How to find the height, given the volume of a cylinder and cone (conjoined together)?

Question

Here is a picture for a diagram:

If water fills this "vase" up to half its capacity (NOT half its height), what will the height of the water be, starting from the bottom?

And if you could explain the steps leading up the answer, that would be great.

Thanks!

EDIT: Sorry, the diagram didn't have the volume, here it is:

Volume of cylinder: 1398cm cubed.
Volume of Cone: 162.32cm cubed.
Total volume: 1560.32cm cubed.

Answer 1

Assuming that "halfway point" means "half of the volume of the object is filled", here are the steps:

1. Find the volume of the object.

   • Find the volume of the cone.
   • Find the volume of the cylinder.
   • Add those numbers together.
     
     

2. Divide the volume of the object by 2 to obtain the volume of water that will be added.

3. Add the water to the object and see how high it goes.

   • The volume of the water will be greater than the volume of the conical part of the object, so subtract the volume of the cone from the volume of the water to obtain the volume of water that reaches the cylindrical part.

   • The volume of the water reaching the cylindrical part will form a cylinder of the same radius as the cylindrical part, but with an unknown height. Solve for this height, knowing the volume and radius.

   • Add the height of the cone.

Answer 2

Here's another method using comparison ratios, which is convenient in this case because the volume formulas for a cone and a cylinder look similar:

$$\frac{V_{cyl}}{V_{cone}} \ = \ \frac{ \pi R^2 \cdot h_{cyl}}{\frac{1}{3} \pi R^2 \cdot h_{cone}} \ = \ 3 \cdot \left(\frac{ h_{cyl}}{ h_{cone}} \right) \ , $$

since the cone and the cylinder for this "vase" have the same "base" radius. Filling the vase to half of its capacity requires a volume of water

$$\frac{1}{2} V_{tot} \ = \ \frac{1}{2} ( \ V_{cyl} + V_{cone} \ ) \ = \ \frac{1}{2} ( \ 3 \cdot \left[ \frac{ h_{cyl}}{ h_{cone}} \right] + \ 1 \ ) \cdot V_{cone} \ . $$

The volume of the cylindrical section that has water in it is then

$$V_{fill} \ = \ \frac{1}{2} V_{tot} \ - \ V_{cone} \ = \ \frac{1}{2} ( \ 3 \cdot \left[ \frac{ h_{cyl}}{ h_{cone}} \right] - \ 1 \ ) \cdot V_{cone} \ . $$

We can now apply the volume ratio we used in the first equation to the filled part of the cylindrical section, since that is also a cylinder. Together with our result immediately above, we have

$$\frac{V_{fill}}{V_{cone}} \ = \ 3 \cdot \left(\frac{ h_{fill}}{ h_{cone}} \right) \ = \ \frac{1}{2} ( \ 3 \cdot \left[ \frac{ h_{cyl}}{ h_{cone}} \right] - \ 1 \ ) $$

$$ \Rightarrow \ \frac{ h_{fill}}{ h_{cone}} = \ \frac{1}{6} ( \ 3 \cdot \left[ \frac{ h_{cyl}}{ h_{cone}} \right] - \ 1 \ ) \ = \ \frac{1}{2} \cdot \left[ \frac{ h_{cyl}}{ h_{cone}} \right] - \ \frac{1}{6} \ . $$

The total height of water in the vase is thus

$$ h_{tot} \ = \ h_{fill} \ + \ h_{cone} \ = \ \left( \frac{1}{2} \cdot \left[ \frac{ h_{cyl}}{ h_{cone}} \right] - \ \frac{1}{6} \right) \cdot h_{cone} \ + \ h_{cone} $$

$$= \ \left( \frac{1}{2} \cdot \left[ \frac{ h_{cyl}}{ h_{cone}} \right] + \ \frac{5}{6} \right) \cdot h_{cone} \ . $$

For $ \ h_{cyl} = 17.8 \ \text{cm.} \ $ and $ \ h_{cone} = 6.2 \ \text{cm.} \ , $ we obtain

$$h_{tot} \ = \ \left( \frac{1}{2} \cdot \left[ \frac{ 17.8}{6.2} \right] + \ \frac{5}{6} \right) \cdot \ 6.2 \ \text{cm.} \ \approx \ 2.269 \ \cdot 6.2 \ \text{cm.} \ \approx \ 14.07 \ \text{cm.}$$

$$ \\ $$

This formula also checks with our geometrical intuition for the situation where the heights of the cone and the cylinder are the same. In this case, the volume of the cylinder would be three times the volume of the cone, so the "vase" would have four times the cone's volume. Filling the vase to half-capacity would put as much water into the cylinder as into the cone, so the cylindrical section would be filled to one-third of its volume, and so to one-third of its height. The total height of water in the vase would therefore be $ \ \frac{4}{3} \ $ of the cone's height. Indeed, the formula produces

$$h_{tot} \ = \ \left( \frac{1}{2} \cdot 1 + \ \frac{5}{6} \right) \cdot \ h_{cone} \ = \ \frac{4}{3} \cdot h_{cone} \ . $$

An interesting feature of this calculation is that the proportionality holds regardless of what the radius of the vase is (so we did not need to spend time calculating $ \ R^2 \ ) \ . $