(a) The curve is sketched with the correct shape, and the section \(\frac{1}{2} \pi \leq \theta \leq \pi\) is correct. The polar coordinates of the point furthest from the pole are \((\sqrt{\pi \arctan \pi}, 0) = (1.99, 0)\).

(b) Using the substitution \(u = \pi - \theta\), the area \(A\) is given by:

\(A = \frac{1}{2} \int_0^\pi (\pi - \theta) \arctan(\pi - \theta) \, d\theta\)

or

\(A = \frac{1}{2} \int_0^\pi u \arctan(u) \, du\)

Integrating by parts:

\(\frac{1}{2} \left[ \frac{1}{2} (u^2) \arctan(u) \right]_0^\pi - \frac{1}{2} \int_0^\pi \frac{u^2}{1 + u^2} \, du\)

\(\int_0^\pi \frac{u^2}{1 + u^2} \, du = \left[ u - \arctan(u) \right]_0^\pi\)

\(A = \frac{1}{4} \pi^2 \arctan(\pi) - \frac{1}{4} \left[ u - \arctan(u) \right]_0^\pi\)

\(A = \frac{1}{4} \pi^2 - \frac{1}{4} \pi = 2.65\)

(c) Let \(y = \sqrt{\pi - \theta} \arctan(\pi - \theta) \sin \theta\).

Find the derivative:

\(\frac{dy}{d\theta} = \sqrt{\pi - \theta} \arctan(\pi - \theta) \cos \theta - \frac{(\pi - \theta)(1 + (\pi - \theta)^2)^{-1} + \arctan(\pi - \theta)}{2\sqrt{\pi - \theta} \arctan(\pi - \theta)} \sin \theta\)

Set \(\frac{dy}{d\theta} = 0\) and form the equation:

\(2(\pi - \theta) \arctan(\pi - \theta) \cot \theta - \frac{\pi - \theta}{1 + (\pi - \theta)^2} - \arctan(\pi - \theta) = 0\)

Check sign change between \(\theta = 1.2\) and \(\theta = 1.3\):

\(2(\pi - 1.2) \arctan(\pi - 1.2) \cot 1.2 - \frac{\pi - 1.2}{1 + (\pi - 1.2)^2} - \arctan(\pi - 1.2) = 0.151\)

and

\(2(\pi - 1.3) \arctan(\pi - 1.3) \cot 1.3 - \frac{\pi - 1.3}{1 + (\pi - 1.3)^2} - \arctan(\pi - 1.3) = -0.395\)

Thus, there is a sign change, confirming a root exists between \(\theta = 1.2\) and \(\theta = 1.3\).