Suppose given n random variables $X_1,\ldots,X_n$, $n > 1$, $X_i \text{ iid } \sim N(0,1)$, and a number $1 < m < n$. I am looking for the distribution of the r.v. $$ R = \frac{X_1^2 + \ldots X_m^2}{X_1^2 + \ldots X_n^2} $$ So $R$ is a ratio between correlated $\chi^2$ variables - hence is not $F$- distributed.
In fact I actually want the distribution on $[0,\pi/2]$ of $$ \Theta = \text{arccos}\left(\sqrt R\right) $$ (positive square root, so $\sqrt R \in [0,1]$), where "arccos" is the inverse cosine function; $\text{arccos}(x) \in [0,\pi/2]$ for $x \in [0,1]$.
($\Theta$ is the angle between a uniform random unit vector in $\mathbb{R}^n$ and the $m$-dimensional hyperplane in $\mathbb{R}^n$ defined by the $1$st $m$ coordinate axes.)
It's a $\text{Beta}(m/2,(n-m)/2)$, which is not surprising, since the numerator is a fraction of the denominator. Let $$A=X_1^2+\ldots + X_m^2$$ and $$B=X_{m+1}^2+\ldots + X_n^2.$$ So $A,B$ are independent $\chi^2$. $A$ has $m$ degrees of freedom and $B$ has $n-m$ degrees of freedom.
You want the distribution of $$R=\frac{A}{A+B}.$$ Let $S=A+B$. Note that $A=RS$ and $B=(1-R)S$. Taking the Jacobian, we find $$\det\begin{pmatrix} \frac{\partial A}{\partial R} & \frac{\partial B}{\partial R} \\ \frac{\partial A}{\partial S} & \frac{\partial B}{\partial S}\end{pmatrix} = \begin{pmatrix} S & -S \\ R & 1-R\end{pmatrix}=S.$$ By change of variables, we have \begin{align*}f_{R,S}(r,s) & = f_A(a)f_B(b)s \propto a^{\frac{m}{2}-1}\exp\Bigl({-\frac{a}{2}}\Bigr) b^{\frac{n-m}{2}-1}\Bigl({-\frac{b}{2}}\Bigr) \\ & = (rs)^{\frac{m}{2}-1}[(1-r)s]^{\frac{n-m}{2}-1} \exp\Bigl(-\Bigl[\frac{rs+(1-r)s}{2}\Bigr]\Bigr)s \\ & = g(s)r^{\frac{m}{2}--1}(1-r)^{\frac{n-m}{2}-1}.\end{align*} Here, $g(s)$ is some function of $s$ that we don't care about, and we don't care about the normalizing constant, either. This is the joint density. To find the marginal, we just integrate out $s$, which will leave $$f_R(r)\propto r^{\frac{m}{2}-1}(1-r)^{\frac{n-m}{2}-1}.$$
Since we ignored normalizing constants, it's worth noting that the support of this pdf is $(0,1)$.