Question

The volume of the solid generated by revolving the region bounded by $y=2x^2+1$ and $y=2x+11$ about $x$ axis is

• A. $\frac{104\mathrm{\pi }}{15}\mathrm{Cu}\mathrm{units}$
• B. $\frac{42\mathrm{\pi }}{15}\mathrm{Cu}\mathrm{units}$
• C. $\frac{52\mathrm{\pi }}{15}\mathrm{Cu}\mathrm{units}$
• D. None of these.

Answer 1

The correct option is D

None of these.

Explanation of the correct option.

Step 1: Compute the intersection points.

Given: Region is bounded by $y=2x^2+1$ and $y=2x+11$.

$$\begin{gathered} 2x^2+1=2x+11 \\ \Rightarrow 2x^2-2x-10=0 \\ \Rightarrow x^2-x-5=0 \end{gathered}$$

Therefore, roots of the quadratic equation is $x=\frac{1+\sqrt{21}}{2},x=\frac{1-\sqrt{21}}{2}$

Hence given curves intersect at two points.

Step 2: Compute the volume.

$V={\int }_{\frac{1-\sqrt{21}}{2}}^{\frac{1+\sqrt{21}}{2}}\pi y^2\mathrm{dx}$

Since, $V=\mathrm{\pi }{\int }_{\frac{1-\sqrt{21}}{2}}^{\frac{1+\sqrt{21}}{2}}\left[{\left(2\mathrm{x}+11\right)}^2-{\left(2{\mathrm{x}}^2+1\right)}^2\right]\mathrm{d}x$
$\Rightarrow$ $V=\mathrm{\pi }{\int }_{\frac{1-\sqrt{21}}{2}}^{\frac{1+\sqrt{21}}{2}}\left[-4{\mathrm{x}}^4+44\mathrm{x}+120\right]\mathrm{d}x$
$\Rightarrow$ $V=\mathrm{\pi }\left[-4\frac{{\mathrm{x}}^5}{5}+44\frac{{\mathrm{x}}^2}{2}+120x\right]\underset{\frac{1-\sqrt{21}}{2}}{\overset{\frac{1+\sqrt{21}}{2}}{}}$
$\Rightarrow$ $V\approx 1,572.11\mathrm{Cu}\mathrm{units}$

Therefore, the volume of the solid generated by the revolving the region is $1,572.11\mathrm{Cu}\mathrm{units}$.

Hence option D is the correct option.