Answer: The area of the surface is \(24\pi\).
For a parametric curve rotated about the \(y\)-axis, the surface area is
\(S=\int 2\pi x\,\mathrm{d}s\),
where \(\mathrm{d}s=\sqrt{\left(\frac{\mathrm{d}x}{\mathrm{d}t}\right)^2+\left(\frac{\mathrm{d}y}{\mathrm{d}t}\right)^2}\,\mathrm{d}t\).
Here \(x=t^2\) and \(y=\frac14 t^4-\ln t\), so
\(\frac{\mathrm{d}x}{\mathrm{d}t}=2t\), \(\frac{\mathrm{d}y}{\mathrm{d}t}=t^3-\frac1t\).
Hence
\(\left(\frac{\mathrm{d}s}{\mathrm{d}t}\right)^2=(2t)^2+\left(t^3-\frac1t\right)^2\)
\(=4t^2+t^6-2t^2+\frac1{t^2}=t^6+2t^2+\frac1{t^2}=\left(t^3+\frac1t\right)^2\).
Since \(t\gt 0\), we have \(\frac{\mathrm{d}s}{\mathrm{d}t}=t^3+\frac1t\).
So
\(S=2\pi\int_1^2 x\,\mathrm{d}s=2\pi\int_1^2 t^2\left(t^3+\frac1t\right)\mathrm{d}t\)
\(=2\pi\int_1^2 (t^5+t)\,\mathrm{d}t\).
Now integrate:
\(S=2\pi\left[\frac16 t^6+\frac12 t^2\right]_1^2\)
\(=2\pi\left(\left(\frac{64}{6}+2\right)-\left(\frac16+\frac12\right)\right)\)
\(=2\pi\left(\frac{32}{3}+2-\frac23\right)=2\pi(12)=24\pi\).
