Answer: The arc length is \(5\sqrt{5}-1\).

The centroid of the region is \(\left(\frac{30}{7},\frac{5}{4}\right)\).

For the curve \(x=\frac{3}{2}t^2\), \(y=t^3\), with \(0\le t\le 2\):

Differentiate:

\(\frac{dx}{dt}=3t,\qquad \frac{dy}{dt}=3t^2.\)

So the arc-length element is

\(\frac{ds}{dt}=\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}=\sqrt{9t^2+9t^4}=3t\sqrt{1+t^2},\)

since \(t\ge 0\).

Hence

\(s=\int_0^2 3t\sqrt{1+t^2}\,dt.\)

Let \(u=1+t^2\). Then \(du=2t\,dt\), so

\(s=\frac32\int_1^5 u^{1/2}\,du=\left[u^{3/2}\right]_1^5=\left[(1+t^2)^{3/2}\right]_0^2.\)

Therefore

\(s=5\sqrt5-1.\)

Now find the centroid of the region enclosed by the curve, the \(x\)-axis and the line \(x=6\). Since \(x=\frac32 t^2\), the value \(x=6\) corresponds to \(t=2\).

For a region under \(y=y(x)\),

\(\bar{x}=\frac{\int x y\,dx}{\int y\,dx},\qquad \bar{y}=\frac{\frac12\int y^2\,dx}{\int y\,dx}.\)

Also, \(dx=\frac{dx}{dt}dt=3t\,dt\).

The area is

\(A=\int_0^6 y\,dx=\int_0^2 t^3(3t)\,dt=3\int_0^2 t^4\,dt=3\left[\frac{t^5}{5}\right]_0^2=\frac{96}{5}.\)

For \(\bar{x}\):

\(\int_0^6 xy\,dx=\int_0^2 \left(\frac32 t^2\right)(t^3)(3t)\,dt=\frac92\int_0^2 t^6\,dt=\frac92\left[\frac{t^7}{7}\right]_0^2=\frac{576}{7}.\)

So

\(\bar{x}=\frac{576/7}{96/5}=\frac{30}{7}.\)

For \(\bar{y}\):

\(\frac12\int_0^6 y^2\,dx=\frac12\int_0^2 t^6(3t)\,dt=\frac32\int_0^2 t^7\,dt=\frac32\left[\frac{t^8}{8}\right]_0^2=24.\)

Hence

\(\bar{y}=\frac{24}{96/5}=\frac54.\)

So the centroid is \(\left(\frac{30}{7},\frac54\right)\).