Changing the order of integration

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Homework Statement
Calculate the following integral (look at the image)
Relevant Equations
Double integrals
Greetings!
As mentionned my aim is to change the order of integral, and I totally agree with the solution I just have one question:
as you can see they have put
0<=y<=1 and 0<=x<=y^2
but would it be wrong if I put
0<=y<=1 and y^2<=x<=1?
Thank you!
1622822248789.png
 
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Yes, it would be wrong. That would be the other part of the square in your figure.
 
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Orodruin said:
Yes, it would be wrong. That would be the other part of the square in your figure.
I got it now, thanks a lot!
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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