqrt{-1} U \mathcal H_n$.
I would like to take $G=\mathcal U_n$ and $K=\mathcal U_{n-1}$; the latter is embedded via $i:K\to G:U\mapsto \begin{pmatrix}1 & 0 \\ 0 & U\end{pmatrix}$ which is isometric, namely $i^*\gamma^{(n)}=\gamma^{(n-1)}$.
The quotient $X=G/K$ is diffeomorphic to $S^{2n-1}\subset\mathbb C^n$ via $f:[U]\mapsto Ue_1$ (i.e. via the map $f$ sending $[U]\in X$ to the first column of $U$, which is a norm 1 vector in $\mathbb C^n$).
It also seems to me that $\Omega_X$, as defined above, is $2^{n-1}\Omega_{S^{2n-1}}$ (namely $\Omega_X=2^{n-1}f^*\Omega_{S^{2n-1}}$), where $\Omega_{S^m}$ is the standard volume form on the sphere.
Finally, the factorization property $(\star)$ above, along with the standard identity $Vol(S^{m-1})=2\pi^{m/2}/\Gamma(m/2)$, would imply that
$$
Vol(\mathcal U_n)=Vol(\mathcal U_{n-1})Vol(X)=Vol(\mathcal U_{n-1})2^{n-1} Vol(S^{2n-1})=Vol(\mathcal U_{n-1})\frac{(2\pi)^n}{(n-1)!}.
$$
Together with the initial value $Vol(\mathcal U_1)=2\pi$ (which can be computed directly) this argument seems to provide the correct value for $Vol(\mathcal U_n)=\frac{(2\pi)^{n(n+1)/2}}{1!\cdots (n-1)!}$ (with respect to the present normalization of the Haar measure), but I am still puzzled by the equality $(\star)$ above.
 A: One way to derive this result is by applying a more general result which allows one to integrate along fibers of a submersion:
Oriented Submersions
Let $M^m$, and $N^n$ be oriented smooth manifolds with $m>n$, and $\pi:M^m\to N^n$ be a smooth submersion. We can define adapted oriented coordinates as a set of oriented coordinate charts such that $\varphi$ has local representative equal to the projection onto the first $n$ coordinates $\pi(x^1,\cdots,x^m)=(x^1\cdots,x^n)$. There is a unique induced orientation on each fiber $\pi^{-1}(p)$ such that $(x^{n+1},\cdots,x^m)$ is an oriented chart on the fiber for any adapted oriented coordinates $x^i$.
Let $\alpha\in\Omega^nN$ and $\beta\in\Omega^{m-n}M$ be compactly supported differential forms. We can separate the following integral:
$$
\int_M\pi^*(\alpha)\wedge\beta=\int_Nf_\beta\alpha \\
\text{where}\ \ \ f_\beta(p):=\int_{\pi^{-1}(p)}\beta|_{\pi^{-1}(p)}
$$
To show this, it suffices to first show it holds in the case that $\beta$ is supported in a single set of adapted oriented coordinates with domain $(0,1)^m\mapsto(0,1)^n$, where it is a straightforward computation. The general case can then be obtained using a partition of unity.
Riemannian Submersions
Let $\pi:M^m\to N^n$ be a Riemannian submersion with $m>n$ ($M$ and $N$ now have metrics, but need not be oriented). Locally, we can find orthonormal coframes $E^1,\cdots,E^m$ on $U\subseteq M$ and $e^1,\cdots,e^n$ on $V(U)\subseteq N$ such that $\pi^*(e^i)=E^i$ for all $p\in U$ and $i\in[1,n]$. Note that $e^1\wedge\cdots\wedge e^n$ is a Riemann volume form on $N$, $E^1\wedge\cdots\wedge E^m$ is a Riemann volume form on $M$, and $E^{n+1}\wedge\cdots\wedge E^m$ restricts a Riemann volume form on each fiber $\pi^{-1}(q)\cap U$, Choosing a smooth $f:M\to \mathbb{R}$ supported in $U$, we can choose local orientations induced by these volume forms, we can apply the above result to the integral of $f$, yielding the following:
$$
\int_Mf\ dV=\int_N\hat{f}dV \\
\text{where}\ \ \ \hat{f}(q)=\int_{\pi^{-1}(q)}f\ dV
$$
Where $dV$ denotes the Riemannian measure on each manifold. Once again this result holds globally, provided $f$ is compactly supported, by a partition of unity argument.
Homogeneous Spaces
In your case, you have defined the metrics so that the quotient map $\pi:G\to X$ is a Riemannian submersion, and each fiber $\pi^{-1}(x)$ is isometric to $H$. Thus we have
$$
\operatorname{vol}(G)=\int_G1dV=\int_X fdV \\
\text{where}\ \ \ f(x):=\int_{\pi^{-1}(x)}1dV=\int_H1dV=\operatorname{vol}(H) \\
\implies \operatorname{vol}(G)=\int_X \operatorname{vol}(H)dV=\operatorname{vol}(X)\operatorname{vol}(H) \\
$$
