Answer:

The polar equation

\(r=a(1-\cos\theta)\)

represents a cardioid. It has a cusp at the origin when \(\theta=0\), is symmetric about the initial line, and has maximum radius \(2a\) when \(\theta=\pi\).

The area between \(\theta=\frac{\pi}{2}\) and \(\theta=\frac{3\pi}{2}\) is

\(A=\frac12\int_{\pi/2}^{3\pi/2}a^{2}(1-\cos\theta)^{2}\,d\theta.\)

Expand the square:

\((1-\cos\theta)^2=1-2\cos\theta+\cos^2\theta.\)

Using \(\cos^2\theta=\frac12(1+\cos2\theta)\),

\(A=\frac{a^2}{2}\int_{\pi/2}^{3\pi/2}\left(\frac32-2\cos\theta+\frac12\cos2\theta\right)d\theta.\)

Hence

\(A=\frac{a^2}{2}\left[\frac32\theta-2\sin\theta+\frac14\sin2\theta\right]_{\pi/2}^{3\pi/2}.\)

This gives

\(A=\frac{a^2}{2}\left(\frac{3\pi}{2}+4\right)=a^{2}\left(\frac{3\pi}{4}+2\right).\)

For arc length in polar coordinates,

\(\left(\frac{ds}{d\theta}\right)^2=r^2+\left(\frac{dr}{d\theta}\right)^2.\)

Here

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

Therefore

\(\left(\frac{ds}{d\theta}\right)^2=a^2(1-\cos\theta)^2+a^2\sin^2\theta.\)

So

\(\left(\frac{ds}{d\theta}\right)^2=a^2(1-2\cos\theta+\cos^2\theta+\sin^2\theta)=2a^2(1-\cos\theta).\)

Since \(1-\cos\theta=2\sin^2\left(\frac{\theta}{2}\right)\),

\(\left(\frac{ds}{d\theta}\right)^2=4a^2\sin^2\left(\frac{\theta}{2}\right).\)

On \(\frac{\pi}{2}\leq\theta\leq\frac{3\pi}{2}\), \(\sin\frac{\theta}{2}\geq0\), so

\(\frac{ds}{d\theta}=2a\sin\frac{\theta}{2}.\)

The arc length is therefore

\(s=\int_{\pi/2}^{3\pi/2}2a\sin\frac{\theta}{2}\,d\theta=\left[-4a\cos\frac{\theta}{2}\right]_{\pi/2}^{3\pi/2}=4\sqrt2\,a.\)