上图显示了对于 $x \\ge 0$，$y=f(x)$ 的函数图像草图。在 $1 \\le x \\le 3$ 范围内，曲线下方画有 $n$ 个宽度均为 $h$ 的矩形。每个矩形绕 x 轴旋转 $2\pi$ 弧度，都会得到一个圆柱形薄片。这 $n$ 个圆柱形薄片的总体积 $V_1$ 可用于估算体积 $V$，其中 $V$ 是由曲线、直线 $x=1$ 和 $x=3$ 以及 x 轴围成的区域绕 x 轴旋转 $2\pi$ 弧度所生成的实际体积。

1. Formulate the sum for the upper rectangles to overestimate the volume.

For $V_2 > V$, the cylindrical discs must overestimate the volume. Since the curve $y=f(x)$ is increasing on the interval $1 \le x \le 3$, we must construct the rectangles using the right endpoints of each sub-interval. The radius of each disc will therefore be $f(1+rh)$ for $r=1, 2, \dots, n$.
The volume of each disc is $\pi [f(1+rh)]^2 h$.
Summing these $n$ discs gives:
$V_2 = \pi h \sum_{r=1}^n \bigl[f(1+rh)\bigr]^2$

1. Relate the limit of the Riemann sum to the definite integral for the volume of revolution.

As $n \to \infty$, the limit of the Riemann sum $V_1$ gives the exact volume of revolution $V$ about the $x$-axis.
$\lim_{n \to \infty} V_1 = V = \pi \int_{1}^{3} [f(x)]^2 \, dx$

2. Substitute the given function and evaluate the integral using a graphing calculator.

Substitute $f(x) = \dfrac{x^2}{\sqrt{1+e^x}}$:
$V = \pi \int_{1}^{3} \left( \frac{x^2}{\sqrt{1+e^x}} \right)^2 \, dx$
$V = \pi \int_{1}^{3} \frac{x^4}{1+e^x} \, dx$
Evaluating this definite integral using a graphing calculator:
$V \approx \pi (3.91262) \approx 12.292$
$\lim_{n \to \infty} V_1 = 12.3 \text{ (to 3 s.f.)}$

1. Express the given equation of the second curve as a function of $x$.

$(x-1)^2 + (y-3)^2 = 9$
$(y-3)^2 = 9 - (x-1)^2$
Since $y \le 3$, we take the negative square root to represent the lower semicircle:
$y - 3 = -\sqrt{9 - (x-1)^2}$
$y = 3 - \sqrt{9 - (x-1)^2}$

2. Find the $x$-coordinates of the intersection points between the two curves using a graphing calculator.

Let $y_1 = \dfrac{x^2}{\sqrt{1+e^x}}$ and $y_2 = 3 - \sqrt{9 - (x-1)^2}$.
Equating $y_1 = y_2$ and solving using a graphing calculator yields the intersection points:
$x_1 \approx 0.33833$
$x_2 \approx 3.8744$

3. Set up the definite integral for the enclosed area.

Observing the graphs on the GC, the curve $y_1$ lies above $y_2$ in the interval $0.33833 \le x \le 3.8744$. The area $A$ is given by:
$A = \int_{0.33833}^{3.8744} \left( y_1 - y_2 \right) dx$
$A = \int_{0.33833}^{3.8744} \left( \frac{x^2}{\sqrt{1+e^x}} - \left( 3 - \sqrt{9 - (x-1)^2} \right) \right) dx$

4. Evaluate the definite integral using the graphing calculator.

Evaluating the definite integral directly with the graphing calculator:
$A \approx 3.0242$
$A = 3.02 \text{ (to 3 s.f.)}$