Find the joint probability density function of transformations, the answer incorrect. What is my mistake? Given joint probability density function:
\begin{align}
f_{X,Y}(x,y)=
\begin{cases}
24x(1-y)&0<x<y<1\\
0&\text{otherwise}
\end{cases}.
\end{align}
Given transformation: $M=\dfrac{X+Y}{2}$ and $W=\dfrac{X}{2}$, find the joint p.d.f. of $M$ and $W$.
I try as follows.
Since $0<x<y<1$ The range of transformation is
$$
M=\dfrac{X+Y}{2}\geq \dfrac{X}{2}=W,
$$
so $0\leq M\leq 1$, $M\geq W$, and $0\leq W\leq 1$.
The invers of transformation is
$X=2W$ and $Y=2M-2W$.
The absolute value of Jacobian is
\begin{align}
|J|=
\begin{vmatrix}
\dfrac{dX}{dM}&\dfrac{dX}{dW}\\
\dfrac{dY}{dM}&\dfrac{dY}{dW}
\end{vmatrix}
=
\begin{vmatrix}
0&2\\
2&-2
\end{vmatrix}
=4.
\end{align}
The p.d.f. of transformation:
\begin{align}
g_{M,W}(m,w)&=f_{X,Y}(x,y)|J|\\
&=f_{X,Y}(2w,2m-2w)|J|\\
&=24(2w)(1-2m+2w)4\\
&=192w(1-2m+2w).
\end{align}
Now, we have
\begin{align}
g_{M,W}(m,w)=
\begin{cases}
192w(1-2m+2w)&0\leq M\leq 1, M\geq W,\text{ and }0\leq W\leq 1\\
0&\text{otherwise}
\end{cases}.
\end{align}
Now I want to check my answer with double integrating joint p.d.f.
\begin{align}
\int\limits_{0}^{1}\int\limits_{0}^{m} 192w(1-2m+2w) \,dw\,dm
\end{align}
and the result is $16$.(I use maple)

So we can conclude $g_{M,W}(m,w)$ is not a p.d.f.
Why this is happen? Am I have a mistake with my answer?

The functional form of the transformed joint density is correct, but the support is incorrect.  All you have to do to rectify this is to first note that the support for $(X,Y)$ is bounded by the lines $$X = 0, \quad X = Y, \quad Y = 1.$$  Because the transformation is linear and invertible, the resulting support in $(M, W)$ space is also bounded by three lines.  We invert the transformation to get $$X = 2W, \quad Y = 2(M-W),$$ hence $$2W = 0, \quad 2W = 2(M-W), \quad 2(M-W) = 1,$$ or $$W = 0, \quad M = 2W, \quad M = W + 1/2.$$  This then becomes the support $$0 < 2W < M < W + 1/2.$$
