Answer
The answer is $4\pi^2r^2$
Work Step by Step
The surface area of the solid obtained by rotating the circle $x^2 + y^2 = r^2$ about the line $y = r$ can be found using Pappus's centroid theorem. According to this theorem, the surface area of a surface of revolution generated by rotating a plane curve about an external axis is equal to the product of the arc length of the curve and the distance traveled by its geometric centroid. In this case, the curve being rotated is a semicircle with radius $r$, so its arc length is $\pi r$. The distance between the centroid of a semicircle and its base is $\frac{4r}{3\pi}$, so the distance between the centroid of the semicircle and the line $y = r$ is $r + \frac{4r}{3\pi}$.
Therefore, according to Pappus's centroid theorem, the surface area of the solid obtained by rotating the circle $x^2 + y^2 = r^2$ about the line $y = r$ is:
$$A = (\text{arc length}) \times (\text{distance traveled by centroid})$$
$$= (\pi r) \times ( 2\pi(r + \frac{4r}{3\pi}))$$
$$= 2\pi^2 r(r + \frac{4r}{3\pi})$$
$$= 2\pi^2 r(\frac{3\pi r + 4r}{3\pi})$$
$$= 2\pi^2 r(\frac{7}{3})r$$
$$= 4\pi^2 r^2$$
So, the surface area of the solid obtained by rotating the circle $x^2 + y^2 = r^2$ about the line $y = r$ is indeed $4\pi^2r^2$. Is there anything else you would like to know?
