The diagram shows a container which consists of a cylinder of radius $r$ cm and height $h$ cm joined to a hemisphere of radius $r$ cm.

1. Write down the volumes of the hemisphere and the cylinder.

Volume of hemisphere
$=\frac{1}{2}\left(\frac{4}{3}\pi r^3\right)=\frac{2}{3}\pi r^3$

Volume of cylinder
$=\pi r^2h$

2. Use the given volume ratio to solve for $h$ in terms of $r$.

Given
$\frac{\text{Volume of hemisphere}}{\text{Volume of cylinder}}=\frac{1}{3},$

$\frac{\frac{2}{3}\pi r^3}{\pi r^2h}=\frac{1}{3}$

Canceling $\pi r^2$,
$\frac{2r}{3h}=\frac{1}{3}$

$6r=3h$

$h=2r$

1. Find the height of the cylindrical part that is in contact with the water.

From part (i), the cylinder has height $h=2r$.
The water depth is $8$ cm from the bottom of the hemisphere.
Since the hemisphere has vertical height $r$, the water rises into the cylinder by
$8-r\text{ cm}.$

2. Add the curved surface area of the hemisphere and the wetted curved surface area of the cylinder.

Surface area in contact with water
$= \text{curved surface area of hemisphere} + \text{curved surface area of wetted cylinder}$
$=2\pi r^2+2\pi r(8-r)$
$=2\pi r^2+16\pi r-2\pi r^2$
$=16\pi r.$