Evaluate the integral from y = 0 to root3/2, of the integral from x = 1 - root(1 - y squared) to root(1 - y squared), of x + xy dx dy.

Step-by-step solution

1. 1

   Understand the region before integrating. The outer limit $x=\sqrt{1-y^2}$ is the right half of the unit circle centred at the origin; the inner limit $x=1-\sqrt{1-y^2}$ is the left arc of the unit circle centred at $(1,0)$. So the region is the horizontal strip between two circular arcs, from $y=0$ up to $y=\tfrac{\sqrt3}{2}$.

2. 2

   Factor the integrand and pull out what is constant in x. $x+xy = x(1+y)$, and $1+y$ does not involve $x$: $I = \int_0^{\sqrt3/2}(1+y)\left(\int_{1-\sqrt{1-y^2}}^{\sqrt{1-y^2}} x\,dx\right)dy.$

3. 3

   Do the inner integral with the difference-of-squares shortcut. Writing $a=\sqrt{1-y^2}$, $\int_{1-a}^{a} x\,dx = \frac{a^2-(1-a)^2}{2} = \frac{2a-1}{2} = a-\frac12 = \sqrt{1-y^2}-\frac12.$ Expanding $(1-a)^2$ and cancelling $a^2$ is what makes this collapse so cleanly.

4. 4

   Split the remaining single integral. $I = \underbrace{\int_0^{\sqrt3/2}\sqrt{1-y^2}\,dy}_{A_1} + \underbrace{\int_0^{\sqrt3/2} y\sqrt{1-y^2}\,dy}_{A_2} - \underbrace{\frac12\int_0^{\sqrt3/2}(1+y)\,dy}_{B}.$

5. 5

   Evaluate the three pieces. $A_1$ uses the standard antiderivative $\tfrac12\left(y\sqrt{1-y^2}+\arcsin y\right)$; with $\arcsin\tfrac{\sqrt3}{2}=\tfrac{\pi}{3}$ this gives $\tfrac{\sqrt3}{8}+\tfrac{\pi}{6}$. $A_2$ uses $u=1-y^2$: $-\tfrac13(1-y^2)^{3/2}$ evaluated from $0$ to $\tfrac{\sqrt3}{2}$ gives $\tfrac13-\tfrac{1}{24}=\tfrac{7}{24}$. And $B = \tfrac12\left(\tfrac{\sqrt3}{2}+\tfrac38\right) = \tfrac{\sqrt3}{4}+\tfrac{3}{16}$.

6. 6

   Combine like terms. The $\sqrt3$ terms give $\tfrac{\sqrt3}{8}-\tfrac{\sqrt3}{4} = -\tfrac{\sqrt3}{8}$, and the rationals give $\tfrac{7}{24}-\tfrac{3}{16} = \tfrac{14-9}{48} = \tfrac{5}{48}$. Hence $I = \frac{\pi}{6}-\frac{\sqrt3}{8}+\frac{5}{48} \approx 0.411259.$

7. 7

   Check by quadrature. Numerically integrating the inner-result function $(1+y)\left(\sqrt{1-y^2}-\tfrac12\right)$ over $[0,\tfrac{\sqrt3}{2}]$ gives $0.4112590913$, agreeing with the closed form to ten decimals.