Kao takva površ torus ima "rupu". Ako označimo sa c radijus od centra "rupe" do centra torusa, a sa a radijus torusa dolazimo do njegove parametarske jednačine :
{ x = ( c + a c o s v ) c o s u y = ( c + a c o s v ) s i n u z = a s i n v {\displaystyle {\begin{cases}x=(c+acosv)cosu\\y=(c+acosv)sinu\\z=asinv\end{cases}}} za u , v ∈ [ 0 , 2 π ) {\displaystyle u,v\in [0,2\pi )} [1]
gdje su u {\displaystyle u} i v {\displaystyle v} uglovi koji čine puni krug, tako da njihove vrijednosti počinju i završavaju u istoj tački
c {\displaystyle c} je udaljenost od centra cijevi do središta torusa,
a {\displaystyle a} je promjer cijevi.
c {\displaystyle c} je glavni radijus, a a {\displaystyle a} sporedni radijus.
Implicitna jednačina u Kartezijevim koordinatama je
( c − x 2 + y 2 ) 2 + z 2 = a 2 , {\displaystyle \left(c-{\sqrt {x^{2}+y^{2}}}\right)^{2}+z^{2}=a^{2},}
Površina torusa je
P = ( 2 π r ) ( 2 π R ) = 4 π 2 c a {\displaystyle P=\left(2\pi r\right)\left(2\pi R\right)=4\pi ^{2}ca} [2] [3]
a zapremina
V = ( π a 2 ) ( 2 π c ) = 2 π 2 c a 2 {\displaystyle V=\left(\pi a^{2}\right)\left(2\pi c\right)=2\pi ^{2}ca^{2}}
V = 2 π 2 R r 2 , {\displaystyle \ V=2\pi ^{2}Rr^{2},}
Dokaz S = S 1 − S 2 , {\displaystyle \ S=S_{1}-S_{2},}
S 1 = π ( R + a ) 2 , {\displaystyle \ S_{1}=\pi (R+a)^{2},}
S 2 = π ( R − a ) 2 {\displaystyle \ S_{2}=\pi (R-a)^{2}}
Prema Pitagorinoj teoremi imamo
a = r 2 − z 2 {\displaystyle \ a={\sqrt {r^{2}-z^{2}}}}
S 1 = π ( R + r 2 − z 2 ) 2 {\displaystyle \ S_{1}=\pi (R+{\sqrt {r^{2}-z^{2}}})^{2}}
S 2 = π ( R − r 2 − z 2 ) 2 {\displaystyle \ S_{2}=\pi (R-{\sqrt {r^{2}-z^{2}}})^{2}}
S = π ( R + r 2 − z 2 ) 2 − π ( R − r 2 − z 2 ) 2 {\displaystyle \ S=\pi (R+{\sqrt {r^{2}-z^{2}}})^{2}-\pi (R-{\sqrt {r^{2}-z^{2}}})^{2}}
S = π [ ( R + r 2 − z 2 ) 2 − ( R − r 2 − z 2 ) 2 ] {\displaystyle \ S=\pi [(R+{\sqrt {r^{2}-z^{2}}})^{2}-(R-{\sqrt {r^{2}-z^{2}}})^{2}]}
S = π ( R 2 + 2 R r 2 − z 2 + r 2 − z 2 − R 2 + 2 R r 2 − z 2 − r 2 + z 2 ) {\displaystyle \ S=\pi (R^{2}+2R{\sqrt {r^{2}-z^{2}}}+r^{2}-z^{2}-R^{2}+2R{\sqrt {r^{2}-z^{2}}}-r^{2}+z^{2})}
S = π ( 2 R r 2 − z 2 + 2 R r 2 − z 2 ) {\displaystyle \ S=\pi (2R{\sqrt {r^{2}-z^{2}}}+2R{\sqrt {r^{2}-z^{2}}})}
S = 4 π R r 2 − z 2 {\displaystyle \ S=4\pi R{\sqrt {r^{2}-z^{2}}}}
V = ∫ − r r S ( z ) d z {\displaystyle \ V=\int \limits _{-r}^{r}S(z)\,dz}
V = ∫ − r r 4 π R r 2 − z 2 d z {\displaystyle \ V=\int \limits _{-r}^{r}4\pi R{\sqrt {r^{2}-z^{2}}}\,dz}
V = 4 π R ∫ − r r r 2 − z 2 d z {\displaystyle \ V=4\pi R\int \limits _{-r}^{r}{\sqrt {r^{2}-z^{2}}}\,dz}
∫ − r r r 2 − z 2 d z {\displaystyle \ \int \limits _{-r}^{r}{\sqrt {r^{2}-z^{2}}}\,dz}
d u d z = d r 2 − z 2 d z {\displaystyle \ {du \over dz}={d{\sqrt {r^{2}-z^{2}}} \over dz}}
d r 2 − z 2 d z d ( r 2 − z 2 ) d ( r 2 − z 2 ) {\displaystyle {d{\sqrt {r^{2}-z^{2}}} \over dz}{d(r^{2}-z^{2}) \over d(r^{2}-z^{2})}}
d r 2 − z 2 d ( r 2 − z 2 ) d ( r 2 − z 2 ) d z {\displaystyle {d{\sqrt {r^{2}-z^{2}}} \over d(r^{2}-z^{2})}{d(r^{2}-z^{2}) \over dz}}
d u d z = 1 2 ( r 2 − z 2 ) ( d r 2 d z 2 − d z 2 d z ) = 1 2 ( r 2 − z 2 ) ( 0 − 2 z ) = − 2 z 2 ( r 2 − z 2 ) = − z ( r 2 − z 2 ) {\displaystyle \ {du \over dz}={1 \over 2(r^{2}-z^{2})}({dr^{2} \over dz^{2}}-{dz^{2} \over dz})={1 \over 2(r^{2}-z^{2})}(0-2z)={-2z \over 2(r^{2}-z^{2})}=-{z \over (r^{2}-z^{2})}}
d u = − z r 2 − z 2 d z {\displaystyle \ du=-{z \over r^{2}-z^{2}}dz}
∫ − r r r 2 − z 2 d z = z r 2 − z 2 | − r r − ∫ − r r − z 2 r 2 − z 2 d z {\displaystyle \ \int \limits _{-r}^{r}{\sqrt {r^{2}-z^{2}}}\,dz=z{\sqrt {r^{2}-z^{2}}}{\bigg |}_{-r}^{r}-\int \limits _{-r}^{r}-{z^{2} \over {\sqrt {r^{2}-z^{2}}}}dz}
∫ − r r r 2 − z 2 d z = z r 2 − z 2 | − r r − ∫ − r r − z 2 − r 2 + r 2 r 2 − z 2 d z {\displaystyle \ \int \limits _{-r}^{r}{\sqrt {r^{2}-z^{2}}}\,dz=z{\sqrt {r^{2}-z^{2}}}{\bigg |}_{-r}^{r}-\int \limits _{-r}^{r}{-z^{2}-r^{2}+r^{2} \over {\sqrt {r^{2}-z^{2}}}}dz}
∫ − r r r 2 − z 2 d z = z r 2 − z 2 | − r r − ∫ − r r r 2 − z 2 r 2 − z 2 d z + ∫ − r r r 2 r 2 − z 2 d z {\displaystyle \ \int \limits _{-r}^{r}{\sqrt {r^{2}-z^{2}}}\,dz=z{\sqrt {r^{2}-z^{2}}}{\bigg |}_{-r}^{r}-\int \limits _{-r}^{r}{r^{2}-z^{2} \over {\sqrt {r^{2}-z^{2}}}}dz+\int \limits _{-r}^{r}{r^{2} \over {\sqrt {r^{2}-z^{2}}}}dz}
∫ − r r r 2 − z 2 d z = z r 2 − z 2 | − r r − ∫ − r r r 2 − z 2 d z + r 2 ∫ − r r 1 r 2 − z 2 d z {\displaystyle \ \int \limits _{-r}^{r}{\sqrt {r^{2}-z^{2}}}\,dz=z{\sqrt {r^{2}-z^{2}}}{\bigg |}_{-r}^{r}-\int \limits _{-r}^{r}{\sqrt {r^{2}-z^{2}}}dz+r^{2}\int \limits _{-r}^{r}{1 \over {\sqrt {r^{2}-z^{2}}}}dz}
∫ − r r r 2 − z 2 d z + ∫ − r r r 2 − z 2 d z = z r 2 − z 2 | − r r + r 2 ∫ − r r 1 r 2 ( 1 − z 2 r 2 ) d z {\displaystyle \ \int \limits _{-r}^{r}{\sqrt {r^{2}-z^{2}}}\,dz+\int \limits _{-r}^{r}{\sqrt {r^{2}-z^{2}}}\,dz=z{\sqrt {r^{2}-z^{2}}}{\bigg |}_{-r}^{r}+r^{2}\int \limits _{-r}^{r}{1 \over {\sqrt {r^{2}(1-{z^{2} \over r^{2}})}}}dz}
2 ∫ − r r r 2 − z 2 d z = z r 2 − z 2 | − r r + r 2 ∫ − r r 1 ( 1 − ( z r ) 2 ) d ( z r ) {\displaystyle \ 2\int \limits _{-r}^{r}{\sqrt {r^{2}-z^{2}}}\,dz=z{\sqrt {r^{2}-z^{2}}}{\bigg |}_{-r}^{r}+r^{2}\int \limits _{-r}^{r}{1 \over {\sqrt {(1-({z \over r})^{2})}}}d({z \over r})}
∫ − r r r 2 − z 2 d z = z r 2 − z 2 + r 2 arcsin ( z r ) 2 | − r r {\displaystyle \ \int \limits _{-r}^{r}{\sqrt {r^{2}-z^{2}}}\,dz={z{\sqrt {r^{2}-z^{2}}}+r^{2}\arcsin({z \over r}) \over 2}{\bigg |}_{-r}^{r}}
∫ − r r r 2 − z 2 d z = r r 2 − r 2 + r 2 arcsin ( r r ) 2 − − r r 2 − ( − r ) 2 + r 2 arcsin ( ( − r ) r ) 2 {\displaystyle \ \int \limits _{-r}^{r}{\sqrt {r^{2}-z^{2}}}\,dz={r{\sqrt {r^{2}-r^{2}}}+r^{2}\arcsin({r \over r}) \over 2}-{-r{\sqrt {r^{2}-(-r)^{2}}}+r^{2}\arcsin({(-r) \over r}) \over 2}}
∫ − r r r 2 − z 2 d z = r 2 arcsin ( 1 ) − r 2 arcsin ( − 1 ) 2 {\displaystyle \ \int \limits _{-r}^{r}{\sqrt {r^{2}-z^{2}}}\,dz={r^{2}\arcsin({1})-r^{2}\arcsin({-1}) \over 2}}
∫ − r r r 2 − z 2 d z = r 2 2 ( π 2 − ( − π 2 ) ) {\displaystyle \ \int \limits _{-r}^{r}{\sqrt {r^{2}-z^{2}}}\,dz={r^{2} \over 2}({\pi \over 2}-(-{\pi \over 2}))}
∫ − r r r 2 − z 2 d z = r 2 π 2 {\displaystyle \ \int \limits _{-r}^{r}{\sqrt {r^{2}-z^{2}}}\,dz={r^{2}\pi \over 2}}
V = 4 π R ∫ − r r r 2 − z 2 d z → V = 4 π R r 2 π 2 {\displaystyle \ V=4\pi R\int \limits _{-r}^{r}{\sqrt {r^{2}-z^{2}}}\,dz\to V=4\pi R{r^{2}\pi \over 2}}
V = 2 π 2 R r 2 , {\displaystyle \ V=2\pi ^{2}Rr^{2},}