Basic objects in space are [color=#0000ff]cylinder, cone [/color]and [color=#0000ff]sphere[/color]. [br][br][br]In the below graph, you can see two diffenrent cylinders.[br]

[br][br]In technology, special cases of cylinders have got their own "nicknames".[br][br][list][*][color=#0000ff]cylinder[/color] refers to circular ends[br][/*][*][color=#0000ff]prism[/color] refers to polygon ends[/*][/list][br]These names are also used when spoken about cylinders, so mathematical definition is quite unknown: [br] [quote][br][br] [color=#0000ff]Cylinder is formed by using two identical simple closed plane curves that are parallel to each other and joining up corresponding points on each of the curves with straight lines[/color][color=#0000ff].[/color][br][br] [/quote][br][br] [br]A cylinder is called [color=#0000ff]a right circular cylinder[/color] if its ends are circles and the line joining their centres is an axis of symmetry of the cylinder. A cylinder is called [color=#0000ff]a regular cylinder[/color] if its ends are polygons and the line joining their midpoints is an axis of symmetry of the cylinder. [br][br]The basic idea in calculating the volume of a cylinder[br][br]    [math]\Large \textcolor{blue}{V=Ah}[/math],[br][br]where [i]A [/i]is the area of the base and [i]h[/i] is the perpendicular height of the cylinder.[br]

[br][br]There was a tubular steel in a classroom. The cross-section of this tubular steel was square with a outer side length of 3 cm and the steel was 1mm thick at every side. Define the weight of this tubular steel, if it has a length of 40 cm.[br][br]The tubular steel is in a shape of cylinder, where the inner part is empty. The area of the cross-section is a subtraction of outer area and empty part.[br][br]  [math]A_u=(3\text{ cm})^2=9\text{ cm}^2\\[br]A_s=(3\text{ cm}-2\,\cdot\,0.1\text{ cm})^2=7.84\text{ cm}^2\\[br] A=A_u-A_s=1.16\text{ cm}^2[/math][br] [br][br]Now, the volume can be solved with the basic formula:[br][br]  [math] V=Ah=1.16\text{ cm}^2\,\cdot\,40\text{ cm}=46.40\text{ cm}^3.[/math][br] [br]The density of a steel is  7850 kg/m[sup]3[/sup]. The weight of the tubular steel can be solved from the formula[br][br]  [math] \rho =\frac{m}{V}\;\;\Leftrightarrow \;\;m=\rho V.[/math][br]  [br]As the density is in different unit than area, we must change the unit for cubic centimetres:[br][br]  [math]\rho =\frac{7850\text{ kg}}{\text{m^3}}=\frac{7850\text{ kg}}{(100\text{cm})^3}=\frac{7850\cdot 1000\text{ g}}{10^6\text{ cm}^3}=7.85\text{ g/cm}^3.[/math][br] [br]The weight is[br][br]  [math]\ m=7.85\text{g/cm}^3\,\cdot \,46.40\text{ cm}^3=364.24\text{ g}\approx 360 \text{ g}.[/math][br]  [br][br]

With Pythagoras' theorem is space, we get quadratic inequality:[br][br][math] \begin{array}{rrl}&\sqrt{l^2+p^2+h^2}&\leq 4.5\\[br]\Rightarrow& 3^2+2^2+h^2&\leq 4.5^2\\[br]\Leftrightarrow& h^2&\leq \frac{29}{4}\\[br]\Leftrightarrow& 0\leq h&\leq 2.7 \\[br] \end{array}[/math][br][br]The lamp can be at most 2.7 metres from the floor.

Define the volume of the hole.[br][br]When solving the volume of a cylinder, we need the cross-section area perpendicular to the height. Now, the circular hole is not perpendicular to the height of the beam but the length of the hole. So, we should think the volume in parts.

The length of the hole in this case is [br][br][math]\ l=\sqrt{40^2+50^2}\approx 64.0[/math] [br][br]The volume of the hole is the sum of right circular cylinder and the ends. The ends together form a circular cylinder with the height [i]y[/i]. [br][br]Because [i]y[/i] would have been subtracted from the original length, but it would be used in volume of the ends, the volume can be solved with the basic formula: [br][br][math] V=\pi r^2 l= \pi \cdot 10^2\cdot 64.0\approx 20116 \text{ mm}^3 = 20.1 \text{ cm}^3[/math][br][br]