The joint pdf of $\underset{\sim}{X}=(X_1, X_2,..., X_p)$ is given by,
$f(\underset{\sim}{x})= k$ if $\sum_{i=1}^{p}x_i^2 \leq c^2, c>0$
Find $k$. Also find the marginal distribution of $\underset{\sim}{X_1}=(X_1, X_2,..., X_q), q<p$.
My attempt:
I have taken the transformation
$x_1=r\cos\theta_1$
$x_2=r\sin\theta_1\cos\theta_2$
$x_3=r\sin\theta_1\sin\theta_2\cos\theta_3$
..............
$x_{p-1}=r\sin\theta_1\sin\theta_2...\sin\theta_{p-2}\cos\theta_{p-1}$
$x_p=r\sin\theta_1\sin\theta_2...\sin\theta_{p-2}\sin\theta_{p-1}$
Then $r^2=\sum_{i=1}^{p}x_i^2 \leq c^2$ and $r \geq 0$
So, $0\leq r \leq c, 0<\theta_i<\pi, i=1,2,...,p-2$ and $0<\theta_{p-1}<2\pi$
The jacobian is,
$|J|=r^{p-1}\sin^{p-2}\theta_1\sin^{p-3}\theta_2...sin\theta_{p-2}$
Using this transformation in $\int_{\sum_{i=1}^{p}x_i^2}f(\underset{\sim}{x})dx_1dx_2...dx_p =1$,
I have got
$k=\frac{\Gamma(\frac{p}{2}+1)}{c^p\pi^{\frac{p}{2}}}$
But I cannot find the marginal distribution of $\underset{\sim}{X_1}=(X_1, X_2,..., X_q), q<p$. To find this, I have to integrate $f(\underset{\sim}{x})$ with respect to $x_{q+1},..., x_p$. But I cannot understand how to use the above transformation to do this integration. Or is there any other way to do it? Please anyone give some hints about how to proceed. Thanks in advance.