Outer surface

As a lateral surface or short coat is called the geometry of a part of the surface of certain body . This article deals with the lateral surface of solids of revolution , including the cylinder , the cone and the truncated cone . For the outer surface of non-rotational solids, reference is made to the respective article (see e.g. pyramid and prism ). "Bottom" ( base area ) and "lid" (top area) of the body, if present, are usually not counted as "shell" (outer surface) and are sometimes referred to as "end faces".

The outer surface of the cylinder, cone and truncated cone can be displayed two-dimensionally by "rolling" or "unwinding". In these cases, simple geometric formulas are sufficient to calculate the area. The general rule for solids of revolution is that their outer surface is created by rotating a graph of a function around a coordinate axis. With this approach, the integral calculus is required to calculate the area.

Lateral surface of the circular cylinder

Straight circular cylinder with a rolled surface

The blue area in the adjacent picture corresponds to the outer surface of the shown circular cylinder . This could be created by rotating a constant function around a coordinate axis.

It is interesting that the outer surface of a cylinder that can hold a sphere in itself (cylinder radius = sphere radius and cylinder height ) corresponds to the surface of the sphere. ${\displaystyle r}$ ${\displaystyle h=2r}$

Lateral surface of the cone

See also cone (geometry) # lateral surface .

Lateral surface of the truncated cone

Truncated cone and its developed lateral surface

The dotted area in the adjacent picture corresponds to the outer surface of the truncated cone shown , viewed from the top . This could be created by rotating a straight line around a coordinate axis.

Derivation

Let it be the surface area of ​​the whole cone, the surface area of ​​the small cone and the surface area of ​​the truncated cone, then the surface area of the truncated cone is calculated ${\displaystyle M_{\mathrm {G} }\ }$ ${\displaystyle M_{\mathrm {H} }\ }$ ${\displaystyle M_{\mathrm {KS} }\ }$ ${\displaystyle M_{\mathrm {KS} }\ }$ ${\displaystyle M_{\mathrm {KS} }=M_{\mathrm {G} }-M_{\mathrm {H} }\ }$

Now we also referred to those already set out in the sketch variable extending the height of the tip with and extending the lateral length of the tip of the cone . ${\displaystyle h}$ ${\displaystyle s}$ ${\displaystyle x}$ ${\displaystyle m\ }$ ${\displaystyle s_{\mathrm {x} }\ }$

You can then use this notation to verify

${\displaystyle (1)~~M_{\mathrm {KS} }=M_{\mathrm {G} }-M_{\mathrm {H} }=\pi r(m+s_{\mathrm {x} })-\pi Rs_{\mathrm {x} }}$

( Note on the formulas for and : The following applies to the area of ​​a segment of a circle and from which follows for the segment arc . The formulas for and result, adapted to the given variables of the cone (see drawing truncated cone on the right, developed lateral surface).) ${\displaystyle M_{\mathrm {G} }\ }$ ${\displaystyle M_{\mathrm {H} }\ }$ ${\displaystyle A=\pi r^{2}{\alpha \over 360^{\circ }}\ }$ ${\displaystyle b=2\pi r{\alpha \over 360^{\circ }}=\pi r{\alpha \over 180^{\circ }}\ }$ ${\displaystyle A={1 \over 2}br\ }$ ${\displaystyle M_{\mathrm {G} }\ }$ ${\displaystyle M_{\mathrm {H} }\ }$

With the help of the beam sets to the following relationship is derived within the cone for ago: . ${\displaystyle s_{\mathrm {x} }\ }$ ${\displaystyle {(m+s_{\mathrm {x} }) \over s_{\mathrm {x} }}={r \over R}\Leftrightarrow R(m+s_{\mathrm {x} })=rs_{\mathrm {x} }\Rightarrow s_{\mathrm {x} }={Rm \over r-R}\ }$

By inserting in one finally gets ${\displaystyle s_{\mathrm {x} }\ }$ ${\displaystyle (1)\ }$

${\displaystyle {\begin{matrix}M_{\mathrm {KS} }&=&\pi (rm+r({Rm \over r-R})-R({Rm \over r-R}))\\&=&\pi (rm+{rRm \over r-R}-{{R}^{2}m \over r-R})\\&=&\pi m(r+{R(r-R) \over r-R})\\&=&\pi (r+R)m\end{matrix}}\ }$

Area calculation with Guldin's rule

With the help of the first Guldin rule , the area can also be easily calculated: ${\displaystyle M=L\cdot 2\pi R}$

${\displaystyle L}$is the length of the generating line ( surface line ) and is the position of its center of gravity ${\displaystyle m}$ ${\displaystyle R}$ ${\displaystyle {\frac {r_{\mathrm {1} }+r_{\mathrm {2} }}{2}}.}$

Insertion results in the outer surface of the truncated cone ${\displaystyle M=\pi \cdot m\cdot (r_{\mathrm {1} }+r_{\mathrm {2} }).}$

Calculation of the lateral surface of a solid of revolution

The graph of a function , the surface line, rotates around the x-axis. Now the surface of this surface line is sought in the area from to . ${\displaystyle f\colon [a,b]\to \mathbb {R} _{0}^{+}}$ ${\displaystyle x_{1}=a}$ ${\displaystyle x_{2}=b}$

Rotation around the x-axis

${\displaystyle M=2\pi \cdot \int _{a}^{b}f(x){\sqrt {1+f'(x)^{2}}}\,\mathrm {d} x}$

Explanation:

One imagines the rotating body before and composed of on the x-axis lined discs, each having a truncated cone to the side length and the radii and represent. The sum of the lateral surfaces of the truncated cones (see above) then forms the entire lateral surface ${\displaystyle \Delta L}$ ${\displaystyle r_{1}}$ ${\displaystyle r_{2}}$

${\displaystyle M=\sum _{i}\pi \cdot \Delta L_{i}\cdot (r_{1i}+r_{2i}).}$

The line element of the rotating function is given as via the Pythagorean Theorem ${\displaystyle \Delta L_{i}}$ ${\displaystyle f(x)}$

${\displaystyle \Delta L_{i}={\sqrt {(\Delta x_{i})^{2}+(\Delta y_{i})^{2}}}={\sqrt {1+\left({\frac {\Delta y_{i}}{\Delta x_{i}}}\right)^{2}}}\Delta x_{i}.}$

At the border crossing to the integral (more and more and at the same time correspondingly thinner truncated cone disks) become and one can write ${\displaystyle r_{1i}=r_{2i}=f(x_{i})}$

${\displaystyle M=\pi \sum _{i}2\cdot f(x_{i})\cdot {\sqrt {1+\left({\frac {\Delta y_{i}}{\Delta x_{i}}}\right)^{2}}}\Delta x_{i}\rightarrow 2\pi \cdot \int _{a}^{b}f(x){\sqrt {1+f'(x)^{2}}}\,\mathrm {d} x.}$

Rotation around the y-axis

The following applies here:

${\displaystyle M=2\pi \cdot \int _{\min(f(a),f(b))}^{\max(f(a),f(b))}x{\sqrt {1+(x')^{2}}}\,\mathrm {d} y}$

with , d. H. after resolved and . ${\displaystyle x=f^{-1}(y)}$ ${\displaystyle x}$ ${\displaystyle x'={\frac {dx}{dy}}}$

See also

• Spherical layer
• Spherical segment
• Surface of revolution