# Thread: Double integral

1. ## Double integral

Let
$f(x)=6x(1-x)$ for $0\leq x\leq 1$
$g(y)=2y$ for $0 < y < 1$

Calculate $F(w) = \iint_{x^2y \leq w}f(x)g(y)dxdy$ for $0\leq w\leq 1$

I'm having trouble figuring out what the limits are since w is also a variable.

2. ## Re: Double integral

Originally Posted by Ridley
Let
$f(x)=6x(1-x)$ for $0\leq x\leq 1$
$g(y)=2y$ for $0 < y < 1$

Calculate $F(w) = \iint_{x^2y \leq w}f(x)g(y)dxdy$ for $0\leq w\leq 1$

I'm having trouble figuring out what the limits are since w is also a variable.
If you let y go from 0 to w, then x goes from 0 to sqrt(w/y)

CB

3. ## Re: Double integral

Originally Posted by Ridley
Let
$f(x)=6x(1-x)$ for $0\leq x\leq 1$
$g(y)=2y$ for $0 < y < 1$

Calculate $F(w) = \iint_{x^2y \leq w}f(x)g(y)dxdy$ for $0\leq w\leq 1$
I suppose you have a typo: on the unbounded region $x^2y\leq w$ the function $F(x,y)=f(x)g(y)$ is not well defined. Take for example the fourth quadrant.

4. ## Re: Double integral

It is implied that f and g are zero everywhere where they are not explicitly defined above (I didn't bother typing it out). I'm not sure if this changes anything.

I was originally trying to solve a probability problem where f and g above are probability density functions for the stochastic variables X and Y. The problem was to find the probability density function for $W = X^2Y$ where X and Y are independent. I figured I could find the distribution for W and then take the derivative of that.

5. ## Re: Double integral

No, w is not a variable. For the purposes of this problem, w is a constant.
When x= 1, y= w. When y= 1, $x= \sqrt{w}$

If you wish to integrate with respect to y first, for each x, y will vary from 0 to $y= w/x^2$. If you wish to integrate with respect to x first, for each y, x will vary from 0 to $x= \sqrt{w}/y$.

Your integral is
$\int_{x=0}^1\int_{y= 0}^{w/x^2} f(x)g(y)dydx$
or
$\int_{y=0}^1\int_{x=0}^{\sqrt{w}/y} fd(x)g(y)dxdy$

6. ## Re: Double integral

I had tried both those integrals before posting the question but they both give the wrong answer. I figured out that you need to split the integral into two different areas to account for all values of w. It becomes obvious if you try to graph the area for different values of w.

$F(w) = \int_{0}^{1}\int_{0}^{w}f(x)g(y)dxdy + \int_{w}^{1}\int_{0}^{\sqrt{w/y}}f(x)g(y)dxdy$

7. ## Re: Double integral

Originally Posted by Ridley
I had tried both those integrals before posting the question but they both give the wrong answer. I figured out that you need to split the integral into two different areas to account for all values of w. It becomes obvious if you try to graph the area for different values of w.

$F(w) = \int_{0}^{1}\int_{0}^{w}f(x)g(y)dxdy + \int_{w}^{1}\int_{0}^{\sqrt{w/y}}f(x)g(y)dxdy$
Just goes to show that what we normally, say and did not do ourselves here, is always sketch the region over which you are integrating!

Anyway just so you can check, below is a plot of the numerical integral as a function of w:

CB