Show that $\int_0^1\int_{1-y}^1\sqrt{(x-1)(y-1)(x+y-1)}\mathrm dx\mathrm dy=\frac{2\pi}{105}$.

Question

How can we show that $\int_0^1\int_{1-y}^1\sqrt{(x-1)(y-1)(x+y-1)}\mathrm dx\mathrm dy=\frac{2\pi}{105}$ ?

Desmos says it's true.

The inner indefinite integral is not nice. And strangely, when I plug in the inner limits of integration $x=1$ and $x=1-y$, I get $0-0=0$, contradicting the Desmos result. (Edit: I should cancel factors in numerator and denominator, then I don't get $0-0$. But then I get $\int_0^1 -\frac14 y^2\sqrt{y-1}\sinh^{-1}(i) \mathrm dy$ and I don't know how to continue.)

Based on another question of mine about a double integral, I tried substituting $a=x-1$ and $b=y-1$, which leads to $\int_{-1}^0\int_{-1-b}^0\sqrt{ab(a+b+1)}\mathrm da\mathrm db$, which has another inner indefinite integral that is not nice.

Context: I asked myself, "A unit stick is broken at two uniformly random points; given that the three pieces form a triangle (which has probability $\frac14$), what is the expected area of the triangle? The answer is $\frac12$ times the integral in this question (which is supported by experimental trials, and also this article). This question was inspired by another question about breaking a stick at random points.

Answer 1

Inner integral does not yield $0-0$. In fact, it is the difference of these two: $$ {\frac {{y}^{2}\arcsin(1)}{16}\sqrt {1-y}} \\ \text{and} \\ {\frac {{y}^{2}\arcsin(-1)}{16}\sqrt {1-y}} $$ They are not the same.

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Added.

The inner integral.
In your attempt, you wrote $\sqrt{y-1}$, square-root of a negative number. Do you really want to do that? You wrote $\sinh^{-1}(i)$ which is $i \sin^{-1}(1)=i\pi/2$.

Using only real nubers, the inner integral is $$ {\frac {{y}^{2}\pi}{8}\sqrt {1-y}} . $$

Are you really saying you don't know how to do $$ \frac {\pi}{8}\int_0^1 {y}^{2}\sqrt {1-y}\;dy $$

Answer 2

Continue with $$I= \int_{-1}^0\int_{-1-b}^0\sqrt{ab(a+b+1)}dadb $$ along with the variable changes $a={u+v}$ and $b={u-v}$, to decouple the double integral

\begin{align} I= \int_{-\frac1{2}}^0\int^{-u}_u 2\sqrt{(u^2-v^2)(2u+1)}dv du \end{align} Then, utilize $\int_{-u}^u \sqrt{u^2–v^2}dv=\frac\pi2 u^2$ to obtain

$$I=\pi \int_{-\frac1{2}}^0 {u^2} \sqrt{2u+1}\ du=\frac{2\pi}{105} $$

Answer 3

Use the following result which is very interested: $$\int_{a}^{b} \sqrt{(b-x)(x-a)} \mathrm{d}x=\frac{\pi(b-a)^2}{8}.\tag 1$$ For detail we can refer Area of semi-disk.

Let $$D=\{(x,y)\mid0\leq y\leq1,1-y\leq x\leq1\},$$ $$f(x,y)=\sqrt{(1-x)(1-y)(x+y-1)},\quad (x,y)\in D,$$ then $$I=\int_0^1\int_{1-y}^1\sqrt{(x-1)(y-1)(x+y-1)}\mathrm dx\mathrm dy =\iint_{D}f(x,y)\mathrm dx\mathrm dy.$$ By $(1)$, we get $$\int_{1-y}^1\sqrt{(1-x)[x-(1-y)]}\ \mathrm dx=\frac{\pi y^2}{8},$$ so $$I=\int_0^1\sqrt{1-y}\cdot\frac{\pi y^2}{8}\mathrm dy =\frac{2\pi}{105}.$$

Answer 4

Self-answering.

With help from Wolfram for the inner anti-derivative, and using $\sinh^{-1}A=\log(A+\sqrt{A^2+1})$, we have:

$\begin{align} \int_{1-y}^1\sqrt{(x-1)(y-1)(x+y-1)}\mathrm dx&=\left[-\frac14y^2\sqrt{y-1}\ln\left({\sqrt{\frac{x-1}{y}}+\sqrt{\frac{x+y-1}{y}}}\right)\right]_{1-y}^1\\ &=-\frac14y^2\sqrt{y-1}\ln{i}\\ &=-\frac{\pi i}{8}y^2\sqrt{y-1}\\ &=\frac{\pi}{8}y^2\sqrt{1-y} \end{align}$

Then the integral in the OP is

$$\frac{\pi}{8}\int_0^1y^2\sqrt{1-y}\mathrm dy=\frac{\pi}{8}\left[-\frac{2}{105}(1-y)^{3/2}(15y^2+12y+8)\right]_0^1=\frac{2\pi}{105}$$

Answer 5

At the begining, if we write the inner integral as $$\sqrt{1-y}\int_{1-y}^1\sqrt{(1-x)(y+x-1)}dx$$ then by $1-x=u$ $$\sqrt{1-y}\int_{0}^{y}\sqrt{yu-u^2}du$$ and by $\frac y2-u=\frac y2\sin v$, we have $$\frac14y^2\sqrt{1-y}\int_{\frac{-\pi}2}^{\frac{\pi} 2}\cos^2vdv=\frac\pi 8 y ^2\sqrt{1-y}$$ Trigonometric way we know, works in real world. I mean no i.

Answer 6

$$ \begin{aligned} I & =\int_0^1 \int_{1-y}^1 \sqrt{(1-x)(1-y)(x+y-1)} d x d y \\ & =\int_0^1 \sqrt{1-y} \int_{1-y}^1 \sqrt{(1-x)(x+y-1)} d x d y \\ & =\int_0^1 \sqrt{1-y} \cdot\frac{y^2}{2} \int_0^{\frac{\pi}{2}} \sin ^2 2 \theta d \theta,\quad \textrm{via the substitution } y\sin \theta=1-x\\&= \frac{\pi}{8} \int_0^1 y^2 \sqrt{1-y}\ d y\\& =\frac{2 \pi}{105} \quad (\textrm{ via integration by parts twice }) \end{aligned} $$