Step 1: Define the Radii
Let the radius of the chocolate ball (without ice-cream) be \( r \) cm.
The thickness of the ice-cream layer is given as 1 cm.
Therefore, the radius of the full sphere (chocolate ball + ice-cream) is:
\( R = r + 1 \) cm
Step 2: Volume of the Ice-cream Layer
The volume of a sphere is given by the formula:
\( V = \frac{4}{3} \pi R^3 \)
So, the volume of the ice-cream layer is the difference between the volume of the bigger sphere (with ice-cream) and the inner sphere (only chocolate ball):
\( V_{\text{ice-cream}} = \frac{4}{3} \pi (r + 1)^3 - \frac{4}{3} \pi r^3 \)
Step 3: Rate of Change Information
We are given the following:
- Rate of decrease of volume of the ice-cream layer: \( \frac{dV}{dt} = -81 \) cm³/min
- Rate of decrease of the outer radius \( R \): \( \frac{dR}{dt} = -\frac{1}{4\pi} \) cm/min
Step 4: Use Derivative of Volume with Respect to Time
We use the formula for the rate of change of volume of a sphere with respect to time:
\( \frac{dV}{dt} = 4\pi R^2 \frac{dR}{dt} \)
Now plug in the values:
\( -81 = 4\pi R^2 \left(-\frac{1}{4\pi} \right) \)
Step 5: Solve for R
Simplify the equation:
\( -81 = -R^2 \)
\( R^2 = 81 \Rightarrow R = \sqrt{81} = 9 \) cm
Step 6: Find the Radius of the Chocolate Ball
We know:
\( R = r + 1 \Rightarrow 9 = r + 1 \Rightarrow r = 8 \) cm
Step 7: Surface Area of the Chocolate Ball
The surface area \( S \) of a sphere is given by:
\( S = 4\pi r^2 \)
Substitute \( r = 8 \):
\( S = 4\pi (8)^2 = 4\pi \times 64 = 256\pi \) cm²
Final Answer:
The surface area of the chocolate ball is \( \boxed{256\pi \text{ cm}^2} \)