So here I got such integral. How to change the order of integration? I am not asking you for the answer, but what should be the first step?

$\displaystyle \int_{-1}^{1}dy \int_{{y^2 \over 2}}^{ \sqrt{3-y^2}} f(x,y)dx$

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- Feb 17th 2010, 01:47 AMjacekchanging the order of integration
So here I got such integral. How to change the order of integration? I am not asking you for the answer, but what should be the first step?

$\displaystyle \int_{-1}^{1}dy \int_{{y^2 \over 2}}^{ \sqrt{3-y^2}} f(x,y)dx$ - Feb 17th 2010, 01:52 AMProve It
I assume this is actually

$\displaystyle \int_{-1}^{1}{\int_{\frac{y^2}{2}}^{\sqrt{3 - y^2}}{f(x, y)\,dx}\,dy}$.

It's important to realise that the terminals actually give you a boundary for $\displaystyle x$ and $\displaystyle y$.

So $\displaystyle \frac{y^2}{2}\leq x \leq \sqrt{3 - y^2}$

and $\displaystyle -1 \leq y \leq 1$.

You should be able to rearrange the inequalities so that you can change the order of integration. - Feb 17th 2010, 12:17 PMjacek
so the 'lines' of this are will be:

$\displaystyle y=-1$

$\displaystyle y=1$

$\displaystyle x=(y^2)/2$

$\displaystyle x= \sqrt{(3-y^2)}$? - Feb 17th 2010, 12:31 PMicemanfan
It really helps to draw a picture for this one. If you change the order of integration, you will need two integrals instead of one.

- Feb 17th 2010, 12:34 PMjacek
Yep, but should I draw it according to these 'lines' above?

- Feb 17th 2010, 12:42 PMicemanfan
- Feb 17th 2010, 12:47 PMjacek
That's ok, my teacher did it once only with integrals.

Could you tell me what's the easiest way to draw $\displaystyle x=\sqrt{(3-y^2)}$ without calculator? - Feb 17th 2010, 12:49 PMicemanfan
Square both sides and you get $\displaystyle x^2 = 3 - y^2$, or $\displaystyle x^2 + y^2 = 3$, which is a circle centered at the origin with radius $\displaystyle \sqrt{3}$. Note that $\displaystyle x=\sqrt{(3-y^2)}$ is only part of this circle. The best way to draw it would be with pencil and compass.

- Feb 17th 2010, 01:01 PMjacek
OK. Is is something like this? http://img692.imageshack.us/img692/3816/kolejnosc.png

- Feb 17th 2010, 01:03 PMicemanfan
- Feb 17th 2010, 01:07 PMjacek
Hehe, no! I can draw much better, but this is just auxiliary drawing, isn't it? :)

So, I need to divide it into three parts - part of circle on the right, rectangle in the middle, and this parabola part on the the left? - Feb 17th 2010, 01:09 PMicemanfan
- Feb 17th 2010, 03:39 PMjacek
Aha, thanks!

What about such integral?

$\displaystyle \int_{-1}^{1}dx \int_{0}^{ \sqrt{1-y^2}} f(y)$

How can I change it? Because it's only a rectangle... - Feb 17th 2010, 05:00 PMjacek
Anyone? :)