The shaded region R bounded by the curve $x = 5 - (y - 1)^2$, the line $y = -x$ and the x-axis is rotated about the x-axis through $360^\circ$. Find the volume of the solid formed, leaving your answer to 2 decimal places.

1. Find the key intersection points between the curve, the line, and the x-axis.

Substitute $x = -y$ into the curve equation $x = 5 - (y - 1)^2$:
$-y = 5 - (y^2 - 2y + 1)$
$y^2 - 3y - 4 = 0$
$(y - 4)(y + 1) = 0$
Since the region is bounded below by the x-axis, $y \ge 0$, so $y = 4$. This gives the top-left intersection point at $x = -4$.

The right boundary curve intersects the x-axis ($y=0$) at $x = 5 - (0-1)^2 = 4$.

2. Express the curve $x = 5 - (y - 1)^2$ in terms of $y$ as a function of $x$.

$(y - 1)^2 = 5 - x$
$y - 1 = \pm \sqrt{5 - x}$
$y = 1 \pm \sqrt{5 - x}$
The enclosed region is bounded on the right by the curve's upper branch $y = 1 + \sqrt{5 - x}$ (for $-4 \le x \le 5$) and the lower branch $y = 1 - \sqrt{5 - x}$ (for $4 \le x \le 5$).

3. Set up the definite integrals for the volume of revolution about the x-axis.

The volume $V$ must be split into three parts based on the top and bottom boundaries:

1. From $x = -4$ to $x = 0$, bounded above by $y = 1 + \sqrt{5 - x}$ and below by $y = -x$.
2. From $x = 0$ to $x = 4$, bounded above by $y = 1 + \sqrt{5 - x}$ and below by the x-axis ($y = 0$).
3. From $x = 4$ to $x = 5$, bounded above by $y = 1 + \sqrt{5 - x}$ and below by $y = 1 - \sqrt{5 - x}$.

$V = \pi \int_{-4}^{0} \left[ (1 + \sqrt{5-x})^2 - (-x)^2 \right] \text{d}x + \pi \int_{0}^{4} (1 + \sqrt{5-x})^2 \text{d}x + \pi \int_{4}^{5} \left[ (1 + \sqrt{5-x})^2 - (1 - \sqrt{5-x})^2 \right] \text{d}x$

4. Evaluate the definite integrals to find the total volume.

Evaluating each integral exactly:
$V_1 = \pi \int_{-4}^{0} (6 - x + 2\sqrt{5-x} - x^2) \text{d}x = \pi \left[ 6x - \frac{x^2}{2} - \frac{4}{3}(5-x)^{3/2} - \frac{x^3}{3} \right]_{-4}^0 = \pi \left( \frac{140 - 20\sqrt{5}}{3} \right)$
$V_2 = \pi \int_{0}^{4} (6 - x + 2\sqrt{5-x}) \text{d}x = \pi \left[ 6x - \frac{x^2}{2} - \frac{4}{3}(5-x)^{3/2} \right]_{0}^4 = \pi \left( \frac{44 + 20\sqrt{5}}{3} \right)$
$V_3 = \pi \int_{4}^{5} 4\sqrt{5-x} \text{d}x = \pi \left[ -\frac{8}{3}(5-x)^{3/2} \right]_4^5 = \pi \left( \frac{8}{3} \right)$

Summing these parts gives a beautifully clean integer multiple:
$V = \pi \left( \frac{140 - 20\sqrt{5} + 44 + 20\sqrt{5} + 8}{3} \right)$
$V = \pi \left( \frac{192}{3} \right) = 64\pi$
$V \approx 201.0619 \dots$

To 2 decimal places, the volume is $201.06$.