A sector of a semi-circle which is $60^\circ$ has an area of $\frac{3\pi}{2}$ units squared.
Show that the curved section is a function of the form $f(x)= \sqrt{9-x^2}$ with domain $[0,\frac{3\sqrt{3}}{2}]$
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Area of sector \begin{align} A&=\frac{60^\circ}{360^\circ}\pi r^2\\ \frac{3\pi}{2}&=\frac{1}{6}\pi r^2\\ r&=3 \end{align} Equation of circle with center on the origin \begin{align} x^2+y^2&=r^2\\ x^2+y^2&=3^2\\ y&=\sqrt{9-x^2} \end{align} then use parametric equation (polar coordinate) $x=r\cos\theta$ and $y=r\sin\theta$ to determine the domain.
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