The area enclosed by $y^2-x^3=4$ and $y=0$ between $x=0$ and $x=3$ is rotated about the $x$-axis. The volume generated is:

1. For rotation about the $x$-axis, $V=\pi\displaystyle\int_a^b y^2\,dx$. 2. From $y^2-x^3=4$ we get $y^2=4+x^3$. 3. So $V=\pi\displaystyle\int_0^3(4+x^3)\,dx=\pi\left[4x+\frac{x^4}{4}\right]_0^3$. 4. At $x=3$: $12+\dfrac{81}{4}=\dfrac{48+81}{4}=\dfrac{129}{4}$; at $x=0$ the bracket is zero. 5. Hence $V=\dfrac{129}{4}\pi$ cubic units. The value $20.25$ integrates $y^2-x^3$ itself and omits $\pi$ entirely. _Source: NECTA ACSEE 2023 Advanced Mathematics 142/1, Question 9: Integration_