# Thread: changing the order of integration

1. ## 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?

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

2. 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?

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

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

4. 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.

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

6. 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.

7. 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?

8. 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 $\displaystyle x=\sqrt{(3-y^2)}$ without calculator?
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.

9. Originally Posted by jacek
Something like that. Is that the best parabola you can draw?

10. 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?

11. 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.

12. 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...

13. Anyone?