Let $R$ be the region bounded by the curves $y = 4\sqrt{x}$ and $y = x^2 + 1$.

1. Find the $x$-coordinates of the points of intersection using a graphing calculator.

To find the points of intersection, equate the two functions:
$4\sqrt{x} = x^2 + 1$

Using a graphing calculator, we solve for $x$:
$x_1 \approx 0.062997$
$x_2 \approx 2.230012$

2. Set up the definite integral for the area of the bounded region.

The area of the region $R$ is the integral of the upper curve minus the lower curve between the points of intersection. By observing the graphs on the calculator, $y = 4\sqrt{x}$ is the upper curve for $0.062997 < x < 2.230012$.

$\text{Area} = \int_{0.062997}^{2.230012} \left( 4\sqrt{x} - (x^2 + 1) \right) \,\mathrm{d}x$

3. Evaluate the definite integral using a graphing calculator.

Using the graphing calculator's numerical integration function directly on the definite integral:
$\text{Area} \approx 2.9749$
$\text{Area} = 2.97 \text{ (to 3 s.f.)}$

1. Concisely retrieve the intersection points and total area from part (i).

From part (i), the $x$-coordinates of the intersections of $y = 4\sqrt{x}$ and $y = x^2 + 1$ are found by solving $4\sqrt{x} = x^2 + 1$.
Using a graphing calculator, $\alpha \approx 0.063230$ and $\beta \approx 2.19799$.
The total area of region $R$ is:
$\text{Area} = \int_{0.063230}^{2.19799} (4\sqrt{x} - x^2 - 1) \,\mathrm{d}x \approx 2.9728$

2. Set up the integral equation for the line $x=c$ dividing the area in half.

Since the line $x=c$ divides region $R$ into two equal areas, the area from $x=\alpha$ to $x=c$ must be half of the total area:
$\int_{0.063230}^{c} (4\sqrt{x} - x^2 - 1) \,\mathrm{d}x = \frac{1}{2} (2.9728)$
$\int_{0.063230}^{c} (4\sqrt{x} - x^2 - 1) \,\mathrm{d}x \approx 1.4864$

3. Evaluate the integral algebraically and solve for $c$ using a GC.

$\left[ \frac{8}{3}x^{\frac{3}{2}} - \frac{1}{3}x^3 - x \right]_{0.063230}^{c} = 1.4864$
Evaluating at the limits gives:
$\left( \frac{8}{3}c^{\frac{3}{2}} - \frac{1}{3}c^3 - c \right) - \left( \frac{8}{3}(0.063230)^{\frac{3}{2}} - \frac{1}{3}(0.063230)^3 - 0.063230 \right) = 1.4864$
$\frac{8}{3}c^{\frac{3}{2}} - \frac{1}{3}c^3 - c - (-0.020891) = 1.4864$
$\frac{8}{3}c^{\frac{3}{2}} - \frac{1}{3}c^3 - c - 1.4655 = 0$
Using a graphing calculator to solve for $c$ within the interval $(0.063, 2.198)$:
$c \approx 1.066 \approx 1.07 \text{ (to 3 s.f.)}$

1. Find the intersection points and the y-limits of integration

Solve $4\sqrt{x}=x^2+1$ (using the GC):
$x \approx 0.0630 \quad\text{and}\quad x \approx 2.2301.$
The corresponding $y$-values (the limits for integration along the $y$-axis) are
$y_1 = 4\sqrt{0.0630} \approx 1.0040, \qquad y_2 \approx 5.9734.$

2. Express the curves as x in terms of y and set up the washer integral

Rearrange each curve to make $x$ the subject:
$y=4\sqrt{x}\ \Rightarrow\ x=\frac{y^2}{16}\ (\text{inner, smaller }x), \qquad y=x^2+1\ \Rightarrow\ x=\sqrt{y-1}\ (\text{outer, larger }x).$
Rotating about the $y$-axis gives a washer cross-section, so
$V=\pi\int_{y_1}^{y_2}\left[\left(\sqrt{y-1}\right)^2-\left(\frac{y^2}{16}\right)^2\right]dy=\pi\int_{1.0040}^{5.9734}\left[(y-1)-\frac{y^4}{256}\right]dy.$

3. Evaluate the integral

Using the GC (or by hand with $\frac{y^2}{2}-y-\frac{y^5}{1280}$ evaluated between the limits):
$V=\pi\int_{1.0040}^{5.9734}\left[(y-1)-\frac{y^4}{256}\right]dy \approx \pi(6.426) \approx 20.2.$