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.

\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

[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;

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?

[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

[Hootenanny]

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?

[Dafe]

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

[Hootenanny]

Ah, darn I should have seen that! Thank you for being so patient!
Nah, its no problem, I didn't spot it at first until I started sketching the region :smile: