Joint density of linearly transformed random variables A random vector (X, Y) is normally distributed with the following joint density:
$$
f^{X,Y}(x,y) = \frac{e^{\frac{1}{2} \left(-5
   \text{x$$}{}^2-2
   \text{x$$}
   \text{y$$}-\text{y$$}{}
   ^2\right)}}{\pi }
$$
I need to calculate the joint density 
$$ f^{U,V} $$
where 
$$
U = X + Y
$$
and
$$
V = 2X
$$
Am I not mistaken that the location of my new distribution is going to be the following vector?
$$ \begin{bmatrix} \mathbb{E} [X + Y]\\ \mathbb{E} [2X] \end{bmatrix}
$$
I've calculated the marginal pdfs for x and y:
$$
f^{X}(x,y) = \sqrt{\frac{2}{\pi }} e^{-2x^2}
$$
$$
f^{Y}(x,y) = \sqrt{\frac{2}{5 \pi }} e^{-\frac{2 y^2}{5}}
$$
and obtained the expected values for x and y:
$$
\mathbb{E} [X] = \int_{}^{} x * f^{X}(x,y) dx = 0
$$
$$
\mathbb{E} [Y] = \int_{}^{} y * f^{Y}(x,y) dy = 0
$$
Now since x and y are normally distributed the location of x + y is going to be 0 as well. 
Does it make any sense? How do I get the covariance matrix? 
Thanks a lot for any hints!
 A: Following @stochastic's comment, I was able to figure it out. This is how it goes:
$$
f^{X,Y}(x,y) = \frac{e^{\frac{1}{2} \left(-5
   \text{x$$}{}^2-2
   \text{x$$}
   \text{y$$}-\text{y$$}{}
   ^2\right)}}{\pi }
$$
$$
U = X + Y
$$
$$
V = 2X
$$
Our transformation function looks then like this:
$$
T(\begin{bmatrix}
           x \\
           y \\
         \end{bmatrix}) = \begin{bmatrix}
           u \\
           v \\
         \end{bmatrix} = \begin{bmatrix}
           x + y \\
           2x \\
         \end{bmatrix}
$$
We need its inverse:
$$
T^{-1}(\begin{bmatrix}
           x \\
           y \\
         \end{bmatrix}) = T(\begin{bmatrix}
           \frac{1}{2}v \\
           u - \frac{1}{2}v \\
         \end{bmatrix}) 
$$
and the jacobian:
$$
DT^{-1} = \begin{bmatrix}
           0 & \frac{1}{2}\\
           1 & -\frac{1}{2} \\
         \end{bmatrix}
$$
The determinant of said jacobian is:
$$
|DT^{-1}| = -\frac{1}{2}
$$
From here we're ready to obtain the transformed joint density by substituting for u and v, and normalizing by the absolute value of the jacobian:
$$
f^{U,V}(u,v) = f^{X,Y}(\frac{1}{2}v,  u - \frac{1}{2}v) * |-\frac{1}{2}| = \frac{1}{\pi} e^{-\frac{u^{2}}{2} -\frac{v^{2}}{2} }
$$
