Answer: (i) The centroid is \(\left(\frac{3e^4+2-5e^{-4}}{e^4-e^{-4}},\frac{e^8+16-e^{-8}}{8(e^4-e^{-4})}\right)\).
(ii) \(\frac{ds}{dx}=\frac12(e^x+e^{-x})\), and the surface area is \(\frac{\pi}{2}\left(\frac{e^8}{2}+8-\frac{e^{-8}}{2}\right)\).
The area of the region is
\(A=\int_0^4 y\,dx=\frac12\int_0^4(e^x+e^{-x})\,dx=\frac12[e^x-e^{-x}]_0^4=\frac12(e^4-e^{-4})\).
For the \(x\)-coordinate of the centroid,
\(\bar{x}=\frac{1}{A}\int_0^4xy\,dx\).
Now
\(\int_0^4xy\,dx=\frac12\int_0^4x(e^x+e^{-x})\,dx\).
Using \(\int xe^x\,dx=e^x(x-1)\) and \(\int xe^{-x}\,dx=-e^{-x}(x+1)\),
\(\int_0^4xy\,dx=\frac12[e^x(x-1)-e^{-x}(x+1)]_0^4=\frac12(3e^4+2-5e^{-4})\).
Therefore
\(\bar{x}=\frac{3e^4+2-5e^{-4}}{e^4-e^{-4}}\).
For the \(y\)-coordinate,
\(\bar{y}=\frac{1}{2A}\int_0^4y^2\,dx\).
Since \(y^2=\frac14(e^{2x}+2+e^{-2x})\),
\(\frac12\int_0^4y^2\,dx=\frac18\int_0^4(e^{2x}+2+e^{-2x})\,dx\).
Thus
\(\frac12\int_0^4y^2\,dx=\frac18\left[\frac12e^{2x}+2x-\frac12e^{-2x}\right]_0^4=\frac1{16}(e^8+16-e^{-8})\).
Hence
\(\bar{y}=\frac{\frac1{16}(e^8+16-e^{-8})}{\frac12(e^4-e^{-4})}=\frac{e^8+16-e^{-8}}{8(e^4-e^{-4})}\).
So the centroid is
\(\left(\frac{3e^4+2-5e^{-4}}{e^4-e^{-4}},\frac{e^8+16-e^{-8}}{8(e^4-e^{-4})}\right)\).
For the arc length,
\(\frac{dy}{dx}=\frac12(e^x-e^{-x})\), so
\(\frac{ds}{dx}=\sqrt{1+\left(\frac{dy}{dx}\right)^2}=\sqrt{1+\frac14(e^x-e^{-x})^2}\).
But
\(1+\frac14(e^x-e^{-x})^2=\frac14(e^x+e^{-x})^2\).
Since \(e^x+e^{-x}\gt0\),
\(\frac{ds}{dx}=\frac12(e^x+e^{-x})\).
The surface area generated by rotation about the \(x\)-axis is
\(S=2\pi\int_0^4y\frac{ds}{dx}\,dx\).
Therefore
\(S=2\pi\int_0^4\frac12(e^x+e^{-x})\cdot\frac12(e^x+e^{-x})\,dx=\frac{\pi}{2}\int_0^4(e^{2x}+2+e^{-2x})\,dx\).
So
\(S=\frac{\pi}{2}\left[\frac12e^{2x}+2x-\frac12e^{-2x}\right]_0^4=\frac{\pi}{2}\left(\frac{e^8}{2}+8-\frac{e^{-8}}{2}\right)\).
