The diagram above shows a sketch of the graph of $y = \text{f}(x)$ for $x \ge 0$. There are $n$ rectangles each of width $h$ drawn under the curve for $1 \le x \le 3$. Each rectangle when rotated through $2\pi$ radians about the $x$-axis, will result in a cylindrical disc. The total volume of the $n$ cylindrical discs $V_1$, can be used to estimate the volume $V$, which is the actual volume generated when the region bounded by the curve, the lines $x = 1$ and $x = 3$, and the $x$-axis is rotated through $2\pi$ radians about the $x$-axis.

1. Express the volume of the cylindrical disc formed by the rectangle at index $r$.

For the rectangle corresponding to index $r$ (where $r = 0, 1, 2, \dots, n-1$), its left edge is at $x = 1 + rh$.
The height of this rectangle is $\text{f}(1 + rh)$.
When rotated $2\pi$ radians about the $x$-axis, it forms a cylindrical disc of radius $\text{f}(1 + rh)$ and thickness $h$.
The volume of this single disc is $\pi [\text{f}(1 + rh)]^2 h$.

2. Sum the volumes of all $n$ discs to show the expression for $V_1$.

To find the total volume $V_1$, we sum the volumes of all $n$ discs from $r = 0$ to $r = n-1$:
$V_1 = \sum_{r=0}^{n-1} \pi [\text{f}(1 + rh)]^2 h$
Factoring out the constants $\pi$ and $h$ gives:
$V_1 = \pi h \sum_{r=0}^{n-1} [\text{f}(1 + rh)]^2 \quad \text{(Shown)}$

3. Determine the value of $h$ in terms of $n$.

The region spans from $x = 1$ to $x = 3$, so the total width of the interval is $3 - 1 = 2$.
Since there are $n$ rectangles of equal width $h$, we have:
$nh = 2$
$h = \frac{2}{n}$

1. Identify the geometrical requirement for an overestimate

From the diagram, the curve $y = \text{f}(x)$ is an increasing function. $V_1$ uses the left-hand endpoint of each subinterval (from $r=0$ to $n-1$), which leaves a gap under the curve, hence $V_1 < V$.
To create an overestimate ($V_2 > V$), we must use rectangles whose heights are determined by the right-hand endpoint of each subinterval.

2. Write the new expression using right-hand endpoints

The right-hand endpoints of the $n$ intervals occur at $x = 1+rh$ for $r=1, 2, \dots, n$.
Replacing the summation limits in $V_1$ to reflect these new evaluation points gives:
$V_2 = \pi h \sum_{r=1}^{n} [\text{f}(1 + rh)]^2$

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

As $n \to \infty$, the width of each cylinder $h \to 0$. The limit of the Riemann sum $V_1$ representing the total volume of the cylindrical discs corresponds to the exact volume $V$ of the solid of revolution.
$\lim_{n \to \infty} V_1 = V = \pi \int_1^3 [\text{f}(x)]^2 \,dx$

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

Substitute $\text{f}(x) = \frac{x^2}{\sqrt{1 + \text{e}^x}}$ into the integral:
$\lim_{n \to \infty} V_1 = \pi \int_1^3 \left( \frac{x^2}{\sqrt{1 + \text{e}^x}} \right)^2 \,dx$
$\lim_{n \to \infty} V_1 = \pi \int_1^3 \frac{x^4}{1 + \text{e}^x} \,dx$
Using a graphing calculator to evaluate the definite integral:
$\lim_{n \to \infty} V_1 \approx 12.291 \approx 12.3 \text{ (to 3 s.f.)}$

1. Express the equation of the second curve with $y$ as the subject.

$(x-1)^2 + (y-3)^2 = 9$\n$(y-3)^2 = 9 - (x-1)^2$\n$y - 3 = \pm\sqrt{9 - (x-1)^2}$\nSince it is given that $y \le 3$, we must take the negative square root to represent the lower semicircle:\n$y = 3 - \sqrt{9 - (x-1)^2}$

2. Find the $x$-coordinates of the points of intersection.

Using a graphing calculator, we find the points of intersection between $y = \text{f}(x)$ and the lower semicircle $y = 3 - \sqrt{9 - (x-1)^2}$ for $x \ge 0$.\n$x \approx 0.33838 \quad \text{or} \quad x \approx 3.8744$

3. Formulate the definite integral for the area and evaluate it using a graphing calculator.

Observe from the graphing calculator that the curve $y = \text{f}(x)$ lies above the semicircle between the intersection points.\n$\text{Area} = \int_{0.33838}^{3.8744} \left( \frac{x^2}{\sqrt{1 + \text{e}^x}} - \left( 3 - \sqrt{9 - (x-1)^2} \right) \right) \text{d}x$\nEvaluating this via the graphing calculator gives:\n$\text{Area} \approx 3.00 \text{ (to 3 s.f.)}$