Answer: \(s=\int_0^1\sqrt{1+x^4}\,\mathrm dx.\)

\(S=\frac{\pi}{9}\left(18s+2\sqrt2-1\right).\)

The curve is

\(y=\frac13x^3+1.\)

Differentiate:

\(\frac{\mathrm dy}{\mathrm dx}=x^2.\)

The arc length from \(x=0\) to \(x=1\) is

\(s=\int_0^1\sqrt{1+\left(\frac{\mathrm dy}{\mathrm dx}\right)^2}\,\mathrm dx.\)

Substituting \(\frac{\mathrm dy}{\mathrm dx}=x^2\),

\(s=\int_0^1\sqrt{1+x^4}\,\mathrm dx.\)

Now rotate the arc about the \(x\)-axis. The surface area is

\(S=2\pi\int_0^1 y\sqrt{1+\left(\frac{\mathrm dy}{\mathrm dx}\right)^2}\,\mathrm dx.\)

Thus

\(S=2\pi\int_0^1\left(\frac13x^3+1\right)\sqrt{1+x^4}\,\mathrm dx.\)

Split the integral:

\(S=2\pi\int_0^1\sqrt{1+x^4}\,\mathrm dx+\frac{2\pi}{3}\int_0^1x^3\sqrt{1+x^4}\,\mathrm dx.\)

The first integral is \(s\), so

\(S=2\pi s+\frac{2\pi}{3}\int_0^1x^3\sqrt{1+x^4}\,\mathrm dx.\)

For the remaining integral, let \(u=1+x^4\). Then \(\mathrm du=4x^3\,\mathrm dx\), so

\(\int x^3\sqrt{1+x^4}\,\mathrm dx=\frac14\int u^{1/2}\,\mathrm du=\frac16u^{3/2}.\)

Hence

\(\int_0^1x^3\sqrt{1+x^4}\,\mathrm dx=\frac16\left[(1+x^4)^{3/2}\right]_0^1.\)

Therefore

\(\int_0^1x^3\sqrt{1+x^4}\,\mathrm dx=\frac16(2^{3/2}-1)=\frac16(2\sqrt2-1).\)

Substitute this into \(S\):

\(S=2\pi s+\frac{2\pi}{3}\cdot\frac16(2\sqrt2-1).\)

So

\(S=2\pi s+\frac{\pi}{9}(2\sqrt2-1)=\frac{\pi}{9}(18s+2\sqrt2-1).\)