Answer: (i) The curve is a cardioid with a cusp at the pole. Since \(r=a(1+\sin\theta)\), it is symmetric about the line \(\theta=\frac{\pi}{2}\). The maximum radius is \(r=2a\) at \(\theta=\frac{\pi}{2}\), so the bulge lies upwards.

(ii) The area enclosed is \(\frac{3\pi a^{2}}{2}\).

(iii) The arc from the pole to the furthest point from the pole has length \(s=\sqrt{2}\,a\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\sqrt{1+\sin\theta}\,\mathrm{d}\theta\).

(iv) With \(u=1+\sin\theta\), this becomes \(s=\sqrt{2}\,a\int_{0}^{2}\frac{1}{\sqrt{2-u}}\,\mathrm{d}u\), so \(s=4a\).

(i) For \(r=a(1+\sin\theta)\), we have \(r=0\) when \(\sin\theta=-1\), i.e. at \(\theta=-\frac{\pi}{2}\). Also \(r\) is largest when \(\sin\theta=1\), so the maximum value is \(r=2a\) at \(\theta=\frac{\pi}{2}\). Hence the curve is a cardioid with its cusp at the pole and opening upwards, symmetric about the vertical axis \(\theta=\frac{\pi}{2}\).

(ii) The area enclosed by a polar curve is \(A=\frac{1}{2}\int r^{2}\,\mathrm{d}\theta\), so

\(A=\frac{1}{2}\int_{-\pi}^{\pi} a^{2}(1+\sin\theta)^{2}\,\mathrm{d}\theta\).

Expand the square:

\(A=\frac{a^{2}}{2}\int_{-\pi}^{\pi}\left(1+2\sin\theta+\sin^{2}\theta\right)\mathrm{d}\theta\).

Use \(\sin^{2}\theta=\frac{1}{2}(1-\cos 2\theta)\):

\(A=\frac{a^{2}}{2}\int_{-\pi}^{\pi}\left(\frac{3}{2}+2\sin\theta-\frac{1}{2}\cos 2\theta\right)\mathrm{d}\theta\).

Now \(\int_{-\pi}^{\pi}\sin\theta\,\mathrm{d}\theta=0\) and \(\int_{-\pi}^{\pi}\cos 2\theta\,\mathrm{d}\theta=0\), so

\(A=\frac{a^{2}}{2}\cdot \frac{3}{2}\cdot 2\pi=\frac{3\pi a^{2}}{2}\).

(iii) For a polar curve, the arc length is given by \(s=\int\sqrt{r^{2}+\left(\frac{\mathrm{d}r}{\mathrm{d}\theta}\right)^{2}}\,\mathrm{d}\theta\).

The arc from the pole to the point furthest from the pole runs from \(\theta=-\frac{\pi}{2}\) to \(\theta=\frac{\pi}{2}\), since \(r=0\) at the pole and \(r=2a\) at the furthest point.

Now

\(r=a(1+\sin\theta),\qquad \frac{\mathrm{d}r}{\mathrm{d}\theta}=a\cos\theta\).

Therefore

\(s=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\sqrt{a^{2}(1+\sin\theta)^{2}+a^{2}\cos^{2}\theta}\,\mathrm{d}\theta\).

Simplifying inside the square root gives

\(s= a\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\sqrt{(1+\sin\theta)^{2}+\cos^{2}\theta}\,\mathrm{d}\theta\)

\(= a\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\sqrt{1+2\sin\theta+\sin^{2}\theta+\cos^{2}\theta}\,\mathrm{d}\theta\)

\(= a\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\sqrt{2(1+\sin\theta)}\,\mathrm{d}\theta\)

\(=\sqrt{2}\,a\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\sqrt{1+\sin\theta}\,\mathrm{d}\theta\).

(iv) Let \(u=1+\sin\theta\). Then \(\mathrm{d}u/\mathrm{d}\theta=\cos\theta\), so

\(\cos\theta=\sqrt{1-\sin^{2}\theta}=\sqrt{1-(u-1)^{2}}=\sqrt{u(2-u)}\)

for \(-\frac{\pi}{2}\le \theta\le \frac{\pi}{2}\), where \(\cos\theta\ge 0\). Also, when \(\theta=-\frac{\pi}{2}\), \(u=0\), and when \(\theta=\frac{\pi}{2}\), \(u=2\).

Since \(\mathrm{d}\theta=\frac{\mathrm{d}u}{\cos\theta}\),

\(s=\sqrt{2}\,a\int_{0}^{2}\sqrt{u}\,\frac{\mathrm{d}u}{\sqrt{u(2-u)}}\)

\(=\sqrt{2}\,a\int_{0}^{2}\frac{1}{\sqrt{2-u}}\,\mathrm{d}u\).

Now evaluate:

\(s=\sqrt{2}\,a\left[-2\sqrt{2-u}\right]_{0}^{2}=\sqrt{2}\,a(0+2\sqrt{2})=4a\).