The diagram shows the curve $C$ with equation $y = \frac{12}{(3x+2)^2 + 4}$ which has a turning point at $\left(-\frac{2}{3}, 3\right)$. The region $R$ is bounded by $C$, the $y$-axis and the line $y = 3$.

1. Set up the definite integral for the area of region $R$.

The region $R$ is bounded above by $y = 3$, below by $y = \frac{12}{(3x+2)^2 + 4}$, and horizontally between the turning point at $x = -\frac{2}{3}$ and the $y$-axis ($x = 0$).

$\text{Area} = \int_{-2/3}^{0} \left( 3 - \frac{12}{(3x+2)^2 + 4} \right) \, dx$

2. Integrate the expression using the standard integral for $\frac{1}{x^2 + a^2}$.

$\begin{aligned} \int \left( 3 - \frac{12}{(3x+2)^2 + 2^2} \right) \, dx &= 3x - 12 \left( \frac{1}{3} \right) \left( \frac{1}{2} \right) \tan^{-1}\left( \frac{3x+2}{2} \right) \\ &= 3x - 2\tan^{-1}\left( \frac{3x+2}{2} \right) \end{aligned}$

3. Evaluate the definite integral using the limits to find the exact area.

$\begin{aligned} \text{Area} &= \left[ 3x - 2\tan^{-1}\left( \frac{3x+2}{2} \right) \right]_{-2/3}^{0} \\ &= \left( 3(0) - 2\tan^{-1}\left(\frac{2}{2}\right) \right) - \left( 3\left(-\frac{2}{3}\right) - 2\tan^{-1}\left(\frac{0}{2}\right) \right) \\ &= \left( 0 - 2\tan^{-1}(1) \right) - \left( -2 - 2\tan^{-1}(0) \right) \\ &= \left( 0 - 2\left(\frac{\pi}{4}\right) \right) - (-2 - 0) \\ &= 2 - \frac{\pi}{2} \end{aligned}$

1. Find the limits of integration for $y$ by calculating the $y$-intercept of the curve $C$.

Substitute $x = 0$ into the equation of $C$ to find the lower bound of $y$:
$y = \frac{12}{(3(0)+2)^2 + 4} = \frac{12}{8} = 1.5$
From the given figure and coordinates of the turning point, the region $R$ is bounded vertically between $y = 1.5$ and $y = 3$.

2. Express $x$ in terms of $y$, selecting the correct square root based on the domain of region $R$.

Rearrange the equation of $C$ to make $x$ the subject:
$y = \frac{12}{(3x+2)^2 + 4}$
$(3x+2)^2 + 4 = \frac{12}{y}$
$(3x+2)^2 = \frac{12}{y} - 4$
Since region $R$ is bounded between the turning point $x = -\frac{2}{3}$ and the $y$-axis $x = 0$, we know that $-\frac{2}{3} \le x \le 0$, which means $3x+2 \ge 0$. Therefore, we must take the positive square root:
$3x+2 = \sqrt{\frac{12}{y} - 4}$
$x = \frac{1}{3}\left( \sqrt{\frac{12}{y} - 4} - 2 \right)$

3. Set up the integral for the volume of revolution about the $y$-axis and evaluate it using a graphing calculator.

The volume of revolution about the $y$-axis is given by $V = \pi \int_{y_1}^{y_2} x^2 \, \mathrm{d}y$:
$V = \pi \int_{1.5}^{3} \left[ \frac{1}{3}\left( \sqrt{\frac{12}{y} - 4} - 2 \right) \right]^2 \, \mathrm{d}y$
Using a graphing calculator to evaluate this definite integral numerically:
$V \approx 0.5125008\dots$
$V = 0.5125 \text{ (to 4 d.p.)}$