length of Gaussian Random Vector Suppose I have a random vector $x=[x_1,...,x_k]$ s.t. $x∼N(\mu,\sum)$. How is the length or magnitude of $x$ distributed?
I know that if $k=2$ and $\sigma_1=\sigma_2$ and $\sigma_{12}=0$ ($x_1$ and $x_2$ are not correlated), it is Rayleigh distribution.
I also know that $\sqrt{\sum_{i=1}^k(\frac{x_i-\mu_i}{\sigma_i})^2}$ is Chi distributed (no correlation). However, the random variables are normalized by its standard deviation, it is just the length of a zero-mean unit variance Gaussian vector. 
If it is not zero mean, we can have noncentral chi distribution. It is non-zero-mean but still unit variance Gaussian vector.
So my question is:


*

*When $\sigma_i$ has different values for all $i=1,...,k$, what is the distribution of vector length/magnitude $|x|=\sqrt{\sum_{i=1}^k x_i^2}$?

*When the random variables are correlated, what is the distribution of the vector length/magnitude $|x|=\sqrt{\sum_{i=1}^k x_i^2}$?
 A: Consider the random vector $X=(X_1,\dots,X_k)$ such that $X\sim N_k(\mu,\Sigma)$.
Set $Q(X)=X^TAX=\sum_{i=1}^kX_i^2$, where $A=\mathrm I_k$. A good reference for the study of $Q(X)$ is "Quadratic forms in random variables" by Mathai and Provost.
Specifically, denote by $P$ an orthogonal matrix that diagonalizes $\Sigma$, and write $P^T\Sigma P=\mathrm{Diag}(\lambda_1,\dots,\lambda_k)$. Also define 
$$
b=P^T\Sigma^{-1/2}\mu.
$$
Then,
$$
Q(X)=\sum_{i=1}^k\lambda_i\left(U_i+b_i\right)^2,\qquad(1)
$$
where $U_i\sim N_k(0,\mathrm I_k)$ (equation $(4.1.1)$ of the reference). This is sometimes called a generalized Chi-squared distribution, and the length of the vector $\sqrt{Q(X)}$ is thus called generalized Chi distribution. The Laplace transform of $Q(X)$ is also obtained in equation $(4.2b.6)$ as
$$
\mathrm L(s)=\exp\left(-\frac12\sum_{i=1}^kb_i^2\right)\exp\left(\frac12\sum_{i=1}^kb_i\frac1{1+2s\lambda_i}\right)\prod_{i=1}^k\frac1{\sqrt{1+2s\lambda_i}},
$$
for $\left|2s\lambda_i\right|<1$. Now, in general $(1)$ does not follow a well-known distribution. As you mention in OP, if $\Sigma=\sigma^2\mathrm I_k$ and $\mu\neq0$, then $Q(X)$ follows a non-central chi-squared distribution. Another case that has a closed-form solution is if $\mu=0$, and $\Sigma$ is diagonal with elements $\sigma_1^2,\dots,\sigma_k^2$. Then (c.f. wikipedia as well as $[5]$ therein), if $\sigma_i\neq\sigma_j$, the density of $Q(X)$ can be computed as
$$
f(x)=\sum_{i=1}^{k} \frac{e^{-\frac{x}{\sigma_i^2}}}{\sigma_i^2 \prod_{j=1, j\neq
i}^{k} (1- \frac{\sigma_j^2}{\sigma_i^2})}1_{x\ge0}.
$$
From this you can deduce the distribution of $\sqrt{Q(X)}$. Similar calculations can be done if $\sigma_i=\sigma_j$ for some $i,j\in\{1,\dots,k\}$.
To answer your question is a general setting is more difficult since to my knowledge, these distributions do not follow other known laws. Therefore, it depends on what application you have in mind. If you want to calculate moments, then you might be able to exploit formula $(1)$. If you want to approximate the pdf of $\sqrt{Q(X)}$, then you can use some of the asymptotic expansions of Chapter $4$ of the book that I mentioned in the beginning of this post.
