该图显示了一个圆柱形固体装饰品，其一端挖去了一个直立的圆锥，另一端连接着一个半球。 圆锥和圆柱的垂直高度分别为 $x$ cm 和 $2x$ cm。

1. Form the ratio of the cone volume to the cylinder volume and simplify.

Volume of cone $= \dfrac{1}{3}\pi r^2 x$

Volume of cylinder $= \pi r^2(2x)$

$\text{Ratio} = \frac{\frac{1}{3}\pi r^2 x}{2\pi r^2 x}=\frac{1}{6}$

1. Substitute the given cylinder volume and height into the cylinder volume formula.

For a cylinder,
$V=\pi r^2h$
Given $V=345$ and $h=8$,
$345=\pi r^2(8)$

2. Solve for the radius.

$r^2=\frac{345}{8\pi}$
$r=\sqrt{\frac{345}{8\pi}}$
$r=3.704\ldots$
So, to 3 significant figures,
$r=3.70$

1. Find the cone height from the given relationship between cylinder height and cone height.

The cylinder height is $2x=8$.

Hence
$x=4$

So the vertical height of the cone is $4\text{ cm}$.

2. Find the slant height of the cone using Pythagoras' theorem.

Using the radius from part (b)(i), $r=3.70\text{ cm}$.

For the cone,
$l=\sqrt{r^2+h^2}$
$l=\sqrt{3.70^2+4^2}$
$l=\sqrt{29.69}=5.448\ldots$

3. Calculate the curved surface area of the cone.

Curved surface area of cone
$=\pi r l$
$=\pi(3.70)(5.448\ldots)$
$=63.3\text{ cm}^2\text{, to 3 s.f.}$

1. Convert the paint thickness into centimetres.

$0.2\text{ mm}=0.02\text{ cm}$

2. Find the total exterior surface area to be painted.

Using $r=3.70\text{ cm}$ from part (b)(i), cylinder height $=8\text{ cm}$, and cone curved surface area $=86.1\text{ cm}^2$ from part (b)(ii),

$\text{curved surface area of cylinder}=2\pi rh$
$=2\pi(3.70)(8)=186\text{ cm}^2\text{ approximately}$

$\text{curved surface area of hemisphere}=2\pi r^2$
$=2\pi(3.70)^2=86.0\text{ cm}^2\text{ approximately}$

Total area to be painted
$=186+86.0+86.1=358.1\text{ cm}^2\text{ approximately}$

3. Multiply surface area by thickness to find the volume of paint.

$\text{Volume of paint}=\text{area}\times\text{thickness}$
$=358.1\times 0.02$
$=7.162$

Using unrounded values gives approximately $7.17\text{ cm}^3$.