The region bounded by the curve y = x+1, the x-axis, and the lines x=0 and x=3 is rotated through 360° about the x-axis. Find the volume of the solid formed, giving your answer as an exact multiple of π. [6]

AS Level Maths 9709: Volume of Revolution About the x-axis

For a region under $y=f(x)$ between $x=a$ and $x=b$ , rotating fully about the $x$ -axis gives

$V = \pi \int_a^b y^2 \, dx$

Here $y = x+1$ , $a=0$ , $b=3$ . Square the function carefully:

$y^2 = (x+1)^2 = x^2 + 2x + 1$

$V = \pi \int_0^3 \left(x^2 + 2x + 1\right) dx$

$\int \left(x^2+2x+1\right) dx = \frac{x^3}{3} + x^2 + x$

At $x=3$ : $\ \dfrac{27}{3} + 9 + 3 = 9 + 9 + 3 = 21$

At $x=0$ : $\ 0$

$\int_0^3 \left(x^2+2x+1\right) dx = 21 - 0 = 21$

$V = 21\pi$

Let $u = x+1$ , so $du = dx$ . When $x=0$ , $u=1$ ; when $x=3$ , $u=4$ . Then $y^2\,dx = u^2\,du$ , so

$V = \pi \int_1^4 u^2\,du = \pi\left[\frac{u^3}{3}\right]_1^4 = \pi\left(\frac{64}{3} - \frac{1}{3}\right) = \pi \times \frac{63}{3} = 21\pi$

This matches the result from Step 2 exactly, confirming the volume is correct.

$\boxed{V = 21\pi} \text{ cubic units}$