Problem with setting the region of integration

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Homework Statement
look at the image below!
Relevant Equations
double integrals
Good day !
I have a problem with the solution of the floowing integrals
Indeed i don't understand why they choose such borders for integral
b/a<c
y<c
doesn't mean that y<b/a !
many thanks in advance!

1614189909864.png
 

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Amaelle said:
doesn't mean that y<b/a !
For ##(x,y)\in\Omega##, yes that is correct. That's why they can set the upper limit on the outer integral to ##b/a## rather than ##c##, even though ##c## may be greater than ##b/a##.

The area of integration is a triangle with vertices (0,0), (0, b/a), (b,0).

Given that c>b/a, the value of c is not used in the calculation. I think they just put it in the question to confuse people!
 
that was an amzing shot! thanks a million!
 

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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