Answer: The arc length is \(4\sqrt2\).

The surface area generated about the \(x\)-axis is

\(\frac{128\sqrt2\pi}{15}.\)

From the parametric equations in the question,

\(\frac{\mathrm dx}{\mathrm dt}=2t,\qquad \frac{\mathrm dy}{\mathrm dt}=-t\sqrt{4-t^2}.\)

Therefore

\(\frac{\mathrm ds}{\mathrm dt}=\sqrt{\left(\frac{\mathrm dx}{\mathrm dt}\right)^2+\left(\frac{\mathrm dy}{\mathrm dt}\right)^2}=\sqrt{4t^2+t^2(4-t^2)}.\)

On \(0\leq t\leq2\), this simplifies to

\(\frac{\mathrm ds}{\mathrm dt}=2\sqrt2\,t.\)

Hence the arc length is

\(s=\int_0^2 2\sqrt2\,t\,\mathrm dt=4\sqrt2.\)

Since

\(\frac{\mathrm dy}{\mathrm dt}=-t\sqrt{4-t^2},\)

integrate with respect to \(t\):

\(y=\frac13(4-t^2)^{3/2},\)

using \(y=0\) at \(t=2\).

The surface area generated by rotating about the \(x\)-axis is

\(S=2\pi\int y\,\mathrm ds=2\pi\int_0^2 y\frac{\mathrm ds}{\mathrm dt}\,\mathrm dt.\)

Substitute \(y=\frac13(4-t^2)^{3/2}\) and \(\frac{\mathrm ds}{\mathrm dt}=2\sqrt2t\):

\(S=2\pi\int_0^2\frac13(4-t^2)^{3/2}(2\sqrt2t)\,\mathrm dt.\)

So

\(S=\frac{4\sqrt2\pi}{3}\int_0^2t(4-t^2)^{3/2}\,\mathrm dt.\)

Let \(u=4-t^2\), so \(\mathrm du=-2t\,\mathrm dt\). Then

\(\int_0^2t(4-t^2)^{3/2}\,\mathrm dt=\frac12\int_0^4u^{3/2}\,\mathrm du=\frac12\cdot\frac25\cdot4^{5/2}=\frac{32}{5}.\)

Therefore

\(S=\frac{4\sqrt2\pi}{3}\cdot\frac{32}{5}=\frac{128\sqrt2\pi}{15}.\)