该图显示了两个敞口容器，一个直圆柱和一个倒置的圆台。直圆柱的底面半径为 10 cm，高为 20 cm。该圆台是通过从一个较大的圆锥底部截去一个较小的圆锥而形成的。被截去的较小圆锥的高为 4 cm。较大圆锥的底面半径为 15 cm，高为 12 cm。 将三个相同的半径为 3 cm 的球形玻璃弹珠放入该圆柱形容器中，然后倒入水至深度为 h cm。

1. Substitute the marble radius into the volume formula for a sphere

Volume of a sphere $=\dfrac{4}{3}\pi r^3$.

For each marble, $r=3$ cm.

$V=\dfrac{4}{3}\pi(3)^3=\dfrac{4}{3}\pi(27)=36\pi$

1. Find the radius of the smaller cone using similar cones

The smaller cone and the larger cone are similar.

$\frac{r}{15}=\frac{4}{12}$

$r=15\times \frac{4}{12}=5\text{ cm}$

2. Find the volume of the larger cone

For a cone,

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

Larger cone:

$V=\frac{1}{3}\pi(15)^2(12)=900\pi\text{ cm}^3$

3. Subtract the smaller cone volume from the larger cone volume

Smaller cone:

$V=\frac{1}{3}\pi(5)^2(4)=\frac{100}{3}\pi\text{ cm}^3$

Frustum volume:

$900\pi-\frac{100}{3}\pi=\frac{2600}{3}\pi\text{ cm}^3$

1. Write an expression for the volume of water in the cylinder

Volume of cylinder up to depth $h$:
$\pi(10)^2h=100\pi h$
Volume of 3 marbles:
$3(36\pi)=108\pi$
So, volume of water in the cylinder is
$100\pi h-108\pi.$

2. Equate the water volume to the frustum volume

Since the water fills the frustum completely without overflow,
$100\pi h-108\pi=\frac{2600}{3}\pi.$

3. Solve for $h$ and round to two decimal places

Cancel $\pi$:
$100h-108=\frac{2600}{3}$
$100h=\frac{2600}{3}+108$
$100h=\frac{2924}{3}$
$h=\frac{2924}{300}=9.7466\ldots$
Therefore,
$h=9.75\text{ cm}$
to two decimal places.

1. Find the exterior surface area of the open cylinder

For the cylinder, radius $r=10$ cm and height $h=20$ cm.

Since it is open at the top, the painted exterior area is
$2\pi rh+\pi r^2=2\pi(10)(20)+\pi(10)^2$
$=400\pi+100\pi=500\pi\text{ cm}^2.$

2. Find the slant height of the frustum

The frustum has height $12-4=8$ cm.

Its radii are $15$ cm and $5$ cm, so the difference in radii is
$15-5=10\text{ cm}.$

Using Pythagoras' theorem on the cross-section,
$l=\sqrt{8^2+10^2}$
$=\sqrt{164}=2\sqrt{41}\text{ cm}.$

3. Find the exterior surface area of the open frustum

For the frustum, $R=15$, $r=5$ and $l=2\sqrt{41}$.

Curved surface area of frustum:
$\pi(R+r)l=\pi(15+5)(2\sqrt{41})$
$=40\sqrt{41}\pi\text{ cm}^2.$

Since the frustum is open at the top, include only the smaller circular base:
$\pi(5)^2=25\pi\text{ cm}^2.$

Exterior area of frustum:
$40\sqrt{41}\pi+25\pi\text{ cm}^2.$

4. Add the painted areas of both containers

Total painted area
$=500\pi+40\sqrt{41}\pi+25\pi$
$=(525+40\sqrt{41})\pi\text{ cm}^2.$

Using a calculator,
$(525+40\sqrt{41})\pi\approx2453.98.$