If the radius of the sphere increased from 10cm to 10.1cm, what is the approximate increase in surface area?

The surface area of a sphere is given by $A = 4\pi r^2$.
To find the approximate change in surface area, we use differentials: $dA = \frac{dA}{dr} \cdot \Delta r$.
First, find the derivative of $A$ with respect to $r$:

The initial radius is $r = 10 \text{ cm}$.
The change in radius is $\Delta r = 10.1 \text{ cm} - 10 \text{ cm} = 0.1 \text{ cm}$.

The approximate increase in surface area is $\boxed{8\pi \text{ cm}^2}$.

The volume of a cylinder is given by $V = \pi r^2 h$.
To find the approximate change in volume, we use differentials: $dV = \frac{dV}{dr} \cdot \Delta r$.
First, find the derivative of $V$ with respect to $r$, treating $h$ as a constant:

The height is $h = 10 \text{ cm}$.
The initial radius is $r = 4 \text{ cm}$.
The change in radius is $\Delta r = 4.02 \text{ cm} - 4 \text{ cm} = 0.02 \text{ cm}$.

The approximate increase in volume is $\boxed{1.6\pi \text{ cm}^3}$.