Calculate the surface area generated by rotating

Calculate the surface area generated by rotating the curve $x^2+y^2=2y$ about the $x$-axis.

Step 1: Identify the curve

\[x^2+(y-1)^2=1\]

A circle of radius $1$ centered at $(0,1)$, so it lies entirely in $y\ge0$ and touches the $x$-axis (the rotation axis) only at the single point $(0,0)$.

Step 2: Both semicircles

$y_\pm=1\pm\sqrt{1-x^2}$, for $x\in[-1,1]$; in both cases $\sqrt{1+(y_\pm')^2}=\dfrac1{\sqrt{1-x^2}}$.

Step 3: Add the two branch integrals

\[S=S_++S_-=2\pi\int_{-1}^1\left[(1+\sqrt{1-x^2})+(1-\sqrt{1-x^2})\right]\frac1{\sqrt{1-x^2}}\,dx = 4\pi\int_{-1}^1\frac{dx}{\sqrt{1-x^2}}\]

Step 4: Evaluate

\[S=4\pi\big[\arcsin x\big]_{-1}^1=4\pi\left[\frac\pi2-\left(-\frac\pi2\right)\right]=4\pi^2\]

\[\boxed{S=4\pi^2\approx39.48}\]

Step 5: Cross-check via Pappus's theorem

The circle has circumference $2\pi(1)=2\pi$ and centroid at its center $(0,1)$, distance $1$ from the axis; since the circle touches the axis only at a single point (measure zero) and does not cross it, Pappus's theorem applies:

\[S=2\pi\cdot(\text{centroid distance})\cdot(\text{circumference}) = 2\pi\cdot1\cdot2\pi=4\pi^2\quad\checkmark\]

This is the degenerate ("horn") torus case where the tube radius equals the distance to the axis.