Week 14

Calculate surface areas and volumes of spheres, pyramids, cones and compound solids (1) 

This is the first of two weekly sets of tasks looking at surface area and volume of various kinds of solids. Here we take a quick glance at spheres, cuboids, cylinders and a compound shape. The focus tends to be qualitative rather than quantitative.

MONDAY: Here we examine the surface area of part of a sphere, in the form of a slice of lime.

It turns out the the plane surface area of the quarter segment(?) of lime is the same as its curved surface area. This is a neat result - and perhaps surprising at first. We can think of the two plane regions as forming a circle of radius R, so their combined area is 𝛑R² cm². Using the formula for the surface area (S) of a sphere (S = 4𝛑R²), we can see that the curved surface area of the quarter piece of lime is also 𝛑R² cm².

If we go back to the ideas developed by Archimedes, then, instead of using the formula, we can state that the curved surface area of a hemisphere is the same as that of a cylinder that envelopes it (below). So if the hemisphere has a radius R, the cylinder has a curved surface area of 2𝛑R × R = 2𝛑R², so half the hemisphere (if cut in the way we cut up the lime) has a curved surface area of 𝛑R².

 

TUESDAY: Many students will be able to discern, from the given diagrams for this task, that the open cylinder formed from strip A has the larger capacity - but can they justify this conclusion?

The cylinder made from strip A is only half the height of the cylinder from strip B, however it is twice as wide as that from strip B, measured from left to right, say, and twice as wide measured from front to back, say. So its capacity is twice that of the other cylinder. 

We can express this more formally: let the cylinder from strip A have a height h and a circular base of radius 2r. Then the other cylinder has a height 2h and a base of radius r; so their capacities are 𝛑(2r)²h = 4𝛑r²h and 𝛑r²2h = 2𝛑r²h.

WEDNESDAY: This is a slightly simpler variant of Tuesday's task, which can be used to reinforce ideas from that task or to help those who struggled with it. The task can be made more grounded still by giving numerical values to the dimensions of the strips, eg strip A could be 2 m by 24 m and strip B could be 4 m by 12 m.

THURSDAY: Another fairly straightforward variant on the previous two tasks.

If we think of the rectangular sheets of paper (say) that would exactly cover the two curved surfaces of the cylinders, they would have the same height but the second rectangle would be twice as wide as the first (the width is equal to the circumference of the cylinder's circular top or base). So the curved surface area of the second cylinder is twice that of the first. 

However, the plane surface area of the larger cylinder, ie the area of its circular top and base, will be four times that of the smaller cylinder. So overall, the total surface area of the larger cylinder is somewhat more than twice that of the smaller cylinder.

The area of the larger cylinder's (circular) base is 4 times that of the smaller cylinder, so its volume is also 4 times that of the smaller cylinder, as they have the same height.

FRIDAY: Here we are asked to find the surface area of two fairly simple composite shapes. The task looks quite tricky as we are not told some of the dimensions of the shapes.

We could solve the task by assigning lengths to the shapes. For example the vertical pieces could be 60 cm long and the horizontal pieces 40 cm and 20 cm long. However, the absence of such information might prompt some students into realising that we don't actually need to know the size of the letters - their surface area is the same as that of the original beam. It turns out that any new faces exposed by sawing through the beams is compensated for by parts of the existing faces being covered when the pieces are joined again.

So the surface area (in cm²) of each letter is 100×(10+10+12+12) + 2×10×12 = 4400 + 240 = 4640.

EXTRA: Having solved Friday's task, students might enjoy this more extreme version.

It turns out that the original beam from which the shape was made was 160 cm long and that it has the same surface area. So its surface area (in cm²) is 160×(10+10+12+12) + 2×10×12 = 7040 + 240 = 7280. 

 

We can also use a method such as this, which is more grounded but still quite efficient:

the four vertical faces that are visible in the drawing can be joined to make a 12 cm by 70 cm rectangle; similarly the four visible horizontal faces make a 12 cm by 100 cm rectangle. the non-visible vertical and horizontal faces can be dealt with in the same way;
the front zig-zag face can be rearranged, for example like this:

So the total surface area (in cm²) is
2×12×70 + 2×12×100 + 2×10×100 + 2×10×60=2×(840+1200+1000+600) = 2×(3640) = 7280.

Note: the diagram above (right) is useful for showing that the original beam was indeed 160 cm long, although the argument still has to be made that in reconstructing the beam we are not changing the total surface area.