SPHERE FORMULAS Formulas for Spherical Radius Machinery's Handbook, 31st Edition
56
To Find
Formula
To Find
Formula
Radius of sphere from volume N v
Radius of section T
3 N v 4 π ----- 3 =
P 2 – 4 H 2 – 8 H ------------------ 2 2 Q
2
P
R N
=
4 + ---
R T
Formulas for Spherical Areas and Volumes
Section
Area Formula
Volume Formula
π 6 -- D 3 4 π 3 = ( )
--- R 3
=
N v
2
Entire sphere N a
4 π R
=
4 π 3 --- R (
1 3
R 2 3
4 π R 1 2
4 π R 2 2
Section G
)
=
–
G v
=
+
G a
2
2 π R
F
2 1 φ 2 – cos -- -
Section K
K a 2 π R =
3 = ---------
K v
2
2
2
E
E 8 ---
F
π F
2
Section S
π F
=
=
4 + ---
6 + ---
S a
S
v
π 6 -- H 2 3 Q 2 4
2
----- 3 P 4 ----- + +
=
2 π RH =
T
H
Section T
T a
v
3 R
2 3 –
) 1 φ 2 – cos --
) 1 φ 2 – cos --
2 R
2 2 +
Section U
=
2 π R 1 (
U
U a 2 π R 1 ( =
v
2 R
1 R 2 2
2 R
W v 2 =
Section W
W a 4 π =
1 R 2
π
φ 360 -----
2 R
φ 360 -----
2 R
1 R 2 2
Section Z Example 1: Find the inside and outside surface area G a and volume G v of wall G , provided that R 1 is 12.5 cm and R 2 is 10.0 cm. Solution: Sectional area G a and sectional volume G v : G a 4 π R 1 2 4 π R Z a 4 π 1 R 2 ( ) = Z v 2 π ( ) = 2 2 + = ( ) Example 2: Find the surface area K a and volume K v of section K of a sphere of radius 4 π 12.5 ( ) 2 4 π 10 2 + 3220.13 cm 2 = 3 --- 12.5 3 10 3 – ( ) 3992.44 cm 3 π = = = G v 4 π 3 --- R 1 3 R 2 3 – ( ) 4 =
15.0 cm, if included angle φ = 90 ° and depth F = 5.0 cm. Solution: Sectional area K a and sectional volume K v : K a 2 π R 2 1 φ 2 – cos -- 2 π 15 2 1 = = ( )
90 ° 2 – cos ----
414.07cm 2 =
2 π R 2 F 3 --------- 2 π 15 =
2 5 --- --- ( ) ( )
2356.19cm 3
=
3 ------- ---
=
K v
Example 3: Find the outside surface area S a and sectional volume S v of section S of a sphere if E = 20.0 cm and F = 5.0 cm. Solution: Sectional area S a and sectional volume S v :
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