Volume of the solid enclosed by $y²+z²=x, x=y, z²=x-y²$ . (Piskunov Chapter XIV problem 39) Is there any change of variables that simplify the calculation of the following double integral ? $$\int_{0}^{1}\int_{x}^{\sqrt{x}}\sqrt{x-y^2}\;dy\,dx$$ This integral arises from exercise 39 of chapter 14 of Piskunov's book Differential and Integral Calculus, in which he says to calculate the volume of the solid bounded by: $y^2+z^2=x \,,\,x=y\,,\,z^2=x-y^2 .$ When performing this integration in cartesian coordinates it takes a very long and tortuous path in which things like $-\frac{\sin \left(2\,\cdot\,\arcsin \left(\sqrt{x}\right)\right)}{4}$ begin to appear.

$$\iint_{0\leq x\leq y \leq \sqrt{x}\leq 1}\sqrt{x-y^2}\,dx\,dy = \int_{0}^{1}\int_{y^2}^{y}\sqrt{x-y^2}\,dx\,dy=\int_{0}^{1}\frac{2}{3}y^{3/2}(1-y)^{3/2}\,dy$$ by Fubini's theorem.
The last integral equals $\frac{2}{3}B\left(\frac{5}{2},\frac{5}{2}\right)=\frac{2\,\Gamma\left(\frac{5}{2}\right)^2}{3\,\Gamma(5)} =\color{red}{\large\frac{\pi}{64}}$ by Euler's Beta function.