曲线 $C$ 由参数方程定义 $x = t - \sin t$，$y = \sin^{2} t$，$\;0 \le t \le \pi$。 曲线 $C$ 上的点 $A$ 对应的参数为 $\dfrac{\pi}{2}$，且是一个极大值点。直线 $l$ 穿过原点和点 $A$。区域 $R$ 由 $C$ 和 $l$ 围成，如下图所示。

1. Substitute the parameter $t = \frac{\pi}{2}$ into the parametric equations to find the coordinates of $A$.

Given $t = \frac{\pi}{2}$,
$x = \frac{\pi}{2} - \sin\left(\frac{\pi}{2}\right) = \frac{\pi}{2} - 1$
$y = \sin^{2}\left(\frac{\pi}{2}\right) = 1^{2} = 1$

The coordinates of $A$ are $\left(\frac{\pi}{2} - 1, 1\right)$.

1. Find the exact coordinates of point $A$.

Given that point $A$ corresponds to parameter $t = \frac{\pi}{2}$, we substitute this into the parametric equations:
$x = \frac{\pi}{2} - \sin\left(\frac{\pi}{2}\right) = \frac{\pi}{2} - 1$
$y = \sin^2\left(\frac{\pi}{2}\right) = 1$
So, the coordinates of $A$ are $\left(\frac{\pi}{2} - 1, 1\right)$.

2. Set up the parametric integral for the area under the curve $C$.

The area under curve $C$ from $x = 0$ to $x = \frac{\pi}{2} - 1$ is given by $\int y \, dx$.
Using parametric integration, we substitute $y$ and $\frac{dx}{dt}$:
$\frac{dx}{dt} = 1 - \cos t$
$\text{Area under } C = \int_{t=0}^{t=\pi/2} y \frac{dx}{dt} \, dt$
$= \int_{0}^{\pi/2} \sin^2 t (1 - \cos t) \, dt$
$= \int_{0}^{\pi/2} (\sin^2 t - \sin^2 t \cos t) \, dt$

3. Evaluate the exact area under the curve using trigonometric identities and standard integrals.

To integrate $\sin^2 t$, we use the double-angle identity: $\sin^2 t = \frac{1 - \cos 2t}{2}$.
The integral becomes:
$\text{Area under } C = \int_{0}^{\pi/2} \left( \frac{1}{2} - \frac{1}{2}\cos 2t - \sin^2 t \cos t \right) \, dt$
Integrating term by term gives:
$= \left[ \frac{1}{2}t - \frac{1}{4}\sin 2t - \frac{1}{3}\sin^3 t \right]_{0}^{\pi/2}$
Substituting the upper limit $t = \frac{\pi}{2}$:
$\frac{1}{2}\left(\frac{\pi}{2}\right) - \frac{1}{4}\sin(\pi) - \frac{1}{3}\sin^3\left(\frac{\pi}{2}\right) = \frac{\pi}{4} - 0 - \frac{1}{3}$
Substituting the lower limit $t = 0$ gives $0$.
So, $\text{Area under } C = \frac{\pi}{4} - \frac{1}{3}$.

4. Calculate the area under the line $l$ (a triangle) and find the exact area of $R$.

The area under the line $l$ forms a right-angled triangle with the $x$-axis. Its base is the $x$-coordinate of $A$ and its height is the $y$-coordinate of $A$.
$\text{Area of triangle} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \left(\frac{\pi}{2} - 1\right) (1) = \frac{\pi}{4} - \frac{1}{2}$
The required exact area of $R$ is the area under $C$ minus the area of the triangle:
$\text{Area of } R = \left( \frac{\pi}{4} - \frac{1}{3} \right) - \left( \frac{\pi}{4} - \frac{1}{2} \right)$
$= \frac{1}{2} - \frac{1}{3}$
$= \frac{1}{6}$

1. Express $\sin t$ in terms of $y$, noting the range of $t$ for the relevant portion of the curve.

From $y = \sin^{2} t$, we have $\sin t = \pm\sqrt{y}$.
For the portion of the curve from $O$ to $A$ (which bounds region $R$), the parameter $t$ lies in the interval $0 \le t \le \frac{\pi}{2}$.
Since $\sin t \ge 0$ for this interval, we take the positive root:
$\sin t = \sqrt{y}$

2. Express $t$ in terms of $y$ and substitute into the parametric equation for $x$.

Since $0 \le t \le \frac{\pi}{2}$, the principal value of the inverse sine function applies, giving:
$t = \sin^{-1}\!\sqrt{y}$
Substitute $t$ and $\sin t$ into the equation for $x$:
$x = t - \sin t$
$x = \sin^{-1}\!\sqrt{y} - \sqrt{y}\ \text{ (shown)}$

1. Identify the equation of the line $l$ passing through the origin and point $A$.

The origin is $O(0,0)$ and from part (i), $A$ has coordinates $\left(\frac{\pi}{2} - 1, 1\right)$.
The gradient of line $l$ with respect to the $y$-axis is $\frac{x}{y} = \frac{\frac{\pi}{2} - 1}{1}$.
The equation of the line $l$ is therefore $x = \left(\frac{\pi}{2} - 1\right)y$.

2. Set up the definite integrals for the volume of revolution about the $y$-axis.

The solid is generated by rotating the region $R$ about the $y$-axis. The volume $V$ is given by $V = \pi \int_{0}^{1} x_{\text{outer}}^2 \, \mathrm{d}y - \pi \int_{0}^{1} x_{\text{inner}}^2 \, \mathrm{d}y$.
From the given coordinates, the line $l$ forms the outer boundary and the curve $C$ forms the inner boundary.
Using the Cartesian equation for $C$ from (iii), $x = \sin^{-1}\!\sqrt{y} - \sqrt{y}$.
$V = \pi \int_{0}^{1} \left[ \left(\frac{\pi}{2} - 1\right)y \right]^2 \mathrm{d}y - \pi \int_{0}^{1} \left( \sin^{-1}\!\sqrt{y} - \sqrt{y} \right)^2 \mathrm{d}y$

3. Evaluate the definite integrals using a graphing calculator to find the volume.

Since an exact algebraic evaluation is not explicitly requested, we can directly evaluate the definite integrals using a graphing calculator:
$V \approx \pi (0.10860) - \pi (0.02819)$
$V \approx 0.25261$
$V \approx 0.253 \text{ (to 3 s.f.)}$