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 \(a\).

From what is given in the question, the area enclosed between the curve and the \(x\)-axis is \[\begin{equation*} 36 = \int_0^a ax - x^2 \:dx = \left[ \frac{ax^2}{2} - \frac{x^3}{3} \right]_0^a = \frac{a^3}{2} - \frac{a^3}{3} = \frac{a^3}{6}. \end{equation*}\]

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.

If we have a curve \(y = f(x)\) with \(p \le x \le q\) and \(f(x) \ge 0\), then the volume enclosed by the surface of revolution obtained by rotating the curve around the \(x\)-axis is given by \[\begin{equation*} \int_{p}^{q} \pi f(x)^2 \:dx. \end{equation*}\]
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 \[\begin{align*} \int_0^6 \pi \left( 6x - x^2 \right)^2 \:dx &= \pi \int_0^6 x^2 (6-x)^2 \:dx \\ &= \pi \int_0^6 x^2 (36 - 12x + x^2) \:dx \\ &= \pi \int_0^6 36x^2 - 12x^3 + x^4 \:dx \\ &= \pi \left[ 12x^3 - 3x^4 + \frac{x^5}{5} \right]_0^6 \\ &= \pi \left( 12 \times 6^3 - 3 \times 6^4 + \frac{6^5}{5} \right) \\ &= \pi \left( \left( 2 - 3 + \frac{6}{5} \right) 6^4 \right) \\ &= \frac{6^4}{5} \pi \\ &= \frac{1296}{5} \pi. \end{align*}\]