The Joint Probability Density Function of X & Y is given by.... Quick question about the limits of integration when solving for joint probability density functions :)
Sorry, about formatting, its difficult to write Stats like this!
F of X Y (x,y) = {c , if X^2 + Y^2  <= 1 }
                 {0, otherwise }
The question I need to solve is this:
- Find the value of the constant c that would make this a valid pdf
My intuition and thoughts:
Well to make this pdf a VALID pdf, the area under the curve needs to be equal to 1 correct? So we can take the double integral of the pdf and set it equal to 1 and solve for the value of c that makes the expression valid.
Where I run into trouble...
1] The limits of integration: How would I know whether to take the integral with respect to x or y first? Does it matter?
2] What are the limits of integration? In this case, I think they are:
{-1, sqrt(1-y^2)} for x
{-1, 1} for y
Is that correct?
Help would be greatly appreciated! I'm not looking for just answers some explanation, especially in simple language would be great!
Thank again and I apologize in advance for the formatting!
 A: Integration is not needed. The area of the circle is $\pi$, so the integral of $c$ over the circleis $\pi c$. It follows that $c=\dfrac{1}{\pi}$.
If you really want to integrate, you could calculate the integral of $c\,dx\,dy$ over the circle in two ways.
(i) Integrate first with respect to $y$, with $y$ going from $-\sqrt{1-x^2}$ to $\sqrt{1-x^2}$. Then integrate with respect to $x$, from $-1$ to $1$.  
This is because the top half of the circle has equation $y=\sqrt{1-x^2}$, and the bottom half has equation $y=\sqrt{1-x^2}$. Alternately, integrate from $0$ to $\sqrt{1-x^2}$ and double the result. Somewhat more simply, let $y$ go from $0$ to $\sqrt{1-x^2}$, and then $x$ from $0$ to $1$, and multiply the result by $4$. 
Alternately, integrate first with respect to $x$, going from $-\sqrt{1-y^2}$ to $\sqrt{1-y^2}$. This fundamentally the same as what we did above, perhaps a little less familiar-feeling. 
(ii) Or else switch to polar coordinates, since circles love polars.  The $c\,dx\,dy$ turns into $cr\,dr\,d\theta$, and we integrate from $r=0$ to $r=1$, then from $\theta=0$ to $\theta=2\pi$. 
