A curve $C$ is defined by the parametric equations $x = \theta^2 - 1$, $y = \tan\frac{\theta}{2}$, $0 \le \theta < \pi$. Find the area of the region which is bounded by $C$, the $y$-axis and the $x$-axis for which $x \le 0$. Give your answer to 4 decimal places.

1. Find the values of $\theta$ corresponding to the boundary intersections with the $x$ and $y$ axes.

To find the intersection with the $x$-axis, we set $y = 0$:
$\tan\frac{\theta}{2} = 0 \implies \theta = 0$
At $\theta = 0$, $x = -1$.

To find the intersection with the $y$-axis, we set $x = 0$:
$\theta^2 - 1 = 0 \implies \theta^2 = 1 \implies \theta = 1 \text{ (since } 0 \le \theta < \pi)$
Thus, the bounded region corresponds to the parameter interval $0 \le \theta \le 1$.

2. Set up the definite integral for the area using the parametric equations.

The area of the region bounded by the curve, the $x$-axis, and the $y$-axis (for $x \le 0$) is given by the integral of $y$ with respect to $x$:
$\text{Area} = \int_{-1}^{0} y \, dx$

Differentiating $x$ with respect to $\theta$:
$\frac{dx}{d\theta} = 2\theta \implies dx = 2\theta \, d\theta$

Substituting the parametric expressions and the corresponding $\theta$ limits:
$\text{Area} = \int_{0}^{1} \tan\left(\frac{\theta}{2}\right) (2\theta) \, d\theta$

3. Evaluate the definite integral using a graphing calculator.

Using a graphing calculator to evaluate the integral numerically (ensuring the calculator is in radian mode):
$\int_{0}^{1} 2\theta \tan\left(\frac{\theta}{2}\right) \, d\theta \approx 0.351288...$
$\text{Area} = 0.3513 \text{ (to 4 d.p.)}$