# Immersion

Consider these solids:

1) A sphere of radius 1cm

2) A solid cylinder with height $\frac{4}{3}$cm and radius $1$cm

3) A solid circular cone with base radius $1$cm and height $4$cm.

4) A solid cylinder of height $\frac{4}{9}$cm with a hole drilled through it, leaving an annular (ring-shaped) cross-section with internal radius $1$cm and external radius $2$cm.

Can you sketch what each solid would look like?

Can you work out the volume of each solid?

Experiments are conducted where a solid is chosen and has a string firmly attached at a fixed point. The solid is then lowered at a rate of 1cm per minute into a beaker of water and the height of displaced water measured, with graphs of height against time drawn.

The results are measured on the following chart:

Can you work out what the two axes represent?

Can you work out which curve corresponds to which solid and in which orientation it is lowered into the beaker? (Note: One solid is used twice, in two different orientations).

Could you sketch the curve for the same solids in other orientations? What about different solids?

*Extension task: Can you find equations which represent the volumes of the immersed parts of the solids? They vary in difficulty; if you cannot find the equation explicity, can you describe clearly what needs to be found? Reproduce as much of the above graph as you can.*

All objects will eventually displace the same volume of fluid.

What are the key features of each solid? What will happen to the curve as these are immersed?

There is no need for lots of algebra in order to match the curves!

To begin, we can work out the volume of each solid:

Solid 1) A sphere of radius $1 \ \mathrm{cm}$ :

$$\textrm{[Volume]} = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi \ \mathrm{cm^3}$$

Solid 2) A solid cylinder of height $\frac{4}{3} \ \mathrm{cm}$ and radius $1 \ \mathrm{cm}$:

$$\textrm{[Volume]} = \pi r^2 h = \frac{4}{3}\pi \ \mathrm{cm^3}$$

Solid 3) A solid circular cone of base radius $1 \ \mathrm{cm}$ and height $4 \ \mathrm{cm}$.

$$\textrm{[Volume]} = \frac{1}{3} \times \textrm{[Volume of a cylinder]} = \frac{1}{3}\pi r^2 h = \frac{4}{3}\pi \ \mathrm{cm^3}$$

Solid 4) A solid cylinder of height $\frac{4}{9}\ \mathrm{cm}$ with a hole drilled through it, leaving an annular cross-section with interal and external radii $2 \ \mathrm{cm}$ and $1 \ \mathrm{cm}$.

$$

\begin{align}

\textrm{[Volume]} & = \textrm{[Volume of the outer cylinder]} - \textrm{[Volume of the inner cylinder]} \\

& = \pi (r_{outer}^{2} - r_{inner}^{2}) h = \frac{4}{3}\pi \mathrm{\ cm^3}\\

\end{align}

$$

We can now work out what the axes represent:

$y$-axis:

The maximum $y$ value reached by all curves is identical at around 4.2, it is in fact equal to $\frac{4}{3}\pi$. This suggests the y axis is a measure of volume, all solids will eventually displace a fluid of $\frac{4}{3}\pi \mathrm{\ cm^3}$.

The $y$ axis represents the volume of fluid displaced (or the volume of the solid immersed) in $\mathrm{cm^3}$.

$x$-axis:

The $x$-axis represents the time elapsed in minutes since lowering began.

Simon now works out which curve matches which solid, with some clear reasoning:

Curve 1:

The time taken to fully immerse the object $\approx 1.3 \mathrm{\ minutes}$

The volume displaced varies linearly with time, this must therefore represent:

Solid 2 (cylinder of height $\frac{4}{3} \mathrm{\ cm}$ and radius $1 \mathrm{\ cm}$) lowered vertically

or

Solid 4 (A solid cylinder of height $\frac{4}{9} \mathrm{\ cm}$ with a hole drilled through it) lowered vertically.

The time taken to fully immerse the object = $\frac{4}{3}\pi \mathrm{\ minutes}$

Curve 1 is therefore a cylinder of height $\frac{4}{3} \mathrm{\ cm}$ and radius $1 \mathrm{\ cm}$ lowered vertically

Curve 2 and Curve 3:

The time taken to fully immerse the object = $2 \mathrm{\ minutes}$

This could therefore be:

Solid 2 lowered sideways or solid 1 lowered in any orientation.

Solid 2 lowered sideways would initially displace a greater fluid than solid 1, it can therefore be seen that curve 2 corresponds to solid 2 lowered sideways and curve 3 to solid 1.

Curve 4 and Curve 5:

The time taken to fully immerse the object = $4 \mathrm{\ minutes}$

This could therefore be:

Solid 3 lowered vertically or solid 4 lowered sideways

Consider solid 3 lowered vertically:

Volume immersed as a function of height ($h$)

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

If we immerse this cone point first the radius varies with the height of object immersed as:

$$r = \frac{h}{4}$$

$$V(h) = \frac{\pi h^3}{48} \textrm{ or } V(t) = \frac{\pi t^3}{48} \ (\textrm{as } 1 = \frac{h}{t})$$

Solid 3 immersed point first must therefore be curve 5 which leaves solid 4 lowered vertically as curve 4.

It is possible to find algebraic forms for the volume displaced at height h as shown below, although this is very involved for some of the solids!

Curve 1:

Cylinder of height $\frac{4}{3} \mathrm{\ cm}$ and radius $1 \mathrm{\ cm}$ lowered vertically.

$$V(h) = \pi h$$

Curve 2:

A solid cylinder of height $\frac{4}{3} \mathrm{\ cm}$ and radius $1 \mathrm{\ cm}$ lowered sideways

The volume immersed equals the length of cylinder multiplied by the area of a segment

From the geometry it can be seen that $\cos(\frac{\theta}{2}) = (1-h)$

$\theta = 2\arccos(1-h)$

$\textrm{[Area of segment]} = 0.5(r^2\theta -r^2 \sin(\theta)) = 0.5(2\arccos(1-h) - \sin(2\arccos(1-h))$

$\sin\theta = 2 \sin\frac{\theta}{2} \cos\frac{\theta}{2}$

$\sin\frac{\theta}{2} = \sqrt{1 - \cos^2\frac{\theta}{2}}$

$\sin(2 \arccos(1-h)) = 2\sqrt{1 - (1-h)^2} (1-h) $

$\textrm{[Area of segment]} = 0.5(r^2\theta -r^2 \sin(\theta)) = 0.5(2\arccos(1-h) - \sin(2\arccos(1-h))$

$\textrm{[Area of segment]} = \frac{1}{2}( 2\arccos(1-h) - 2\sqrt{1 - (1-h)^2}) (1-h)$

The volume immersed equals the length of the cylinder multiplied by the area of the segment

$\textrm{[Volume at h]} = \frac{2}{3}( 2\arccos(1-h) - 2\sqrt{1 - (1-h)^2}) (1-h)$

Curve 3:

A sphere of radius $1 \mathrm{\ cm}$

The volume immersed as a function of h is equal to the volume generated when we rotate the equation of a circle about the $x$-axis by 360 degrees and evaluate this integral between the limits $r$ and $(r-h)$.

Equation of circle: $y^2 + x^2 = r^2 = 1$

$$\textrm{[Volume]} = \int_{1-h}^{1} \pi f(x)^2 \ \mathrm{d}x = \int_{1-h}^{1} \pi (1-x^2) \ \mathrm{d}x =\pi \left[ x - \frac{x^3}{3} \right]^{1} _{1-h} = \pi h^2(1-\frac{h}{3})$$

Curve 4:

A solid cylinder of height $\frac{4}{9} \mathrm{\ cm}$ with a hole drilled through it, leaving an annular cross-section with interal and external radii $2 \mathrm{\ cm}$ and $1 \mathrm{\ cm}$ lowered sideways.

When $h$ is less than $1 \mathrm{\ cm}$ the volume immersed takes the same form as curve 2 but simply changing the radius from $1 \mathrm{\ cm}$ to $2 \mathrm{\ cm}$ and changing the length of the solid from $\frac{4}{3} \mathrm{\ cm}$ to $\frac{4}{9} \mathrm{\ cm}$.

From the geometry we see that:

$\cos(\frac{\theta}{2}) = \frac{2 - h}{2}$

$\theta = 2 \arccos(1-0.5h)$

$A(h) = 0.5(r^2\theta -r^2 \sin\theta) = 0.5(8 \arccos(1- 0.5h) - 4 \sin (2 \arccos(1-0.5h))$

$\sin\theta = 2\sin\frac{\theta}{2} \cos\frac{\theta}{2}$

$\sin\frac{\theta}{2} = \sqrt{1 - \cos^2\frac{\theta}{2}}$

$\sin\theta = 2\sqrt{1 - (1-0.5h)^2} (1-0.5h) $

$\therefore \ A(h) = 4 \arccos(1- 0.5h) - 4(1-0.5h)\sqrt{1 - (1-0.5h)^2}$

$\textrm{[Volume]} = \frac{4}{9} \textrm{[area]} = \frac{16}{9}(\arccos(1- 0.5h) - (1-0.5h)\sqrt{1 - (1-0.5h)^2}$

(when $h$ is less than 1)

When h is greater than $1\mathrm{\ cm}$ the volume immersed can be found by subtracting the area of the inner segment from the outer segment and then multiplying by the length of the cylinder.

Volume of the outer segment is as above:

$\textrm{[Volume]} = \frac{4}{9} \textrm{[area]} = \frac{16}{9}(\arccos(1- 0.5h) - (1-0.5h)\sqrt{1 - (1-0.5h)^2} )$

Volume of inner segment:

From the geometry it can be seen that:

$\theta = 2\arccos(2-h)$

$\textrm{[Volume]} = \frac{4}{9} (0.5(r^2\theta -r^2 \sin\theta)) = \frac{2}{9} ( 2\arccos(2-h) - \sqrt{1 -(2-h)^2})$

Therefore:

$

\begin{align}

\textrm{[Volume immersed]} & = \frac{4}{9}(4\arccos(1- 0.5h) - 4(1-0.5h)\sqrt{1 - (1-0.5h)^2)} \\

& -\arccos(2-h) + 0.5 \sqrt{1 -(2-h)^2} )\\

\end{align}

$

(when $h$ is greater than 1)

Therefore

(when $h$ is less than 1)

$V(h) = \frac{4}{9} \textrm{[area]} = \frac{16}{9}(\arccos(1- 0.5h) - (1-0.5h)\sqrt{1 - (1-0.5h)^2} )$

(when $h$ is greater than 1)

$V(h) = \frac{4}{9}(4\arccos(1- 0.5h) - 4(1-0.5h)\sqrt{1 - (1-0.5h)^2)} -\arccos(2-h) + 0.5 \sqrt{1 -(2-h)^2} )$

Curve 5:

A solid circular cone of base radius $1 \mathrm{\ cm}$ and height $4 \mathrm{\ cm}$ lowered point first.

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

$r(h) = \frac{h}{4}$

$V(h) = \frac{\pi h^3}{48} $

### Why do this problem?

This problem provides an opportunity to think about what the shape of a graph means, by relating the graph to the situation which created it. Students need to reason which graph matches with which solid, and clearly communicate their ideas. Along the way, there is the chance to revisit formulas for working out the volumes of solids.

### Possible approach

Start by asking students to sketch the four solids described in the problem:

1) A sphere of radius $1$cm

2) A solid cylinder with height $\frac{4}{3}$cm and radius $1$cm

3) A solid circular cone with base radius $1$cm and height $4$cm.

4) A solid cylinder of height $\frac{4}{9}$cm with a hole drilled through it, leaving an annular (ring-shaped) cross-section with internal radius $1$cm and external radius $2$cm.

Then for each of the solids, students could work out the volume - this is a good opportunity to discuss the advantages of leaving answers in terms of $\pi$.

Now hand out this worksheet with the five graphs on. Challenge students to label the axes and the five graphs, annotating each curve along the way to show key points where interesting things are happening, and how they relate to the shape of the solid.

Finally, bring the class together to discuss their answers and how they pieced together which graph is which.

### Key questions

What does a straight line section of graph tell you about the cross section of the solid being immersed?

### Possible support

Maths Filler offers an opportunity to work with graphs showing changing volumes based on simple surface areas and graphs made up of straight lines rather than curves.

### Possible extension

See extension task at the end of the problem.