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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:
A$20.25$ cubic units
B$\frac{129}{4}\pi$ cubic units
C$\frac{129}{2}\pi$ cubic units
D$12\pi$ cubic units
Answer & Solution
Correct answer: B. $\frac{129}{4}\pi$ cubic units
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_
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