Answer: The sketch is a smooth curve above the initial line, starting at the origin when \(\theta=-\frac{\pi}{2}\), passing through \((r,\theta)=(1,0)\), and ending at \((2,\frac{\pi}{2})\).

The required areas are found using the polar area formula \(A=\frac12\int r^2\,d\theta\).

For the positive part of the curve, \(0\le \theta\le \frac{\pi}{2}\):

\(A_1=\frac12\int_0^{\pi/2}(1+\sin\theta)^2\,d\theta=\frac{3\pi}{8}+1\).

For the negative part of the curve, \(-\frac{\pi}{2}\le \theta\le 0\):

\(A_2=\frac12\int_{-\pi/2}^{0}(1+\sin\theta)^2\,d\theta=\frac{3\pi}{8}-1\).

So the ratio is \(A_1:A_2=\left(\frac{3\pi}{8}+1\right):\left(\frac{3\pi}{8}-1\right)\approx 12.2:1\).

We first describe the curve \(r=1+\sin\theta\) for \(-\frac{\pi}{2}\le \theta\le \frac{\pi}{2}\).

At \(\theta=-\frac{\pi}{2}\), \(r=1-1=0\), so the curve starts at the pole. At \(\theta=0\), \(r=1\), so it crosses the initial line one unit from the pole. At \(\theta=\frac{\pi}{2}\), \(r=2\), so it ends two units above the pole on the positive \(y\)-axis. Since \(r\ge 0\) on this interval, the whole arc lies above the initial line.

Now use the polar area formula

\(A=\frac12\int r^2\,d\theta\).

Area \(A_1\) is the region enclosed by the initial line, the half-line \(\theta=\frac{\pi}{2}\), and the part of the curve with \(\theta\gt 0\). Hence

\(A_1=\frac12\int_0^{\pi/2}(1+\sin\theta)^2\,d\theta\).

Expand the square:

\(A_1=\frac12\int_0^{\pi/2}\left(1+2\sin\theta+\sin^2\theta\right)d\theta\).

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

\(A_1=\frac12\int_0^{\pi/2}\left(1+2\sin\theta+\frac{1-\cos 2\theta}{2}\right)d\theta\)

\(=\frac12\int_0^{\pi/2}\left(\frac32+2\sin\theta-\frac12\cos 2\theta\right)d\theta\).

Integrating gives

\(A_1=\frac12\left[\frac{3\theta}{2}-2\cos\theta-\frac14\sin 2\theta\right]_0^{\pi/2}\).

Substitute the limits:

\(A_1=\frac12\left(\frac{3\pi}{4}+2\right)=\frac{3\pi}{8}+1\).

Area \(A_2\) is the region enclosed by the initial line and the part of the curve with \(\theta\lt 0\). So

\(A_2=\frac12\int_{-\pi/2}^{0}(1+\sin\theta)^2\,d\theta\).

Using the same antiderivative,

\(A_2=\frac12\left[\frac{3\theta}{2}-2\cos\theta-\frac14\sin 2\theta\right]_{-\pi/2}^{0}\)

\(=\frac12\left(2-\frac{3\pi}{4}\right)=\frac{3\pi}{8}-1\).

Therefore

\(A_1:A_2=\left(\frac{3\pi}{8}+1\right):\left(\frac{3\pi}{8}-1\right)\).

Numerically,

\(\frac{\frac{3\pi}{8}+1}{\frac{3\pi}{8}-1}\approx 12.2\).

So the required ratio is \(12.2:1\) to 1 decimal place.