Find the surface area of the solid of revolution obtained by rotating the curve y=2x^3 from x=1 to x=6 about the x-axis?

May 25, 2018

$\textcolor{b l u e}{{S}_{A} = 2 \pi {\int}_{1}^{6} \left(2 {x}^{3}\right) \cdot \sqrt{1 + {\left(6 {x}^{2}\right)}^{2}} \cdot \mathrm{dx} = 5.863 \cdot {10}^{5}}$

Explanation:

The surface area of the solid revolution around the $\text{x-axis}$ is given by:

color(red)[S_A=2piint_a^by*sqrt[1+(y')^2]*dx

we will find the surface area bounded x=1 , x=6 and the x-axis

${S}_{A} = 2 \pi {\int}_{1}^{6} \left(2 {x}^{3}\right) \cdot \sqrt{1 + {\left(6 {x}^{2}\right)}^{2}} \cdot \mathrm{dx}$

${S}_{A} = 2 \pi {\int}_{1}^{6} \left(2 {x}^{3}\right) \cdot \sqrt{1 + \left(36 {x}^{4}\right)} \cdot \mathrm{dx}$

${S}_{A} = \frac{2 \pi}{72} {\int}_{1}^{6} \left(144 {x}^{3}\right) \cdot \sqrt{1 + \left(36 {x}^{4}\right)} \cdot \mathrm{dx}$

$= \frac{\pi}{36} {\left[\frac{2}{3} {\left(1 + \left(36 {x}^{4}\right)\right)}^{\frac{3}{2}}\right]}_{1}^{6} = {\left[\frac{\pi \cdot {\left(36 \cdot {x}^{4} + 1\right)}^{\frac{3}{2}}}{54}\right]}_{1}^{6}$

$= \left[4 \cdot \left({46657}^{\frac{3}{2}} / 216 - {37}^{\frac{3}{2}} / 216\right) \cdot \pi\right] = \left[\frac{\left({46657}^{\frac{3}{2}} - {37}^{\frac{3}{2}}\right) \cdot \pi}{54}\right]$

$= 5.863 \cdot {10}^{5}$