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},}
![{\displaystyle \ S=\pi [(R+{\sqrt {r^{2}-z^{2}}})^{2}-(R-{\sqrt {r^{2}-z^{2}}})^{2}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6d71fd025ef998f16c5d3249f2e3858178a36eb7)
