Changing Order of Integration: A Refresher for Finals

[rocomath]

Ok, need refresher b4 finals!

Change order of integration

$$\int_{-2}^0\int_0^{\sqrt{4-x^2}}f(x,y)dydx$$

to

$$\int_0^2\int_{-\sqrt{4-y^2}}^0f(x,y)dxdy$$ going from right to left, and the fact that my original integral is the part of the circle in quadrant 2, it should be $$x=-\sqrt{4-y^2}$$

or

$$\int_0^2\int_0^{\sqrt{4-y^2}}f(x,y)dxdy$$ this is my original answer, but now I'm thinking it's the 1st one :-\

 

[Hurkyl]

Your geometric argument looks correct. You can argue in a purely algebraic fashion too -- you need to argue in two steps:

(1) Starting with the inequations $-2 \leq x \leq 0$ and $0 \leq y \leq \sqrt{4 - x^2}$, derive the inequations $0 \leq y \leq 2$ and $-\sqrt{4 - y^2} \leq x \leq 0$.

(2) Starting with the inequations $0 \leq y \leq 2$ and $-\sqrt{4 - y^2} \leq x \leq 0$, derive the inequations $-2 \leq x \leq 0$ and $0 \leq y \leq \sqrt{4 - x^2}$

If you can do that, then you will have (algebraically!) shown that those two pairs of inequations describe the same shape.

 

[rocomath]

Hurkyl, my brain is fried ... lol. What do you mean by inequations?

My original integral

$$0\leq y\leq\sqrt{4-x^2}$$ is the positive semi circle

$$-2\leq x\leq 0$$ so i now have just a quarter of a circle in quadrant 2

 

[Defennder]

He means the inequalities. If you can show by means of inequality maniupulations that you can start with one form and arrive at the other, as well as vice versa, then you would have have shown that the region under consideration is the same, ie. you have sucessfully obtained the correct limits for the change of order in integration. Your method is purely graphical, which is also fine.

 

[rocomath]

So both answers are correct? :O :O :O

 

[Defennder]

Your original answer is correct.