Change the order of Integrtation

Dafe

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

[tex]\int_0^2\int_{2x}^{4x-x^2} dydx[/tex]
2. Relevant equations



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

[tex]0<=y<=2, \sqrt{4-y}+2 <=x<=\frac{1}{2}y[/tex]

Hootenanny

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

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 :/

Hootenanny

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;

[tex]0\leq x \leq 2[/tex]

[tex]0\leq y \leq 4[/tex]

[tex]x\leq \frac{1}{2}y \equiv y \geq 2x[/tex]

[tex]y \leq 4x-x^2[/tex]

Would you agree?

Dafe

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

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

Hootenanny

You've chosen the wrong solution to your equation, [itex]y=4x-x^2[/itex] has two solutions;

[tex]x = 2\pm\sqrt{4-y}[/tex]

Now, your original equation [itex]y=4x-x^2[/itex] 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?

Dafe

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

Hootenanny

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