Question

The volume of the solid formed by rotating the area enclosed between the curve $y^2=4x$,$x=4$ and $x=5$ about the $x-$axis is (in cubic units)

• A. $18\mathrm{\pi }$
• B. $36\mathrm{\pi }$
• C. $9\mathrm{\pi }$
• D. $24\mathrm{\pi }$

Answer 1

The correct option is A

$18\mathrm{\pi }$

Explanation for the correct option:

If $y$ is given as a function of $x$, the volume of the solid obtained by rotating the portion of the curve between $x=a$ and $x=b$ about the $x-$axis is given by

$V={\int }_a^b{\mathrm{\pi y}}^2\mathrm{dx}$

Here the function $y$ is given as $y^2=4x$ and the values of $a$ and $b$ are $4$ and $5$ respectively.

Substituting these values in the formula for volume we get,

$V={\int }_4^5\pi \left(4x\right)\mathrm{dx}$

$\Rightarrow$ $V=4\mathrm{\pi }{\int }_4^5\mathrm{xdx}$

$\Rightarrow$ $V=4\mathrm{\pi }{\left[\frac{{\mathrm{x}}^2}{2}\right]}_4^5$ $\left[\because {\int }_a^{b}xdx={\left[\frac{x^2}{2}\right]}_a^b\right]$

$\Rightarrow$ $V=2\mathrm{\pi }{\left[x^2\right]}_4^5$

$\Rightarrow$ $V=2\mathrm{\pi }\left(5^2-4^2\right)$

$\Rightarrow$ $V=2\mathrm{\pi }\left(25-16\right)$

$\Rightarrow$$V=18\mathrm{\pi }\mathrm{cu}\mathrm{units}$

Thus the volume of the solid formed is $18\mathrm{\pi }\mathrm{cu}\mathrm{units}$.

Hence, option $\left(A\right)$ i.e. $18\mathrm{\pi }$ is the correct option.