Question

Find the exact area of the surface obtained by rotating the curve about the x-axis.

y=x³, 0≤x≤2

Short answer

Area obtained by rotating the curve\(y = {x^3}\)about x-axis is \(\frac{\pi }{{27}}(145\sqrt {145} - 1)\).

Step-by-step solution

Area of a surface of a revolution

If a function is positive and has a continuous derivative, the surface area defines by rotating the surface obtained by curve \({\rm{y = f(x)}}\)

About x-axis as:

\({S_x} = \int_a^b 2 \pi y\sqrt {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} dx\)

Given parameters

Equation of the curve is \(y = {x^3}\) and interval is \((0,2)\).

Calculating surface area

Surface area of the curve obtained after rotation about x-axis is:

\(\begin{aligned}{S_x} &= \int_a^b 2 \pi y\sqrt {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} dx\\ &= \int_0^2 2 \pi {x^3}\sqrt {1 + {{\left( {\frac{d}{{dx}}{x^3}} \right)}^2}} dx\\ &= \int_0^2 2 \pi {x^3}\sqrt {1 + {{\left( {3{x^2}} \right)}^2}} dx\\ &= 2\pi \int_0^2 {\sqrt {1 + 9{x^4}} } dx\end{aligned}\)

Substitute \(u = 1 + 9{x^4}\)

\(\begin{aligned}\frac{{du}}{{dx}} &= 36{x^3}\\\frac{{du}}{{36}} &= {x^3}dx\end{aligned}\)

Now new limit of integration for u is

\(u = \left\{ {\begin{aligned}{*{20}{l}}1&{,{\rm{ }}if{\rm{ }}x = 0}\\{145}&{,{\rm{ }}if{\rm{ }}x = 2}\end{aligned}} \right.\)

\(\begin{aligned}{S_x} &= 2\pi \int_1^{145} {\sqrt u } \frac{{du}}{{36}}\\ &= \frac{\pi }{{18}}\int_1^{145} {{u^{1/2}}} du\\ &= \frac{\pi }{{18}}\left( {\frac{2}{3}{u^{3/2}}} \right)_1^{145}\\ &= \frac{\pi }{{18}}\left( {\frac{2}{3}{{(145)}^{3/2}} - \frac{2}{3}} \right)\\ &= \frac{\pi }{{18}} \cdot \frac{2}{3}(145\sqrt {145} - 1)\\ &= \frac{\pi }{{27}}(145\sqrt {145} - 1)\end{aligned}\)

Therefore, surface area of the curve obtained after rotation about x-axis is \(\frac{\pi }{{27}}(145\sqrt {145} - 1)\).