The curve \(y = ax - x^2\) cuts the \(x\)-axis at the points \((0,0)\) and \((a,0)\). The area enclosed between the curve and the \(x\)-axis is \(36\) square units. Calculate the value of \(a\).

Thus, \(a^3 = 216\) and so \(a = 6\).

With this value of \(a\) find, as a multiple of \(\pi\), the volume obtained when the above area is rotated through \(4\) right angles about the \(x\)-axis.

How does this arise? We can approximate the enclosed solid with a series of cylinders of thickness \(\delta x\) (corresponding to a small increase in \(x\)) and radius \(f(x)\) (corresponding to the height of the function). The volume of the approximation is
\[\begin{equation*}
\sum \pi f(x)^2 \delta x \xrightarrow{\delta x \to 0} \int_{p}^{q} \pi f(x)^2 \:dx.
\end{equation*}\]

In our case, then, the volume required is
\[\begin{align*}
\int_0^6 \pi \left( 6x - x^2 \right)^2 \:dx &= \pi \int_0^6 x^2 (6-x)^2 \:dx \\
&= \pi \int_0^6 x^2 (36 - 12x + x^2) \:dx \\
&= \pi \int_0^6 36x^2 - 12x^3 + x^4 \:dx \\
&= \pi \left[ 12x^3 - 3x^4 + \frac{x^5}{5} \right]_0^6 \\
&= \pi \left( 12 \times 6^3 - 3 \times 6^4 + \frac{6^5}{5} \right) \\
&= \pi \left( \left( 2 - 3 + \frac{6}{5} \right) 6^4 \right) \\
&= \frac{6^4}{5} \pi \\
&= \frac{1296}{5} \pi.
\end{align*}\]