Solid Geometry

Solid Geometry Sphere

Volume =4/3πr3 Surface Area = 4πr2 Great circle is always in the center of the sphere

Spherical Segment with One Base 𝜋ℎ2 Volume = (3𝑅 − ℎ) 3 Surface Area = 2πRh

Spherical Segment with two Bases

Volume =

𝜋ℎ (3𝑎2 + 3𝑏 2 + ℎ2 ) 6

Calc – tech for volume A salad bowl is in the form of a spherical segment with two bases. The upper-base is a great circle of radius 5”. The altitude of the bowl is 2”. Find its capacity. Ans. 148.7 in3 Solution: Mode 3,3 1. 2. 3. 4. 5.

AC Shift 1 ,5 ,1, shift ,rcl ,(-) Shift 1 ,5 ,2, shift, rcl ,(˚’”) Shift 1 ,5 ,3, shift, rcl ,(HYP) MODE 1

𝑢𝑝𝑝𝑒𝑟 𝑙𝑖𝑚𝑖𝑡

∫ 𝑙𝑜𝑤𝑒𝑟 𝑙𝑖𝑚𝑖𝑡

𝐶𝑥 2 + 𝐵𝑥 + 𝐴

RECTANGULAR PARALLELEPIPED Volume = 𝑎𝑏𝑐 c

b

Lateral Area = 2(ac+bc) Total Surface Area = 2(ab+bc+ac)

a

PYRAMID Volume =

1 𝐴 𝐻 3 𝐵𝐴𝑆𝐸

FRUSTUM OF A PYRAMID 𝐻

Volume = 3 (𝐵1 + 𝐵2 + √𝐵1 𝐵2)

CONE 1 Volume = 𝜋𝑟 2 ℎ 3 Lateral Area = πrL Slant Height: L2=r2+h2

Frustum of Cone Volume =

𝜋ℎ 2 (𝑟 + 𝑅 2 + 𝑟𝑅) 3

Lateral Area = ΠL(r+R) Slant Height: L2=(R-r)2+h2

Right Circular Cylinder Volume = 𝜋𝑟 2 ℎ Lateral Area = 2Πrh Slant Height: 2πr2+2πrh

Larc=(πθr)/180 VolumeTruncated prism=Ave. h(Abase)

Example: The Lateral edge of a frustum of a regular pyramid is 1.8m long. It upper base is a square, 1m x 1m, while its base, also a square, 2.4m x 2.4m. what is the volume of the frustum? Answer: 5.6m2

1.8m h=?

by Pythagorean theorem : 1.82 – 0.72 =h2 h= 1.658m 0.7m

Mode 3,3 x 0 h/2 h 1.658

∫ 0

𝐶𝑥 2 + 𝐵𝑥 + 𝐴

y area lower base midsection area area upper base

x 0 1.658/2 1.658

y 5.76 (5.76+1)/2 =3.38

1