如图所示，一个高为 2 单位、半径为 $r$ 且母线长为 $l$ 的直圆锥内接于一个半径为 $R$ 的球体中。

1. Express the base radius $r$ in terms of the sphere's radius $R$ using the Pythagorean theorem.

The vertical distance from the centre of the sphere to the base of the cone is $|2 - R|$.
Considering the right-angled triangle formed by the sphere's radius $R$, the base radius $r$, and this vertical distance:
$r^2 + (2 - R)^2 = R^2$
$r^2 + 4 - 4R + R^2 = R^2$
$r^2 = 4R - 4$

2. Express the slant height $l$ in terms of $R$ using the Pythagorean theorem on the cone.

Considering the right-angled triangle formed by the cone's height ($h = 2$), base radius $r$, and slant height $l$:
$l^2 = 2^2 + r^2$
$l^2 = 4 + (4R - 4)$
$l^2 = 4R$
$l = \sqrt{4R} = 2\sqrt{R}$

3. Substitute $r$ and $l$ into the curved surface area formula to obtain the final expression.

The curved surface area of the cone is $A = \pi r l$.
Substituting $r = \sqrt{4R - 4}$ and $l = 2\sqrt{R}$:
$A = \pi (\sqrt{4R - 4})(2\sqrt{R})$
$A = \pi (2\sqrt{R - 1})(2\sqrt{R})$
$A = 4\pi\sqrt{(R - 1)R}$
$A = 4\pi\sqrt{R^2 - R} \quad \text{(shown)}$

1. Set up the rate of change equation for the volume of the sphere to find $\frac{dR}{dt}$ when $R = 2$.

Given $V = \frac{4}{3}\pi R^3$, we differentiate with respect to $t$:
$\frac{dV}{dt} = 4\pi R^2 \frac{dR}{dt}$
Substitute $\frac{dV}{dt} = 8$ and $R = 2$:
$8 = 4\pi (2)^2 \frac{dR}{dt}$
$8 = 16\pi \frac{dR}{dt} \implies \frac{dR}{dt} = \frac{1}{2\pi}$

2. Differentiate the given expression for $A$ with respect to $R$.

Given $A = 4\pi(R^2 - R)^{\frac{1}{2}}$, we use the chain rule to differentiate with respect to $R$:
$\frac{dA}{dR} = 4\pi \left(\frac{1}{2}\right)(R^2 - R)^{-\frac{1}{2}}(2R - 1)$
$\frac{dA}{dR} = \frac{2\pi(2R - 1)}{\sqrt{R^2 - R}}$

3. Evaluate $\frac{dA}{dR}$ at $R = 2$.

Substitute $R = 2$ into the derivative:
$\frac{dA}{dR} = \frac{2\pi(2(2) - 1)}{\sqrt{2^2 - 2}}$
$\frac{dA}{dR} = \frac{2\pi(3)}{\sqrt{2}} = \frac{6\pi}{\sqrt{2}} = 3\pi\sqrt{2}$

4. Use the chain rule to compute the exact rate of change of $A$.

By the chain rule:
$\frac{dA}{dt} = \frac{dA}{dR} \times \frac{dR}{dt}$
$\frac{dA}{dt} = 3\pi\sqrt{2} \times \frac{1}{2\pi}$
$\frac{dA}{dt} = \frac{3\sqrt{2}}{2}$