# Change the order of Integrtation

1. May 15, 2007

### Dafe

1. The problem statement, all variables and given/known data
Change the order of integration and perform the integration.

$$\int_0^2\int_{2x}^{4x-x^2} dydx$$
2. Relevant equations

3. The attempt at a solution
I've tried changing it to this but I end up with the wrong answer..

$$0<=y<=2, \sqrt{4-y}+2 <=x<=\frac{1}{2}y$$

2. May 15, 2007

### Hootenanny

Staff Emeritus
Sketch the region, what does it look like to you? Have you met polar coordinates yet?

3. May 15, 2007

### Dafe

I've sketched it, it's looks a little cut of a parabola.. I can't see how I can describe it using polar coordinates :/

4. May 15, 2007

### Hootenanny

Staff Emeritus
Nevermind, it seems that polar coordinates are not nesscary, but you should reconsider your limits. Lets look at how the region is bounded. We have;

$$0\leq x \leq 2$$

$$0\leq y \leq 4$$

$$x\leq \frac{1}{2}y \equiv y \geq 2x$$

$$y \leq 4x-x^2$$

Would you agree?

Last edited: May 15, 2007
5. May 15, 2007

### Dafe

Yes, I do agree. What I did was solve the last equation and so I got
$$0<=y<=2, \sqrt{4-y}+2 <=x<=\frac{1}{2}y$$ ..

Still can't see what I'm doing wrong. I'm kinda slow

6. May 15, 2007

### Hootenanny

Staff Emeritus
You've chosen the wrong solution to your equation, $y=4x-x^2$ has two solutions;

$$x = 2\pm\sqrt{4-y}$$

Now, your original equation $y=4x-x^2$ goes through the point (x,y)=(0,0), therefore, your new solution must also go through the same point if it is to describe the same region. Which of your solutions goes through the origin?

7. May 15, 2007

### Dafe

Ah, darn I should have seen that! Thank you for being so patient!

8. May 15, 2007

### Hootenanny

Staff Emeritus
Nah, its no problem, I didn't spot it at first until I started sketching the region