Calculating P(Y<$\sqrt{X}$) for Joint PDF of X and Y

[mynameisfunk]

Homework Statement

Suppose that the joint pdf of $X$ and $Y$ is
$$f(x,y)= (8/3)xy , 0<x<1,0<y<2, x<y<2x$$
Compute $$P(Y<\sqrt{X})$$

Homework Equations

The Attempt at a Solution

$$\int_0^1 \int_x^{\sqrt{x}} (8/3)xy dy dx = (4/3) \int_0^1 x - x^2 dx = 2/9$$

 

[Ray Vickson]

Homework Statement

Suppose that the joint pdf of $X$ and $Y$ is
$$f(x,y)= (8/3)xy , 0<x<1,0<y<2, x<y<2x$$
Compute $$P(Y<\sqrt{X})$$

Homework Equations

The Attempt at a Solution

$$\int_0^1 \int_x^{\sqrt{x}} (8/3)xy dy dx = (4/3) \int_0^1 x - x^2 dx = 2/9$$

Wrong answer: start over. Be very careful about the integration region or regions---always draw a picture first, before writing down your integrals.

 

[mynameisfunk]

I want to add to this post. Sorry for the double posting. I know this solution I posted above can't be right. I tried switching the order of integration from dydx to dxdy and I get 1/3. One of my classmates suggested doing a bivariate transformation, which I haven't tried but I am a little confused as to why I wouldn't be able to just go ahead and compute this directly.

 

[mynameisfunk]

Here is the picture I drew. Am I not drawing this right?

 

[Ray Vickson]

Here is the picture I drew. Am I not drawing this right?

Not quite right: for small x > 0 you have the wrong upper limit on y (but it is OK for larger x).