JAMB 2000 · UME · Q37

Calculus (JAMB bridge)

A bowl is designed by revolving completely the area enclosed by $y = x^2 - 1$ , $y = 0$ , $y = 3$ and $x \ge 0$ around the $y$ -axis. What is the volume of this bowl?

Worked solution (try it first)

About the $y$ -axis the volume is $\pi\int x^2\,dy$ .
From $y = x^2 - 1$ , $x^2 = y + 1$ .
The limits are $y = 0$ to $y = 3$ : $V = \pi\int_0^3 (y + 1)\,dy$
$= \pi\left[\frac{y^2}{2} + y\right]_0^3$ .
That is $\pi\left(\frac92 + 3\right) = \frac{15\pi}{2}$ cubic units, option B.

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