finding the limits of integration. joint probability A candy company distributes boxes of chocolates
with a mixture of creams, toffees, and cordials.
Suppose that the weight of each box is 1 kilogram, but
the individual weights of the creams, toffees, and cordials
vary from box to box. For a randomly selected
box, let X and Y represent the weights of the creams
and the toffees, respectively, and suppose that the joint
density function of these variables is
f(x, y) =

24xy, 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, x+ y ≤ 1,
0, elsewhere.
(a) Find the probability that in a given box the cordials
account for more than 1/2 of the weight.

In letter a, it means x+y < 1/2, now how can we find the limits of the double integral? proper approach to solve this problem? thanks
PS: there would be no problem if the limits can be easily deciphered. but for this one its asking P(X + Y < 1/2)
 A: The first thing you have to do is sketch the region of interest
$$
X+Y<\textstyle{1\over 2}, \quad X,Y\ge 0
$$
This will be be the region of all points in  in quadrant 1 whose coordinates $(x,y)$ satisfy: $$y<-x+\textstyle{1\over 2}:$$

The region is shown in pink above (note the density is 0 above the dashed line). If we call this region $A$, the integral is
$$
\int\kern-5pt\int_A f(x,y) dA.
$$
To set up the double integral as an iterated integral, you may think of the region as being generated by the vertical lines $\color{darkgreen}{\ell_x}$ as $x$ ranges from $x=0$ to $x=1/2$.
You first fix $x$ and "integrate along $\ell_x$" in the vertical direction. Then you integrate the $\ell_x$ integrals from $x=0$ to $x=1/2$.
So, fix $x$ and consider $\ell_x$.  The  limits of the  inner integral are from the bottom of $\ell_x$ to the top.  The bottom of $\ell_x$ is $y=0$ and the top is $y=-x+{1\over 2}$.
Note the inner integral will be with respect to $y$.  
So the inner integral is
$$
\int_0^{-x+{1\over 2}} xy \,dy
$$
Now set up the outer integral.  As mentioned, we integrate the above expression from $x=0$ to $x=1/2$:
$$
\int_0^{1/2}\int_0^{-x+{1\over 2}} xy\, \,dy\,dx.
$$

Alternatively, you can think of the region as being generated by horizontal lines $\color{maroon}{\ell_y}$ that range from $y=0$ to $y=1/2$.
Here, you'd integrate along a horizontal line first, from its left endpoint, 0, to its right endpoint $-y+{1\over2}$. Then, integrate with respect to $y$ from $y=0$ to $y=1/2$:
$$
\int_0^{1/2}\int_0^{-y+{1\over 2}} xy\, \,dx\,dy.
$$
