changing the order of integration

**jacek** · February 17th 2010, 01:47 AM

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?

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

**Prove It** · February 17th 2010, 01:52 AM

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.

**jacek** · February 17th 2010, 12:17 PM

so the 'lines' of this are will be:
$y=-1$
$y=1$
$x=(y^2)/2$
$x= \sqrt{(3-y^2)}$ ?

**icemanfan** · February 17th 2010, 12:31 PM

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.

**jacek** · February 17th 2010, 12:34 PM

Yep, but should I draw it according to these 'lines' above?

**icemanfan** · February 17th 2010, 12:42 PM

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.

**jacek** · February 17th 2010, 12:47 PM

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?

**icemanfan** · February 17th 2010, 12:49 PM

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.

**jacek** · February 17th 2010, 01:01 PM

OK. Is is something like this? http://img692.imageshack.us/img692/3816/kolejnosc.png

**icemanfan** · February 17th 2010, 01:03 PM

Originally Posted by jacek

OK. Is is something like this? http://img692.imageshack.us/img692/3816/kolejnosc.png

Something like that. Is that the best parabola you can draw?

**jacek** · February 17th 2010, 01:07 PM

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?

**icemanfan** · February 17th 2010, 01:09 PM

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.

**jacek** · February 17th 2010, 03:39 PM

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...

**jacek** · February 17th 2010, 05:00 PM

Anyone?

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