Joint Density Function Let
$$
f =
\begin{cases}
 cx& \text{if } 1> x > y > 0\\
0  &\text{otherwise},
\end{cases}
$$
let this function be the joint density function for $X$ and $Y$, I want to determine the value of $c$ such that it is a joint denisty function, I keep getting $c=6$ rather than $3$ which is the answer. Im more concerned regarding how Im integrating to in the region $1>x>y>0$, i must be taking the integration limits wrong i presume, if someone could kindly show me how to integrate a region with constraints like this that would be great. 
Thanks
 A: Either


*

*$x$ runs from $0$ to $1$ and for each fixed value of $x$, the other variable $y$ runs from $0$ to $x$; or

*$y$ runs from $0$ to $1$ and for each fixed value of $y$, the other variable $x$ runs from $y$ to $1$.


So you have either
$$
\int_0^1\left(\int_0^x\cdots\cdots\, dy\right) \, dx
$$
or
$$
\int_0^1 \left(\int_y^1\cdots\cdots\,dx\right) \, dy.
$$
In the first case you have
$$
\int_0^x cx \, dy = cx^2,
$$
and then
$$
\int_0^1 cx^2\,dx = \frac c 3.
$$
In the second case you have
$$
\int_y^1 cx\,dx = c\frac{1-y^2}{2}
$$
and then
$$
\int_0^1 c\frac{1-y^2}{2}\,dy = \left.c\left(\frac y 2 - \frac{y^3}6\right) \right|_0^1 = \frac c 3.
$$
A: The region is the triangle formed by the points $(0,0)$, $(1,0)$ and $(1,1)$. A good way to think about it is to fix an $0 < x < 1$. Then you want to go over all $y$ that is between $0$ and $x$, so it's considering all cross-sections of the triangle and then adding them all up. This gives us the integral equation:
$$
\int_{0}^1 \int_0^x c x \, dydx = 1.
$$
The integral on the right hand side is
$$
\int_{0}^1 \int_0^x c x \, dydx = \int_0^1 c x^2 \, dx = \dfrac{c}{3}
$$
so that setting this equal to 1 gives $c = 3$ as in the book.
A: I am unable to solve the problem without drawing a picture. We have $0\lt y\lt x\lt 1$. In particular, $y\lt x$, so our density function lives below the line $y=x$ (which I drew). Also, $x$ is between $0$ and $1$. 
So the region where our density function lives is the triangle with corners $(0,0)$, $(1,0)$, and $(1,1)$. 
The integral of $cx$ over this triangle should be $1$. We express the double integral as an iterated integral in two ways: (i) integrate with respect to $y$, and then $x$, or (ii) the other way.
Method 1: We look at the triangle, and note that $y$ goes from $0$ to $x$, and then $x$ goes from $0$ to $1$. So we get
$$\int_{x=0}^1 \left(\int_{y=0}^x cx \,dy\right)\,dx.$$
Integrate. The inner integral is trivial, since $cx$ does not involve $y$. So we simply get $cx^2$, Now integrate with respect to $x$. We get $\frac{c}{3}$.
Method 2: We look at the triangle. At the beginning, $x=y$, and at the end it is $1$. So our integral is 
$$\int_{y=0}^1 \left(\int_{x=y}^1 cx \,dx\right)\,dy.$$
The inner integral is $\frac{c}{2}(1-y^2)$. Now integrate with respect to $y$. We get $\frac{c}{3}$.
