What volume is generated when $y = ax - x^2$ is rotated about the x-axis?

What volume is generated when $y = ax - x^2$ is rotated about the x-axis?

Question

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 $a$.

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.

Solution

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 $a = 6$.

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

$$\begin{align*} \int_0^6 \pi \left( 6x - x^2 \right)^2 \:dx &= \pi \int_0^6 x^2 (6-x)^2 \:dx \\ &= \pi \int_0^6 x^2 (36 - 12x + x^2) \:dx \\ &= \pi \int_0^6 36x^2 - 12x^3 + x^4 \:dx \\ &= \pi \left[ 12x^3 - 3x^4 + \frac{x^5}{5} \right]_0^6 \\ &= \pi \left( 12 \times 6^3 - 3 \times 6^4 + \frac{6^5}{5} \right) \\ &= \pi \left( \left( 2 - 3 + \frac{6}{5} \right) 6^4 \right) \\ &= \frac{6^4}{5} \pi \\ &= \frac{1296}{5} \pi. \end{align*}$$