Double integral

**acevipa** · November 6th 2010, 04:03 PM

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

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

Would it be like this

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

**Ackbeet** · November 6th 2010, 04:57 PM

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

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

**acevipa** · November 6th 2010, 05:24 PM

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.

**Ackbeet** · November 6th 2010, 06:16 PM

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.

**acevipa** · November 6th 2010, 06:24 PM

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

**Ackbeet** · November 6th 2010, 06:26 PM

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

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