Evaluate the double integral over the square 0 <= x <= 1, 0 <= y <= 1 of the maximum of sqrt(2x - x^2) and 1 - y.

Step-by-step solution

1. 1

   Recognise the radical as a circle. Complete the square: $2x-x^2=1-(x-1)^2$, so $\sqrt{2x-x^2}=\sqrt{1-(x-1)^2}$ is the upper unit semicircle centred at $(1,0)$. Over $0\le x\le 1$ it rises from $0$ to $1$.

2. 2

   Substitute to centre the picture. Let $u=1-x$ and $v=1-y$. The Jacobian has absolute value $1$ and the square maps to itself, so the integral becomes $\iint_{[0,1]^2}\max\left\{\sqrt{1-u^2},\,v\right\}du\,dv$.

3. 3

   Split the square along the quarter circle. The two candidates are equal when $v=\sqrt{1-u^2}$, the quarter circle. Below it the radical wins; above it $v$ wins. Splitting there is what removes the max.

4. 4

   Integrate the lower piece. For each $u$, $v$ runs from $0$ to $\sqrt{1-u^2}$ and the integrand is the constant $\sqrt{1-u^2}$, contributing $\int_0^1\left(1-u^2\right)du=1-\frac13=\frac23$.

5. 5

   Integrate the upper piece. For each $u$, $v$ runs from $\sqrt{1-u^2}$ to $1$ with integrand $v$, giving $\int_0^1\frac{1-\left(1-u^2\right)}{2}du=\int_0^1\frac{u^2}{2}du=\frac16$.

6. 6

   Add and verify. The total is $\frac23+\frac16=\frac56$. A midpoint-rule grid of $4000\times4000$ cells on the original integrand returns $0.83333333$, matching $\frac56$.