Answer: \(\displaystyle \frac{\mathrm dr}{\mathrm d\theta}=-\frac{a}{(1+\theta)^2}\lt 0\), so \(r\) decreases as \(\theta\) increases.

At the point furthest from the initial line, \(\tan\theta=1+\theta\), and this root lies between \(1.1\) and \(1.2\).

The curve starts at \((a,0)\), ends on the positive \(y\)-axis at \(r=\frac{2a}{\pi+2}\), and lies in the first quadrant.

The area is

\(\displaystyle \frac{\pi a^2}{2(\pi+2)}.\)

Since

\(r=\frac{a}{1+\theta},\)

we have

\(\frac{\mathrm dr}{\mathrm d\theta}=-\frac{a}{(1+\theta)^2}.\)

Because \(a\gt 0\) and \((1+\theta)^2\gt 0\),

\(\frac{\mathrm dr}{\mathrm d\theta}\lt 0.\)

Thus \(r\) decreases as \(\theta\) increases.

The distance from a point on the curve to the initial line is its \(y\)-coordinate:

\(y=r\sin\theta=\frac{a\sin\theta}{1+\theta}.\)

Differentiate:

\(\frac{\mathrm dy}{\mathrm d\theta}=a\frac{(1+\theta)\cos\theta-\sin\theta}{(1+\theta)^2}.\)

At the point furthest from the initial line, \(\frac{\mathrm dy}{\mathrm d\theta}=0\), so

\((1+\theta)\cos\theta-\sin\theta=0.\)

Hence

\((1+\theta)\cos\theta=\sin\theta,\)

and therefore

\(\tan\theta=1+\theta.\)

Let

\(f(\theta)=\tan\theta-1-\theta.\)

Then

\(f(1.1)\approx1.9648-2.1\lt 0,\)

whereas

\(f(1.2)\approx2.5722-2.2\gt 0.\)

So \(f(\theta)=0\) has a root between \(1.1\) and \(1.2\).

For the sketch: at \(\theta=0\), \(r=a\), so the curve starts at \((a,0)\). At \(\theta=\frac{\pi}{2}\),

\(r=\frac{a}{1+\pi/2}=\frac{2a}{\pi+2}.\)

So it ends on the positive \(y\)-axis, lying in the first quadrant and curving inward as \(r\) decreases.

The required area is

\(A=\frac12\int_0^{\pi/2}\left(\frac{a}{1+\theta}\right)^2\,\mathrm d\theta.\)

Thus

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

Since

\(\int(1+\theta)^{-2}\,\mathrm d\theta=-(1+\theta)^{-1},\)

we get

\(A=\frac{a^2}{2}\left[-\frac{1}{1+\theta}\right]_0^{\pi/2}.\)

Therefore

\(A=\frac{a^2}{2}\left(1-\frac{1}{1+\pi/2}\right)=\frac{a^2}{2}\left(1-\frac{2}{\pi+2}\right)=\frac{\pi a^2}{2(\pi+2)}.\)