A joint PDF question: $\displaystyle f(x,y)=1-\frac x3-\frac y3$

Question

Really stuck on this problem:

If $\displaystyle f(x,y)=1-\frac x3-\frac y3$ for $0 \le x \le 2$ and $0 \le y \le h,$ the find $h$.

I know I need to integrate but confused how to set it up. Thanks.

Answer 1

$\int_0^2 \int_0^h(1-\frac{x}{3}-\frac{y}{3})dx dy=1 $

This gives us $2h-(4h/6)-2*h^2/6=1$.

Which means, $4h/3-h^2/3=1$, or $h^2-4h+3=0$.

Hence $h=1$ or $h=3$.But $f(x,y)$ can't be negative, hence $h=1$.

Answer 2

Hint :

If $f(x,y)$ is a joint pdf, where $f(x,y)>0$, then use property $$ \iint_{X,Y\in\ \Omega} f(x,y)\ dx\ dy=1. $$ In our case, we have $$ \int_{y=0}^h\int_{x=0}^2\left(1-\frac x3-\frac y3\right)\ dx\ dy=1. $$ The integral itself is easy to be evaluated.