Discussion
The semi-circle \(y = \sqrt{6x - 5 - x^{2}}\) is rotated clockwise by the angle \(\frac{\pi}{2}\) in the plane about the origin.
What is the area of the planar region traced out by the semi-circle?
Answer: A
The area traced out is \(6\pi\).
Squaring \(y = \sqrt{6x - 5 - x^{2}}\) gives \((x - 3)^{2} + y^{2} = 4\) with \(y \geq 0\): the upper half of the circle of radius 2 centred at \((3,0)\), running from \((1,0)\) to \((5,0)\). Its distance from the origin ranges over \([1,5]\).
Each circle \(x^{2} + y^{2} = r^{2}\) with \(1 \leq r \leq 5\) meets the semi-circle at exactly one point, where \(x = \frac{r^{2} + 5}{6}\). A rotation about the origin keeps every point on its own circle, so that point sweeps an arc of angle \(\frac{\pi}{2}\) and length \(\frac{\pi}{2}r\). Adding up the arcs:
\[ \begin{aligned} \text{Area} &= \int_{1}^{5} \frac{\pi}{2}\,r\,dr \\ &= \frac{\pi}{2} \cdot \frac{25 - 1}{2} = 6\pi \end{aligned} \]
This is the area of a quarter of the annulus between radii 1 and 5. \(2\pi\) is the area of the half-disc under the semi-circle, and \(4\pi\) is the area of the whole disc of radius 2; neither accounts for the sweep.
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