Find the volume of the solid generated by rotating one

Find the volume of the solid generated by rotating one arch of the cycloid $x = a(t - \sin t)$, $y = a(1 - \cos t)$ about: 1° the $x$-axis, 2° the $y$-axis, 3° the line $x = a\pi$.

One arch corresponds to $t\in[0,2\pi]$, running from $(0,0)$ up to $(a\pi,2a)$ at the crown and back down to $(2a\pi,0)$; the area under one arch is the classical result $3\pi a^2$.

Part 1: Rotation about the $x$-axis

For a parametric curve, $V_x=\pi\int y^2\,dx = \pi\int_0^{2\pi}[a(1-\cos t)]^2\cdot a(1-\cos t)\,dt = \pi a^3\int_0^{2\pi}(1-\cos t)^3\,dt$.

Expanding $(1-\cos t)^3$ and using $\int_0^{2\pi}\cos t\,dt=\int_0^{2\pi}\cos2t\,dt=\int_0^{2\pi}\cos^3t\,dt=0$, only the constant term $\frac52$ survives:

\[\int_0^{2\pi}(1-\cos t)^3\,dt = 5\pi\]

\[\boxed{V_x = 5\pi^2a^3 \text{ cubic units}}\]

Part 2: Rotation about the $y$-axis

Since the whole arch lies at $x\ge0$ (touching $x=0$ only at the single point $t=0$), it does not cross the axis, so Pappus's theorem applies directly to the full region: $V_y=2\pi\bar x\,A$. By the arch's left-right symmetry about $x=a\pi$, the centroid satisfies $\bar x=a\pi$ exactly, and $A=3\pi a^2$:

\[V_y = 2\pi\cdot a\pi\cdot3\pi a^2 = 6\pi^3a^3\]

\[\boxed{V_y = 6\pi^3a^3 \text{ cubic units}}\]

(This can also be confirmed directly via the shell-method integral $V_y=2\pi\int_0^{2\pi}xy\,\frac{dx}{dt}\,dt$, which evaluates to the same $6\pi^3a^3$.)

Part 3: Rotation about the line $x=a\pi$

Here the axis $x=a\pi$ passes straight through the middle of the region, which is exactly mirror-symmetric about it — the same situation as Problem 69's rotation about $x=y$. Revolving the entire region about $x=a\pi$ therefore sweeps out only what one symmetric half alone would sweep (the other half is its mirror image and traces an identical solid, not an additional one); so we integrate over just the left half, $t\in[0,\pi]$ (where $x\le a\pi$), using the shell method with radius $a\pi-x$, without an extra doubling factor for "both halves":

\[V_{x=a\pi} = 2\pi\int_0^{\pi}(a\pi-x)\,y\,\frac{dx}{dt}\,dt\]

Carrying out this integral (a polynomial-in-$\cos t,\sin t,t$ computation) gives:

\[V_{x=a\pi} = \frac{\pi a^3(9\pi^2-16)}{6}\]

This is confirmed independently via Pappus's theorem applied to just the left half of the region: with half-area $A_{\text{half}}=\frac{3\pi a^2}{2}$ and half-centroid $\bar x_{\text{half}}=\dfrac{a(16+9\pi^2)}{18\pi}$, the distance to the axis is $a\pi-\bar x_{\text{half}}$, and $V=2\pi(a\pi-\bar x_{\text{half}})A_{\text{half}}$ reproduces the same value exactly.

\[\boxed{V_{x=a\pi} = \frac{\pi a^3(9\pi^2-16)}{6} \text{ cubic units}}\]

Summary

\[\boxed{ \begin{array}{ll} 1^{\circ} \text{ About } x\text{-axis:} & V_x = 5\pi^2a^3 \approx 49.35\,a^3 \\[0.5em] 2^{\circ} \text{ About } y\text{-axis:} & V_y = 6\pi^3a^3 \approx 186.0\,a^3 \\[0.5em] 3^{\circ} \text{ About } x=a\pi: & V_{x=a\pi} = \dfrac{\pi a^3(9\pi^2-16)}{6} \approx 37.15\,a^3 \end{array} }\]

Key insight: Rotations about the $x$-axis and $y$-axis are handled by the ordinary parametric disk/Pappus formulas since the region never crosses either axis; but rotating about the arch's own vertical axis of symmetry $x=a\pi$ requires recognizing that the region straddles that axis in mirror-image halves, so only one half should be integrated — exactly the same subtlety that appears in Problem 69's rotation about $x=y$.