Answer: The surface area is \(\frac{11\pi}{9}\).
The centroid of the region is \(\left(\frac{4}{7},\frac{55}{192}\right)\).
The curve is given by \(x=t^2\), \(y=t-\frac13t^3\), where \(0\leq t\leq 1\).
First find the surface area when the curve is rotated about the \(x\)-axis:
\(S=2\pi\int_0^1 y\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}\,dt\).
Now \(\frac{dx}{dt}=2t\) and \(\frac{dy}{dt}=1-t^2\), so
\(\sqrt{(2t)^2+(1-t^2)^2}=\sqrt{4t^2+1-2t^2+t^4}=\sqrt{(1+t^2)^2}=1+t^2\).
Therefore
\(S=2\pi\int_0^1 \left(t-\frac13t^3\right)(1+t^2)\,dt\).
Expand the integrand:
\(\left(t-\frac13t^3\right)(1+t^2)=t+\frac23t^3-\frac13t^5\).
Thus
\(S=2\pi\left[\frac{t^2}{2}+\frac{t^4}{6}-\frac{t^6}{18}\right]_0^1=2\pi\left(\frac12+\frac16-\frac1{18}\right)\).
So
\(S=2\pi\cdot\frac{11}{18}=\frac{11\pi}{9}\).
Now find the centroid of the region bounded by \(C\), the \(x\)-axis and \(x=1\).
The area is
\(A=\int y\,dx=\int_0^1 \left(t-\frac13t^3\right)(2t)\,dt\).
So
\(A=\int_0^1 \left(2t^2-\frac23t^4\right)dt=\left[\frac{2t^3}{3}-\frac{2t^5}{15}\right]_0^1=\frac{8}{15}\).
For \(\bar{x}\),
\(\int x y\,dx=\int_0^1 t^2\left(t-\frac13t^3\right)(2t)\,dt=\int_0^1 \left(2t^4-\frac23t^6\right)dt\).
Hence
\(\int x y\,dx=\left[\frac{2t^5}{5}-\frac{2t^7}{21}\right]_0^1=\frac{32}{105}\).
Therefore
\(\bar{x}=\frac{32/105}{8/15}=\frac47\).
For \(\bar{y}\),
\(\frac12\int y^2\,dx=\frac12\int_0^1 \left(t-\frac13t^3\right)^2(2t)\,dt\).
This is
\(\int_0^1 \left(t^3-\frac23t^5+\frac19t^7\right)dt=\left[\frac{t^4}{4}-\frac{t^6}{9}+\frac{t^8}{72}\right]_0^1=\frac{11}{72}\).
Thus
\(\bar{y}=\frac{11/72}{8/15}=\frac{55}{192}\).
So the centroid is \(\left(\frac47,\frac{55}{192}\right)\).
