Find the exact value of $\pi \int_{1}^{6} x^3 \text{ d}x$. The diagram shows a sketch of the curve $y = x^{\frac{3}{2}}$, where $x \ge 0$. The region under the curve between $x = 1$ and $x = 6$, shown shaded in the diagram, is $R$. It is required to estimate the volume of the solid formed when $R$ is rotated through $2\pi$ radians about the $x$-axis. The region $R$ is split into $n$ vertical strips of equal width as shown in the diagram below. Each vertical strip can be approximated by a rectangle. Each rectangle, when rotated through $2\pi$ radians about the $x$-axis, will result in a circular disc. The sum of the volumes of the $n$ circular discs with equal thickness is denoted by $V$. Suppose $V$ gives an overestimation of the volume of revolution of the solid formed when $R$ is rotated through $2\pi$ radians about the $x$-axis. Show that $V = \frac{5\pi}{n} \sum_{r=1}^{n} \left[ 1 + r\left(\frac{5}{n}\right) \right]^3$. Find an expression for $V$, leaving your answers in the form $\frac{5\pi}{4} \left( 259 + \frac{a}{n} + \frac{b}{n^2} \right)$, where $a$ and $b$ are constants to be determined. You may use the results
$\sum_{r=1}^{n} r^2 = \frac{1}{6}n(n+1)(2n+1) \text{ and } \sum_{r=1}^{n} r^3 = \frac{1}{4}n^2(n+1)^2.$ Using your answer in part (c), state the limit of $V$ as $n \to \infty$. Explain how this could be verified by the answer in part (a).

1. Integrate the polynomial $x^3$ with respect to $x$.

$\int x^3 \text{ d}x = \frac{x^4}{4}$

2. Evaluate the definite integral using the limits 1 to 6 and multiply by $\pi$.

$\pi \left[ \frac{x^4}{4} \right]_{1}^{6} = \pi \left( \frac{6^4}{4} - \frac{1^4}{4} \right) = \pi \left( \frac{1296}{4} - \frac{1}{4} \right) = \frac{1295\pi}{4}$

1. Calculate the width of each vertical strip.

The region is bounded between $x = 1$ and $x = 6$.
Width of each of the $n$ strips $= \frac{6-1}{n} = \frac{5}{n}$.

2. Determine the radius of the $r$-th circular disc.

Since the curve $y = x^{\frac{3}{2}}$ is increasing for $x \ge 0$, an overestimation is given by using the right endpoints of each strip to form the approximating rectangles.
The $x$-coordinate of the right endpoint of the $r$-th strip is $x_r = 1 + r\left(\frac{5}{n}\right)$.
The radius of the $r$-th disc is the $y$-value at $x_r$, so $y_r = \left[ 1 + r\left(\frac{5}{n}\right) \right]^{\frac{3}{2}}$.

3. Formulate the total volume $V$ of the $n$ discs and show the final result.

The volume of the $r$-th disc is $\pi y_r^2 \Delta x = \pi \left( \left[ 1 + r\left(\frac{5}{n}\right) \right]^{\frac{3}{2}} \right)^2 \left(\frac{5}{n}\right) = \pi \left[ 1 + r\left(\frac{5}{n}\right) \right]^3 \left(\frac{5}{n}\right)$.
Summing the volumes of all $n$ discs gives the total estimated volume:
$V = \sum_{r=1}^{n} \pi \left[ 1 + r\left(\frac{5}{n}\right) \right]^3 \left(\frac{5}{n}\right)$
Since $\frac{5\pi}{n}$ is independent of $r$, we can factor it out of the summation:
$V = \frac{5\pi}{n} \sum_{r=1}^{n} \left[ 1 + r\left(\frac{5}{n}\right) \right]^3$

1. Expand the cubic expression inside the summation.

$\left( 1 + \frac{5r}{n} \right)^3 = 1 + 3\left(\frac{5r}{n}\right) + 3\left(\frac{5r}{n}\right)^2 + \left(\frac{5r}{n}\right)^3$
$= 1 + \frac{15r}{n} + \frac{75r^2}{n^2} + \frac{125r^3}{n^3}$

2. Distribute the sum and substitute the standard summation formulas.

$V = \frac{5\pi}{n} \left[ \sum_{r=1}^n 1 + \frac{15}{n}\sum_{r=1}^n r + \frac{75}{n^2}\sum_{r=1}^n r^2 + \frac{125}{n^3}\sum_{r=1}^n r^3 \right]$

$V = \frac{5\pi}{n} \left[ n + \frac{15}{n}\left(\frac{n(n+1)}{2}\right) + \frac{75}{n^2}\left(\frac{n(n+1)(2n+1)}{6}\right) + \frac{125}{n^3}\left(\frac{n^2(n+1)^2}{4}\right) \right]$

3. Simplify the individual fractional terms.

$V = \frac{5\pi}{n} \left[ n + \frac{15(n+1)}{2} + \frac{25(2n^2+3n+1)}{2n} + \frac{125(n^2+2n+1)}{4n} \right]$

4. Combine the terms over a common denominator of $4n$.

$V = \frac{5\pi}{n} \left[ \frac{4n^2 + 30n(n+1) + 50(2n^2+3n+1) + 125(n^2+2n+1)}{4n} \right]$

$V = \frac{5\pi}{n} \left[ \frac{4n^2 + 30n^2 + 30n + 100n^2 + 150n + 50 + 125n^2 + 250n + 125}{4n} \right]$

$V = \frac{5\pi}{n} \left( \frac{259n^2 + 430n + 175}{4n} \right)$

5. Extract the common factor and rearrange the expression into the requested format.

$V = \frac{5\pi}{4n^2} \left( 259n^2 + 430n + 175 \right)$

$V = \frac{5\pi}{4} \left( \frac{259n^2}{n^2} + \frac{430n}{n^2} + \frac{175}{n^2} \right)$

$V = \frac{5\pi}{4} \left( 259 + \frac{430}{n} + \frac{175}{n^2} \right)$

$\text{Thus, } a = 430 \text{ and } b = 175.$

1. State the limit of $V$ as $n \to \infty$ using the expression from part (c).

From part (c), $V = \frac{5\pi}{4} \left( 259 + \frac{a}{n} + \frac{b}{n^2} \right)$.

As $n \to \infty$, the terms $\frac{a}{n} \to 0$ and $\frac{b}{n^2} \to 0$.

$\lim_{n\to\infty} V = \frac{5\pi}{4} (259) = \frac{1295\pi}{4}$

2. Evaluate the integral from part (a) to explain how it verifies the limit.

In part (a), the exact volume of revolution of the curve $y = x^{\frac{3}{2}}$ about the $x$-axis is given by the integral:
$\pi \int_{1}^{6} \left(x^{\frac{3}{2}}\right)^2 \text{ d}x = \pi \int_{1}^{6} x^3 \text{ d}x$

Evaluating this integral yields:
$\pi \left[ \frac{x^4}{4} \right]_1^6 = \pi \left( \frac{1296}{4} - \frac{1}{4} \right) = \frac{1295\pi}{4}$

Since the sum $V$ represents a Riemann sum approximation for the volume of revolution by dividing it into $n$ discs, taking the limit as $n \to \infty$ gives the exact volume of revolution. The limit perfectly matches the exact value obtained by integration in part (a), thereby verifying the result.