Answer: The cartesian equation is \(y^2=1+2x\).

The sketch is a right-opening parabola, symmetric about the \(x\)-axis, with intercepts \(\left(-\dfrac12,0\right)\) and \((0,\pm1)\). The shaded region is between the parabola and the \(y\)-axis.

The required area is \(\dfrac{2}{3}\).

Use \(x=r\cos\theta\) and \(r^2=x^2+y^2\). From \(r=\dfrac{1}{1-\cos\theta}\),

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

Since \(r\cos\theta=x\), this becomes

\(r-x=1.\)

Therefore \(r=x+1\), so

\(\sqrt{x^2+y^2}=x+1.\)

Squaring gives

\(x^2+y^2=(x+1)^2=x^2+2x+1,\)

and hence

\(y^2=1+2x.\)

This is a parabola opening to the right, symmetric about the \(x\)-axis. It crosses the \(x\)-axis at \(\left(-\dfrac12,0\right)\) and meets the \(y\)-axis at \((0,\pm1)\).

The required polar area corresponds to the region between this parabola and the \(y\)-axis. Using the upper and lower halves of the curve,

\(A=2\int_{-1/2}^{0}\sqrt{1+2x}\,dx.\)

Let \(u=1+2x\), so \(du=2\,dx\). Then

\(A=2\cdot\frac12\int_0^1 u^{1/2}\,du=\int_0^1 u^{1/2}\,du=\left[\frac23u^{3/2}\right]_0^1=\frac23.\)