# changing the order of integration

• Feb 17th 2010, 02:47 AM
jacek
changing 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?

$\int_{-1}^{1}dy \int_{{y^2 \over 2}}^{ \sqrt{3-y^2}} f(x,y)dx$
• Feb 17th 2010, 02:52 AM
Prove It
Quote:

Originally Posted by jacek
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?

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

I assume this is actually

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

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

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

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

You should be able to rearrange the inequalities so that you can change the order of integration.
• Feb 17th 2010, 01:17 PM
jacek
so the 'lines' of this are will be:
$y=-1$
$y=1$
$x=(y^2)/2$
$x= \sqrt{(3-y^2)}$?
• Feb 17th 2010, 01:31 PM
icemanfan
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, 01:34 PM
jacek
Yep, but should I draw it according to these 'lines' above?
• Feb 17th 2010, 01:42 PM
icemanfan
Quote:

Originally Posted by jacek
Yep, but should I draw it according to these 'lines' above?

Yes. A slight correction: if you use only integrals to describe the area and you switch the order of integration, you will need three integrals, not two, although one of these is a rectangle.
• Feb 17th 2010, 01:47 PM
jacek
That's ok, my teacher did it once only with integrals.

Could you tell me what's the easiest way to draw $x=\sqrt{(3-y^2)}$ without calculator?
• Feb 17th 2010, 01:49 PM
icemanfan
Quote:

Originally Posted by jacek
That's ok, my teacher did it once only with integrals.

Could you tell me what's the easiest way to draw $x=\sqrt{(3-y^2)}$ without calculator?

Square both sides and you get $x^2 = 3 - y^2$, or $x^2 + y^2 = 3$, which is a circle centered at the origin with radius $\sqrt{3}$. Note that $x=\sqrt{(3-y^2)}$ is only part of this circle. The best way to draw it would be with pencil and compass.
• Feb 17th 2010, 02:01 PM
jacek
• Feb 17th 2010, 02:03 PM
icemanfan
Quote:

Originally Posted by jacek

Something like that. Is that the best parabola you can draw?
• Feb 17th 2010, 02:07 PM
jacek
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, 02:09 PM
icemanfan
Quote:

Originally Posted by jacek
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?

Yes, that is correct.
• Feb 17th 2010, 04:39 PM
jacek
Aha, thanks!

What about such integral?

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

How can I change it? Because it's only a rectangle...
• Feb 17th 2010, 06:00 PM
jacek
Anyone? :)