The diagram above shows a sketch of the graph of $y=f(x)$ 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. It is now given that $f(x)=\dfrac{x^2}{\sqrt{1+e^x}}$ for $x\ge 0$.

1. Write the volume of the r-th cylindrical disc

The $r$-th rectangle ($r=0,1,\dots,n-1$) has left edge at $x=1+rh$, so its height (the radius when rotated) is $f(1+rh)$ and its thickness is $h$. Rotating it through $2\pi$ gives a cylinder of volume
$V_r=\pi\,[\,\text{radius}\,]^2\times\text{thickness}=\pi\,[f(1+rh)]^2\,h.$

2. Sum the n discs to obtain V1

Adding the volumes of all $n$ discs:
$V_1=\sum_{r=0}^{n-1}\pi\,[f(1+rh)]^2\,h=\pi h\sum_{r=0}^{n-1}\left[f(1+rh)\right]^2,$
since $\pi$ and $h$ are common to every term.

3. State h in terms of n

The $n$ rectangles of equal width $h$ exactly cover $1\le x\le 3$, a total width of $3-1=2$. Hence
$nh=2\ \Rightarrow\ h=\dfrac{2}{n}.$

1. Formulate the upper sum (right Riemann sum) by shifting the index to the right endpoints.

Since the function $f(x)$ is increasing for $1 \le x \le 3$, the left-endpoint rectangles are inscribed ($V_1 < V$, underestimating the volume). The right-endpoint rectangles will be circumscribed, giving an overestimation ($V_2 > V$).

The volume of each right-endpoint cylindrical disc is $\pi \left[ f(1 + rh) \right]^2 h$.
To sum these $n$ right-endpoint discs, we change the summation indices to run from $r=1$ to $n$ instead of $r=0$ to $n-1$.

$V_2 = \pi h \sum_{r=1}^{n} [f(1+rh)]^2$

1. Recognise that the limit of the Riemann sum as $n \to \infty$ is the definite integral representing the actual volume

As $n \to \infty$, the total volume of the cylindrical discs $V_1$ approaches the actual volume $V$ generated by rotating the region under the curve about the $x$-axis.

$\lim_{n \to \infty} V_1 = V = \pi \int_{1}^{3} [f(x)]^2 \, dx$

2. Substitute $f(x)$ into the integral and evaluate using a graphing calculator

$V = \pi \int_{1}^{3} \left( \frac{x^2}{\sqrt{1+e^x}} \right)^2 \, dx$
$= \pi \int_{1}^{3} \frac{x^4}{1+e^x} \, dx$
Using a graphing calculator to evaluate the definite integral:
$V \approx 12.2907$
$V = 12.3 \text{ (to 3 s.f.)}$

1. Express the equation of the given curve explicitly in terms of $x$ for the condition $y \le 3$.

$(x-1)^2 + (y-3)^2 = 9$
$(y-3)^2 = 9 - (x-1)^2$
Taking the square root on both sides, and given that $y \le 3$ (which means $y-3 \le 0$):
$y - 3 = -\sqrt{9 - (x-1)^2}$
$y = 3 - \sqrt{9 - (x-1)^2}$

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

Equating the two functions to find their intersection points:
$\frac{x^2}{\sqrt{1+e^x}} = 3 - \sqrt{9 - (x-1)^2}$
Using a graphing calculator to solve for $x$:
$x \approx 0.33842 \quad \text{and} \quad x \approx 3.8742$

3. Set up the definite integral for the area of the bounded region and evaluate it using a graphing calculator.

From the graphs, the curve $y = f(x)$ lies above the semicircle $y = 3 - \sqrt{9 - (x-1)^2}$ between the points of intersection. The bounded area $A$ is given by:
$A = \int_{0.33842}^{3.8742} \left[ \frac{x^2}{\sqrt{1+e^x}} - \left( 3 - \sqrt{9 - (x-1)^2} \right) \right] \, dx$
Using a graphing calculator to evaluate the definite integral:
$A \approx 3.02 \text{ (to 3 s.f.)}$