# Thread: Double integral

1. ## Double integral

$\displaystyle A=\int_{-1}^0\int_0^{(x+1)^2}f(x,y) \ dy \ dx + \int_0^1\int_0^{(x-1)^2} f(x,y) \ dy \ dx
$

By reversing the order of integration, express A as a single iterated integral.

Would it be like this

$\displaystyle A=\int_{0}^1\int_{\sqrt{y}-1}^{0}f(x,y) \ dx \ dy + \int_0^1\int_0^{\sqrt{y}+1} f(x,y) \ dx \ dy$

2. On the second integral, I think you need to take the negative square root. That is, I think you should have

$\displaystyle\int_{0}^{1}\int_{0}^{-\sqrt{y}+1}f(x,y)\,dx\,dy.$

3. Originally Posted by Ackbeet
On the second integral, I think you need to take the negative square root. That is, I think you should have

$\displaystyle\int_{0}^{1}\int_{0}^{-\sqrt{y}+1}f(x,y)\,dx\,dy.$
The answer is apparently

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

I don't understand why.

4. If the book had said

$\displaystyle\int_{0}^{1}\int_{\sqrt{y}-1}^{-\sqrt{y}+1}f(x,y)\,dx\,dy,$

I would have believed it.

5. Originally Posted by Ackbeet
If the book had said

$\displaystyle\int_{0}^{1}\int_{\sqrt{y}-1}^{-\sqrt{y}+1}f(x,y)\,dx\,dy.$

I would have believed it.
Yeah Ackbeet, I managed to now get the same answer as you. Maybe the answer at the back of the book is wrong. Thanks anyways

6. I think the book is wrong. You're welcome!