Answer: The area of one loop is \(\displaystyle \frac{\pi a^2}{12}\).

The maximum distance from the initial line is \(\displaystyle \frac{9a}{16}\).

The loop lies in the first quadrant and is symmetric about \(\theta=\frac{\pi}{6}\).

For \(r=a\sin3\theta\), one loop is traced for

\(0\leq\theta\leq\frac{\pi}{3}.\)

The area of this loop is

\(A=\frac12\int_0^{\pi/3}a^2\sin^2 3\theta\,\mathrm d\theta.\)

Using \(\sin^2u=\frac12(1-\cos2u)\),

\(A=\frac{a^2}{2}\int_0^{\pi/3}\frac12(1-\cos6\theta)\,\mathrm d\theta.\)

So

\(A=\frac{a^2}{4}\left[\theta-\frac16\sin6\theta\right]_0^{\pi/3}=\frac{\pi a^2}{12}.\)

The distance from the initial line is

\(y=r\sin\theta=a\sin3\theta\sin\theta.\)

At a maximum,

\(\frac{\mathrm dy}{\mathrm d\theta}=a(3\cos3\theta\sin\theta+\sin3\theta\cos\theta)=0.\)

Dividing by \(\cos3\theta\cos\theta\), where appropriate, gives

\(3\tan\theta+\tan3\theta=0.\)

Using

\(\tan3\theta=\frac{3t-t^3}{1-3t^2},\qquad t=\tan\theta,\)

we obtain

\(\frac{3t-t^3}{1-3t^2}+3t=0.\)

This simplifies to

\(t(6-10t^2)=0.\)

The interior maximum has

\(\tan^2\theta=\frac35.\)

Using this value in \(y=a\sin3\theta\sin\theta\) gives

\(y_{\max}=\frac{9a}{16}.\)

The sketch is a single loop in the first quadrant, starting and ending at the pole at \(\theta=0\) and \(\theta=\frac\pi3\), symmetric about \(\theta=\frac\pi6\).