Answer: Integral of a Max with a Circle

$\boxed{\dfrac{\pi}{3} - \dfrac{\sqrt{3}}{4}}$

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Intuition First

The function $y = \sqrt{1-x^2}$ is the upper unit semicircle. The integrand $\max(0,\, \sqrt{1-x^2} - \tfrac{1}{2})$ is zero wherever the circle dips below the horizontal line $y = \tfrac{1}{2}$, and equals the height above that line wherever the circle is above it.

So the integral is simply the area of the circular segment of the unit disk that lies above the line $y = \tfrac{1}{2}$.

$\text{(integral)} = \text{(sector area)} - \text{(triangle area)}$

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Geometric Solution

Step 1 — Find where the circle crosses $y = \tfrac{1}{2}$.

$\sqrt{1-x^2} = \frac{1}{2} \implies x = \pm\frac{\sqrt{3}}{2}$

In polar terms, these are the points at angles $\theta = \tfrac{\pi}{6}$ and $\theta = \tfrac{5\pi}{6}$. The arc above $y = \tfrac{1}{2}$ subtends an angle of $\tfrac{5\pi}{6} - \tfrac{\pi}{6} = \tfrac{2\pi}{3}$.

Step 2 — Sector area.

$A_{\text{sector}} = \frac{1}{2} r^2 \cdot \theta = \frac{1}{2}(1)^2 \cdot \frac{2\pi}{3} = \frac{\pi}{3}$

Step 3 — Triangle area.

The sector contains a triangle with vertices at the origin and the two intersection points $\bigl(\pm\tfrac{\sqrt{3}}{2},\, \tfrac{1}{2}\bigr)$.

• Base: distance between the two points $= \sqrt{3}$
• Height: distance from origin to the chord $y = \tfrac{1}{2}$ is $\tfrac{1}{2}$

$A_{\text{triangle}} = \frac{1}{2} \cdot \sqrt{3} \cdot \frac{1}{2} = \frac{\sqrt{3}}{4}$

Step 4 — Circular segment area.

$\int_{-1}^{1} \max\!\left(0, \sqrt{1-x^2} - \frac{1}{2}\right)dx = A_{\text{sector}} - A_{\text{triangle}} = \frac{\pi}{3} - \frac{\sqrt{3}}{4}$

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Verification via Calculus

The integrand is nonzero only where $\sqrt{1-x^2} \geq \tfrac{1}{2}$, i.e. $|x| \leq \tfrac{\sqrt{3}}{2}$.

$\int_{-\sqrt{3}/2}^{\sqrt{3}/2} \!\left(\sqrt{1-x^2} - \frac{1}{2}\right)dx$

Using $\displaystyle\int \sqrt{1-x^2}\,dx = \frac{x\sqrt{1-x^2}}{2} + \frac{\arcsin x}{2} + C$:

$\left[\frac{x\sqrt{1-x^2}}{2} + \frac{\arcsin x}{2}\right]_{-\sqrt{3}/2}^{\sqrt{3}/2} - \frac{1}{2}\cdot\sqrt{3}$

$= 2\!\left(\frac{\frac{\sqrt{3}}{2}\cdot\frac{1}{2}}{2} + \frac{1}{2}\cdot\frac{\pi}{3}\right) - \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{4} + \frac{\pi}{3} - \frac{\sqrt{3}}{2} = \frac{\pi}{3} - \frac{\sqrt{3}}{4} \checkmark$