Evaluate $\int_{0}^{1} \int_{0}^{1-x} \int_{0}^{1-x-y} \sqrt{\frac{x}{y z}} \,d z \,d y \,d x$ $$
\text { Evaluate } \int_{0}^{1} \int_{0}^{1-x} \int_{0}^{1-x-y} \sqrt{\frac{x}{y z}} \,d z\, d y\, d x
$$
How to solve this ?
I tried this
$\begin{aligned} & \int_{0}^{1} \int_{0}^{1-x} \int_{0}^{1-x-y} \sqrt{\frac{x}{y z}} d z d y d x \\
=& \int_{0}^{1} \int_{0}^{1-x} \frac{\sqrt{x}}{\sqrt{y}}[2 \sqrt{z}]{0}^{1-x-y} d y d x \\
=& \int_{0}^{1} \int_{0}^{1-x} \frac{\sqrt{x}}{\sqrt{y}} 2 \sqrt{1-x-y} d y d x \\
=& \int_{0}^{1} \int_{0}^{1-x} 2 \sqrt{x} \frac{\sqrt{1-x-y}}{\sqrt{y}} d y d x \\=& \int_{0}^{1} \int_{0}^{1-x} 2 \sqrt{x} \sqrt{\frac{1-x-y}{y}} d y d x \end{aligned}$
 A: First make a substitution $1-y\to y$ to get that this is equivalent to
$$\int_0^1\int_x^1 2\sqrt{x}\sqrt{\frac{y-x}{1-y}}\, dy\, dx$$
Next, switch the order of integetration,
$$\int_0^1\int_0^y \frac{2}{\sqrt{1-y}}\sqrt{xy-x^2}\, dx\, dy$$
$$=\int_0^1\int_0^y \frac{1}{\sqrt{1-y}}\sqrt{4xy-4x^2-y^2+y^2}\, dx\, dy$$
$$=\int_0^1\int_0^y \frac{1}{\sqrt{1-y}}\sqrt{y^2-(2x-y)^2}\, dx\, dy$$
$$=\int_0^1\frac{1}{\sqrt{1-y}}\int_0^y \sqrt{y^2-(2x-y)^2}\, dx\, dy$$
You can make a quick sketch of the graph of $\sqrt{y^2-(2x-y)^2}$ where y is a parameter ranging from $0$ to $1$ to see that geometrically
$$\int_0^y \sqrt{y^2-(2x-y)^2}\, dx=\frac{\pi y^2}{4}$$
Hence, our integral is equivalent to
$$\frac{\pi}{4}\int_0^1 \frac{y^2}{\sqrt{1-y}}\, dy$$
Make a substitution of $u=\sqrt{1-y}$ to get
$$=\frac{\pi}{2}\int_0^1(1-u^2)^2\, du$$
$$=\frac{\pi}{2}\int_0^1 u^4-2u^2+1\, du$$
$$=\frac{\pi}{2}\left(\frac{1}{5}-\frac{2}{3}+1\right)$$
$$=\frac{4\pi}{15}$$
I'm not sure how much of this is necessary, but I believe that switching the order of integration is an essential step.
