TITLE: What is the area of $y^2=\sqrt x-x$ (Guitar Pick)
QUESTION [8 upvotes]: I made a typo while experimenting on Desmos and typed $y^2=\sqrt x-x$. It drew a shape, one that I've never seen before:
 With my very limited knowledge of calculus, I know the area would be equal to: $$2\int_0^1 \sqrt{\sqrt x - x} \,dx$$ However, I have no clue how to evaluate this integral. Using Desmos, I can get a decimal approximation (it's about 0.785), and Wolfram Alpha can give me the final result ($\pi/4$). No site I can think of has the solution and steps to solve it, so I figured I'd ask it here. How would you evaluate this integral?

REPLY [9 votes]: Start by substituting $x=t^2$. This gives
$$\begin{align}2\int_0^1\sqrt{\sqrt{x}-x}\,\mathrm dx &= 4 \int_0^1 t\sqrt{t-t^2}\,\mathrm dt \\ &= 4 \int_0^1 t\sqrt{\frac14-\Big(t-\frac12\Big)^2}\,\mathrm dt \\ &\overset{t-\frac12=u}{=} 4 \int_{-1/2}^{1/2} \Big(u+\frac12\Big)\sqrt{\frac14-u^2}\,\mathrm du \\ &= 4 \int_0^{1/2} \sqrt{\frac14-u^2} \,\mathrm du \quad \text{ odd and even integrand} \\ &= \left. 2u\sqrt{\frac14-u^2}+\frac12 \arcsin\Big(\frac u{1/2}\Big) \right|_0^{1/2}\\ &= \frac \pi4\end{align}$$
In the last step, I used the formula
$$\int \sqrt{a^2-x^2}\, dx = \frac x2\sqrt{a^2-x^2}+\frac{a^2}2\arcsin \frac xa+C$$