Determine $k$ in PDF. 
Let $X$ and $Y$ random variables with PDF given by
\begin{eqnarray*}
f(x,y)=\left\{\begin{aligned}kxy, \quad x\geq 0, y\geq 0, x+y\leq 1\\ 0, \quad \text{elsewhere} \end{aligned} \right.
\end{eqnarray*}
$1.$ Calculate $k$.
$2.$ Calculate the marginals density of $X$ and $Y$.
$3.$ Are $X$ and $Y$ independent random variables?
$4.$ Calculate $$\mathbb{P}(X\geq Y), \quad \mathbb{P}(X\geq 1/2 | X+Y\leq 3/4), \quad \text{and} \quad \mathbb{P}(X^{2}+Y^{2}\leq 1)$$
$5.$ Calcule the joint density function of $U=X+Y$ and $V=X-Y$ with their respective marginals.

My approach:

*

*Since that $$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f(x,y)=1 \implies \int_{0}^{1}\int_{0}^{1-x}kxydydx=1 \implies k=24.$$


*By definition, we have that $$f_{X}(x)=\int_{0}^{1-x}24xydy=12x(x-1)^{2}, \quad x\in [0,1]$$and $$f_{Y}(y)=\int_{0}^{1-y}24xydx=12y(y-1)^{2}, \quad y \in [0,1]$$


*Since that $$f(x,y)\not=f_{X}(x)f_{Y}(y)$$so $X$ and $Y$ are not independents.


*Now, here I'm problem in how can I calculate $\mathbb{P}(X\geq 1/2| X+Y\leq 3/4)$ and $\mathbb{P}(X^{2}+Y^{2}\leq 1)$ Can you help me here?. I could find that $\mathbb{P}(X\geq Y)=1/2$.


*Here, I don't know how can I solve this part. I was trying to apply the Jacobian transformation, but I don't know how to use this method well. So, I'm not getting a good solution focus.
 A: Indeed, $k=24$, and the marginals are as you calculated: $f_{\small X}(x)=12x(1-x)^2\mathbf 1_{0\leqslant x\leqslant 1}$ and $f_{\small Y}(x)=f_{\small X}(x)$ by symmetry.
Also by symmetry, $\Bbb P(X\leq Y)=\Bbb P(X\geq Y)$, so (since $\Bbb P(X=Y)=0)$ then $\Bbb P(X\geq Y)=1/2$.

Then you just need to use the definition of conditional probability:$$\begin{align}\Bbb P(X\geq 1/2\mid X+Y\leq 3/4)&=\dfrac{\Bbb P(1/2\leq X, X+Y\leq 3/4)}{\Bbb P(X+Y\leq 3/4)}\\[1ex]&=\dfrac{\Bbb P(1/2\leq X\leq 3/4-Y)}{\Bbb P(X\leq 3/4-Y)}\\[1ex]&=\dfrac{\displaystyle\int_0^{1/4}\int_{1/2}^{3/4-y}24xy\,\mathrm dx\,\mathrm d y}{\displaystyle\int_0^{3/4}\int_0^{3/4-y}24 xy\,\mathrm d x\,\mathrm d y}\end{align}$$
For the next problem, notice that for $0\leq x\leq 1$ that $1-x\leq \surd(1-x^2)$

We have $U=X+Y$ and $V=X-Y$, so $U+V=2X$ and $U-V=2Y$.
Thus the Jacobian matrix is: $$\begin{align}\mathsf J_{\small U,V}^{\small X,Y}(u,v)&=\begin{bmatrix}\dfrac{\partial (u+v)/2}{\partial u}&\dfrac{\partial (u+v)/2}{\partial v}\\\dfrac{\partial (u-v)/2}{\partial u}&\dfrac{\partial (u-v)/2}{\partial v}\end{bmatrix}\\[1ex]&=\begin{bmatrix}1/2&1/2\\1/2&-1/2\end{bmatrix}\end{align}$$
Now just apply the Jacobian Transformation: $$ f_{\small U,V}(u,v)=\lVert\mathsf J_{\small U,V}^{\small X,Y}(u,v)\rVert\, f_{\small X,Y}\left(\tfrac {u+v}{2},\tfrac {u-v}2\right)$$
$\blacksquare$
A: 
I have a question, can you explain a little more why $\mathbb{P}(X^{2}+Y^{2}\leq 1)$

Considering that, by assumption of the exercise
$$\mathbb{P}(X,Y \in A)=1$$
I think that the answer is self evident looking at the following drawing

