Answer: The surface area is \(\frac{11\pi}{9}\).

The centroid of the region is \(\left(\frac{4}{7},\frac{55}{192}\right)\).

For the parametric curve \(x=t^2\), \(y=t-\frac{1}{3}t^3\), \(0\le t\le 1\), we have

\(\frac{\mathrm{d}x}{\mathrm{d}t}=2t\), and \(\frac{\mathrm{d}y}{\mathrm{d}t}=1-t^2\).

Surface area

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

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

Now

\(\left(\frac{\mathrm{d}x}{\mathrm{d}t}\right)^2+\left(\frac{\mathrm{d}y}{\mathrm{d}t}\right)^2=4t^2+(1-t^2)^2=4t^2+1-2t^2+t^4=(1+t^2)^2\).

Since \(1+t^2\gt 0\) on \([0,1]\),

\(\sqrt{\left(\frac{\mathrm{d}x}{\mathrm{d}t}\right)^2+\left(\frac{\mathrm{d}y}{\mathrm{d}t}\right)^2}=1+t^2\).

So

\(S=2\pi\int_0^1\left(t-\frac{1}{3}t^3\right)(1+t^2)\,\mathrm{d}t\).

Expand the integrand:

\(\left(t-\frac{1}{3}t^3\right)(1+t^2)=t+\frac{2}{3}t^3-\frac{1}{3}t^5\).

Hence

\(S=2\pi\int_0^1\left(t+\frac{2}{3}t^3-\frac{1}{3}t^5\right)\mathrm{d}t\)

\(=2\pi\left[\frac{t^2}{2}+\frac{t^4}{6}-\frac{t^6}{18}\right]_0^1\)

\(=2\pi\left(\frac{1}{2}+\frac{1}{6}-\frac{1}{18}\right)=2\pi\cdot\frac{11}{18}=\frac{11\pi}{9}.\)

Centroid of the region

The region is bounded by the curve, the \(x\)-axis, and \(x=1\). Since \(x=t^2\), the interval \(0\le t\le 1\) covers \(0\le x\le 1\).

For a region under a curve \(y\) from \(x=0\) to \(x=1\),

\(A=\int y\,\mathrm{d}x\), \(\bar{x}=\frac{1}{A}\int x y\,\mathrm{d}x\), and \(\bar{y}=\frac{1}{2A}\int y^2\,\mathrm{d}x\).

Using the parameter \(t\), with \(\mathrm{d}x=\frac{\mathrm{d}x}{\mathrm{d}t}\mathrm{d}t=2t\,\mathrm{d}t\):

\(A=\int_0^1 y\frac{\mathrm{d}x}{\mathrm{d}t}\,\mathrm{d}t=\int_0^1\left(t-\frac{1}{3}t^3\right)(2t)\,\mathrm{d}t\)

\(=\int_0^1\left(2t^2-\frac{2}{3}t^4\right)\mathrm{d}t=\left[\frac{2t^3}{3}-\frac{2t^5}{15}\right]_0^1=\frac{8}{15}.\)

Next,

\(\int x y\,\mathrm{d}x=\int_0^1 x y\frac{\mathrm{d}x}{\mathrm{d}t}\,\mathrm{d}t\)

\(=\int_0^1 t^2\left(t-\frac{1}{3}t^3\right)(2t)\,\mathrm{d}t=\int_0^1\left(2t^4-\frac{2}{3}t^6\right)\mathrm{d}t\)

\(=\left[\frac{2t^5}{5}-\frac{2t^7}{21}\right]_0^1=\frac{32}{105}.\)

So

\(\bar{x}=\frac{\frac{32}{105}}{\frac{8}{15}}=\frac{32}{105}\cdot\frac{15}{8}=\frac{4}{7}.\)

Finally,

\(\frac{1}{2}\int y^2\,\mathrm{d}x=\frac{1}{2}\int_0^1 y^2\frac{\mathrm{d}x}{\mathrm{d}t}\,\mathrm{d}t\)

\(=\frac{1}{2}\int_0^1\left(t-\frac{1}{3}t^3\right)^2(2t)\,\mathrm{d}t\)

\(=\int_0^1\left(t^3-\frac{2}{3}t^5+\frac{1}{9}t^7\right)\mathrm{d}t\)

\(=\left[\frac{t^4}{4}-\frac{t^6}{9}+\frac{t^8}{72}\right]_0^1=\frac{11}{72}.\)

Therefore

\(\bar{y}=\frac{\frac{11}{72}}{\frac{8}{15}}=\frac{11}{72}\cdot\frac{15}{8}=\frac{55}{192}.\)

Hence the centroid is \(\left(\frac{4}{7},\frac{55}{192}\right)\).