Question

To determine: The area of the surface \(y = {x^3},0 \le x \le 2\) rotating about \(x\)-axis.

Short answer

The area of the surface is approximately \(0.99\pi \).

Step-by-step solution

Formula used:

The area of the surface of revolution can be obtain by \(A = 2\pi \int_a^b f (x)\sqrt {1 + {{\left( {{f^\prime }(x)} \right)}^2}} dx\).

Consider the given surface as \(f(x) = y = {x^3}\) over \(a = 0\) and \(b = 2\).

Now substitute the above values in the area formula as follows.

\(\begin{aligned}A = 2\pi \int_a^b f (x)\sqrt {1 + {{\left( {{f^\prime }(x)} \right)}^2}} dx\\ &= 2\pi \int_0^2 y \sqrt {1 + {{\left( {{y^\prime }} \right)}^2}} dx\\ &= 2\pi \int_0^2 {{x^3}} \sqrt {1 + {{\left( {3{x^2}} \right)}^2}} dx\\ &= 2\pi \int_0^2 {{x^3}} \sqrt {1 + 9{x^4}} dx\end{aligned}\)

Consider a variable \(u = 1 + 9{x^4}\).

Then, \(du = 36{x^3}dx\).

Substitute \(1 + 9{x^4} = u\) and \(dx = \frac{{du}}{{36{x^3}}}\) in the area as follows.

\(\begin{aligned}dA &= 2\pi \int_{1 + 9{{(0)}^4}}^{1 + 9{{(2)}^4}} {\left( {{x^3}\sqrt u } \right)} \frac{{du}}{{36{x^3}}}\\ &= \frac{\pi }{{18}}\int_1^{145} {{u^{\frac{1}{2}}}} du\\ &= \frac{\pi }{{18}}\left( {\frac{2}{3}{u^{\frac{2}{3}}}} \right)_1^{145}\\ &= \frac{\pi }{{27}}\left( {{{(145)}^{\frac{2}{3}}} - 1} \right) \approx 0.99\pi \end{aligned}\)

Thus, the area of the surface is approximately \(0.99\pi \).