Quick cal question (probably simple

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Homework Statement


The base of a soild is the region bounded by y=x^2 and y=x. Find the volume if cross sections perpendicular to the x-axis are squares


Homework Equations


Integral of A(x)dx


The Attempt at a Solution



Ok, so I have to take the integral from 0 to 1 after solving for X

since it is a square, the setup is side^2dx... For some reason i cannot figure out the value used for side using 2 functions...

I believe it should be something similar to 2(side)^2dx evaluating from 0 to 1, but what is the value for side and how was it found?
 
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y=x^2 is below y=x between x=0 and x=1. So isn't 'side' x-x^2, the distance between them??
 

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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