Answer: (i) The curve starts at the pole when \(\theta=0\), since then \(r^2=\ln 1=0\). As \(\theta\) increases from \(0\) to \(2\pi\), \(r^2=\ln(1+\theta)\) increases steadily, so the curve moves continuously away from the pole in the direction of increasing \(\theta\), crossing the initial line again only as part of the full sweep through angles. A correct sketch is an outward spiral-like curve beginning at the pole on the initial line and expanding as \(\theta\) increases to \(2\pi\).
(ii) The area enclosed by a polar curve from \(\theta=\alpha\) to \(\theta=\beta\) is \(\frac12\int_{\alpha}^{\beta} r^2\,d\theta\). Here the region bounded by \(C\) and the initial line corresponds to \(0\le \theta\le 2\pi\), so
\(A=\frac12\int_0^{2\pi}\ln(1+\theta)\,d\theta.\)
Use \(u=1+\theta\), so that \(du=d\theta\). When \(\theta=0\), \(u=1\); when \(\theta=2\pi\), \(u=1+2\pi\). Hence
\(A=\frac12\int_1^{1+2\pi}\ln u\,du.\)
Now integrate by parts, or use \(\int \ln u\,du=u\ln u-u\):
\(A=\frac12\Big[ u\ln u-u\Big]_1^{1+2\pi}.\)
So
\(A=\frac12\Big((1+2\pi)\ln(1+2\pi)-(1+2\pi) - (1\cdot 0 -1)\Big)\)
\(=\frac12\Big((1+2\pi)\ln(1+2\pi)-2\pi\Big).\)
Therefore the area is \(\frac12\big((1+2\pi)\ln(1+2\pi)-2\pi\big)\).
(i) Since \(r^2=\ln(1+\theta)\), we have \(r=0\) when \(\theta=0\). Also, as \(\theta\) increases, \(\ln(1+\theta)\) increases, so \(r\) increases continuously. The curve therefore begins at the pole on the initial line and spirals steadily outwards as \(\theta\) goes from \(0\) to \(2\pi\).
(ii) The area in polar coordinates is
\(A=\frac12\int r^2\,d\theta.\)
Hence
\(A=\frac12\int_0^{2\pi}\ln(1+\theta)\,d\theta.\)
Let \(u=1+\theta\). Then \(du=d\theta\), and the limits change from \(\theta=0\) to \(u=1\), and from \(\theta=2\pi\) to \(u=1+2\pi\). So
\(A=\frac12\int_1^{1+2\pi}\ln u\,du.\)
Now
\(\int \ln u\,du=u\ln u-u,\)
so
\(A=\frac12\Big[u\ln u-u\Big]_1^{1+2\pi}.\)
Therefore
\(A=\frac12\left((1+2\pi)\ln(1+2\pi)-(1+2\pi)- (0-1)\right)\)
\(=\frac12\left((1+2\pi)\ln(1+2\pi)-2\pi\right).\)
So the exact area is \(\frac12\left((1+2\pi)\ln(1+2\pi)-2\pi\right)\).
