Find the surface area generated by rotating one arch

Find the surface area generated by rotating one arch of the cycloid $x=a(t-\sin t),\ y=a(1-\cos t)$: a) about the $x$-axis, b) about the line $x=\pi a$.

Part a: About the $x$-axis

Step 1: Set up

With $\sqrt{(dx/dt)^2+(dy/dt)^2}=2a\sin\frac t2$ (from Problem 93):

\[S_x=2\pi\int_0^{2\pi}a(1-\cos t)\cdot2a\sin\frac t2\,dt\]

Step 2: Simplify using $1-\cos t=2\sin^2\frac t2$

\[S_x=8\pi a^2\int_0^{2\pi}\sin^3\frac t2\,dt = 16\pi a^2\int_0^\pi\sin^3u\,du\qquad(u=t/2)\]

Step 3: Evaluate $\int_0^\pi\sin^3u\,du$

\[\int_0^\pi\sin^3u\,du=\int_0^\pi\sin u(1-\cos^2u)\,du = 2-\frac23=\frac43\]

Step 4: Result

\[S_x=16\pi a^2\cdot\frac43=\frac{64\pi a^2}3\]

\[\boxed{S_x=\frac{64\pi a^2}3\approx67.02a^2}\]

Part b: About the line $x=\pi a$

Step 1: Distance from the axis

The distance from a point on the cycloid to the line $x=\pi a$ is $r(t)=|\pi a-a(t-\sin t)|=a|\pi-t+\sin t|$. This is nonnegative on $[0,\pi]$ and nonpositive on $[\pi,2\pi]$; by the arch's symmetry about $t=\pi$, the two halves contribute equally, so

\[S_{x=\pi a}=2\cdot2\pi\int_0^\pi a(\pi-t+\sin t)\cdot2a\sin\frac t2\,dt = 8\pi a^2\int_0^\pi(\pi-t+\sin t)\sin\frac t2\,dt\]

Step 2: Split into three integrals

\[8\pi a^2\left[\pi\int_0^\pi\sin\frac t2\,dt - \int_0^\pi t\sin\frac t2\,dt + \int_0^\pi\sin t\sin\frac t2\,dt\right]\]

Step 3: Evaluate each

First: $\displaystyle\int_0^\pi\sin\frac t2\,dt=\left[-2\cos\frac t2\right]_0^\pi=2$.

Second (by parts, $u=t,\ dv=\sin\frac t2\,dt$): $\displaystyle\int_0^\pi t\sin\frac t2\,dt=\left[-2t\cos\frac t2\right]_0^\pi+2\int_0^\pi\cos\frac t2\,dt = 0+2\left[2\sin\frac t2\right]_0^\pi=4$.

Third (using $\sin t=2\sin\frac t2\cos\frac t2$, $u=\sin\frac t2$): $\displaystyle\int_0^\pi2\sin^2\frac t2\cos\frac t2\,dt=2\left[\frac{2u^3}3\right]_0^1=\frac43$.

Step 4: Combine

\[S_{x=\pi a}=8\pi a^2\left[2\pi-4+\frac43\right]=8\pi a^2\left[2\pi-\frac83\right]=16\pi^2a^2-\frac{64\pi a^2}3\]

\[\boxed{S_{x=\pi a}=\frac{16\pi a^2(3\pi-4)}3}\]

Step 5: Numerical value

\[S_{x=\pi a}=\frac{16\pi a^2(3\pi-4)}3\approx90.89a^2\]