Question

Calculate the surface area of the solid of revolution obtained by revolving the region between $f\left(x\right)=2x^3$and the $x$-axis on $\left[0,3\right]$around the $x$-axis.

Short answer

The surface area of the solid of revolution obtained by revolving the region between $f\left(x\right)=2x^3$and the $x$-axis on $\left[0,3\right]$around the $x$-axis is, $1526.81$(approximately) square units.

Step-by-step solution

Step 1. Given information

The region between$f\left(x\right)=2x^3$and the x-axis on$\left[0,3\right]$.

Step 2. The formula for the surface area is,

Surface area$=2\mathrm{\pi }{\int }_{\mathrm{a}}^{\mathrm{b}}\mathrm{f}\left(\mathrm{x}\right)\sqrt{1+{\left({\mathrm{f}}^{\text{'}}\left(\mathrm{x}\right)\right)}^2}\mathrm{dx}$.

Let $f\left(x\right)=2x^3,f^{\text{'}}\left(x\right)=6x^2$.

Substitute the known values in the formula.

Surface area$=2\mathrm{\pi }{\int }_0^{3}2{\mathrm{x}}^3\sqrt{1+{\left(6{\mathrm{x}}^2\right)}^2}\mathrm{dx}$

role="math" localid="1648890749974" $=2\mathrm{\pi }{\int }_0^{3}2{\mathrm{x}}^3\sqrt{1+36{\mathrm{x}}^4}\mathrm{dx}$

Let $u=1+36{\mathrm{x}}^4,\mathrm{du}=144{\mathrm{x}}^3$.

role="math" localid="1648892010258" $$\begin{gathered} 2\mathrm{\pi }{\int }_{x=0}^{x=3}2{\mathrm{x}}^3\sqrt{1+36{\mathrm{x}}^4}\mathrm{dx}=2\mathrm{\pi }\left(\frac{2}{144}\right){\int }_{\mathrm{x}=0}^{\mathrm{x}=3}\sqrt{\mathrm{u}}\mathrm{du} \\ =\left(\frac{\mathrm{\pi }}{36}\right){\left[\frac{2}{3}{\mathrm{u}}^{\frac{3}{2}}\right]}_0^3 \\ =\left(\frac{\mathrm{\pi }}{36}\right)\left(\frac{2}{3}\right){\left[{\left(1+36{\mathrm{x}}^4\right)}^{\frac{3}{2}}\right]}_0^3 \\ =\left(\frac{\mathrm{\pi }}{54}\right){\left[\left(1+36{\mathrm{x}}^6\right)\right]}_0^3 \\ =\left(\frac{\mathrm{\pi }}{54}\right)\left[\left(1+36{\left(3\right)}^6\right)-\left(1+36{\left(0\right)}^6\right)\right] \\ =\left(\frac{\mathrm{\pi }}{54}\right)\left[1+26,244-1\right] \\ =486\mathrm{\pi } \\ \cong 1526.81 \end{gathered}$$