Really simple question, but I'm having trouble grasping this

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Why is the area of the cardioid r = 2sinθ-2 in the first quadrant represented with limits from π to 3π/2 and not with limits from 0 to π/2? Isn't the first quadrant 0 to π/2?
 
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r is negative (or zero) for all values of theta, so when theta is between pi and 3 pi / 2, (r, theta) is in the first quadrant.
 
Actually, never mind, I'm still not understanding this :(
 
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Try plotting the point for theta = 5 pi / 4. Maybe you will see it then.
 
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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