Gillesania Engineering Review & Training Center

APRIL 2026 PRE-REVIEW

Instructor: Engr. Irah Angellie B. Doroy

DAILY REMINDER

Photo Credits: Franklin Mano| Facebook

 SOLID GEOMETRY

Solid Geometry
deals with the three-dimensional figures called solids.

Solids
are three-dimensional figures bounded by surfaces or plane figures.

Common examples:
 SOLID GEOMETRY Example #1

How many cubic yards of concrete are needed for a driveway 18.0 ft wide, 33.0 ft long, and
4.00 in thick?
Solution:
This driveway is a rectangular prism, whose
volume is:
𝑉 = 𝑙𝑤ℎ

1𝑓𝑡
4 𝑖𝑛 𝑉 = (33.0 𝑓𝑡) 18.0 𝑓𝑡 4.0 𝑖𝑛 ×
12𝑖𝑛
33 𝑓𝑡 𝑉=
18 𝑓𝑡
3
1 𝑦𝑑
𝑉 = 198 𝑓𝑡 3 ×
3 𝑓𝑡

𝑽 = 𝟕. 𝟑𝟑 𝒚𝒅𝟑
SOLID GEOMETRY Example #2
INTRODUCTION
A tank in an ice plant is to contain 3000 liters of brine. It is constructed to be 4 m long
and 1.5 m wide. Find the following:
A. Volume in cu. meters Solution:
B. Height of the tank A:
1𝑚3
C. Base area 𝑉 = 3000 𝑙𝑖𝑡𝑒𝑟𝑠 ×
1000 𝑙𝑖𝑡𝑒𝑟𝑠
D. Lateral Area
𝑽 = 𝟑 𝒎𝟑

B:
𝑉 = 𝑙𝑤ℎ
3 = (4 × 1.5 × 𝐻)

𝑯 = 𝟎. 𝟓 𝒎
1.5 𝑚
4𝑚
 SOLID GEOMETRY Example #2
INTRODUCTION
A tank in an ice plant is to contain 3000 liters of brine. It is constructed to be 4 m long
and 1.5 m wide. Find the following:
A. Volume in cu. meters Solution:
B. Height of the tank C:
𝐴𝑏 = 𝐿𝑊
C. Base area
D. Lateral Area = 4 × 1.5
𝑨𝒃 = 𝟔𝒎𝟐

D: 𝐴𝑙 = 2𝑊𝐻 + 2𝐿𝐻

0.5 𝑚 = 2(1.5 × 0.5) + 2(4 × 0.5)

𝑨𝒍 = 𝟓. 𝟓 𝒎𝟐

1.5 𝑚
4𝑚
SOLID GEOMETRY Example #3

Find the volume of the figure shown in terms of 𝜸

Solution:
Volumes of containers (or solid) with uniform cross-sectional
area such as prism:
𝑉 = 𝐴ℎ

𝑉 = 𝐴𝑡𝑟𝑎𝑝𝑒𝑧𝑜𝑖𝑑 × ℎ

1
𝑉= 𝑏1 + 𝑏2 × 𝐻 × ℎ
2

1
𝑉= 2𝛾 + 5𝛾 × 3 × 1.5
2

𝑽 = 𝟏𝟓. 𝟕𝟓𝜸
SOLID GEOMETRY Example #4
INTRODUCTION
Three dimensions of a rectangular parallelepiped tank are in the ratio 3:2:5. If its
volume is 810 cu.m., find the following:
a. Find the dimensions
b. Find the surface area
c. If the tank is being filled up with water, what is the pressure at the bottom?
Solution:
A.
𝑉 = 𝑙𝑤ℎ Dimensions:

810 = (3𝑥)(2𝑥)(5𝑥) 𝑙 = 3𝑥 = 3 3 = 𝟗𝒎

𝟓𝒙 𝑤 = 2𝑥 = 2 3 = 𝟔𝒎
810 = 30𝑥 3
ℎ = 𝟏𝟓𝒎
ℎ = 5𝑥 = 5 3 = 𝟏𝟓𝒎
3 810
𝑥=
𝟑𝒙 𝟐𝒙
30
𝑙 = 𝟗𝒎 𝑤 = 𝟔𝒎
𝑥=3
SOLID GEOMETRY Example #4
INTRODUCTION
Three dimensions of a rectangular parallelepiped tank are in the ratio 2:3:5. If its
volume is 810 cu.m., find the following:
a. Find the dimensions
b. Find the surface area
c. If the tank is being filled up with water, what is the pressure at the bottom?
Solution:
B. C.
𝐴𝑠 = 2(𝑤ℎ + 𝑙ℎ + 𝑤𝑙) 𝑃 = 𝛾ℎ
𝑘𝑁
= 2(6 × 15 + 9 × 15 + 6 × 9) = 9.81 3
(15)
𝟓𝒙 𝑚
ℎ = 𝟏𝟓𝒎 𝑨𝒔 = 𝟓𝟓𝟖 𝒎𝟐 𝑷 = 𝟏𝟒𝟕. 𝟏𝟓 𝒌𝑷𝒂

𝟑𝒙 𝟐𝒙
𝑙 = 𝟗𝒎 𝑤 = 𝟔𝒎
 SOLID GEOMETRY Example #5
Gordon has three silos on his farm. The silos are each in the shape of a right circular cylinder. One silo has
a 12-ft diameter and is 40 ft tall. The second silo has a 14 ft-diameter and is 50 ft tall. The third silo has an
18-ft diameter and is 60 ft tall.
a. What is the total capacity of all three silos in cubic feet?
b. If Gordon fills all three of his silos and then feeds his cattle 150 𝑓𝑡 3 of silage per day, in how many
days will all three silos be empty?
Solution:
Volumes of containers (or solid) with uniform cross-sectional
area such as rectangular containers, prisms and cylinders.
50′
60′ 𝑉 = 𝐴𝑏 ℎ
40′
For right circular cylinders: 𝜋 2
𝑉= 𝑑 ℎ
𝑉 = 𝜋𝑟 2 ℎ 4
12′ 14′ 18′
𝜋
𝑉1 = 122 40 = 4523.893 𝑐𝑢. 𝑓𝑡
4
𝜋
𝑉2 = 142 50 = 7696.902 𝑐𝑢. 𝑓𝑡
4
𝜋
𝑉3 = 182 60 = 15268.14 𝑐𝑢. 𝑓𝑡
4
 SOLID GEOMETRY Example #5
Gordon has three silos on his farm. The silos are each in the shape of a right circular cylinder. One silo has
a 12-ft diameter and is 40 ft tall. The second silo has a 14 ft-diameter and is 50 ft tall. The third silo has an
18-ft diameter and is 60 ft tall.
a. What is the total capacity of all three silos in cubic feet?
b. If Gordon fills all three of his silos and then feeds his cattle 150 𝑓𝑡 3 of silage per day, in how many
days will all three silos be empty?
Solution:
A:
60′ 𝑉𝑇 = 𝑉1 + 𝑉2 + 𝑉3
50′
40′
𝑽𝑻 = 𝟐𝟕𝟒𝟖𝟖. 𝟗𝟑𝟔 𝒇𝒕𝟑

12′ 14′ 18′

B:
𝑉1 = 4523.893 27488.936𝑓𝑡 3
𝑑=
150 𝑓𝑡 3 /𝑑𝑎𝑦
𝑉2 = 7696.902
𝑑 = 183.26 ≈ 𝟏𝟖𝟑 𝒅𝒂𝒚𝒔
𝑉3 = 15268.14 𝑇ℎ𝑢𝑠, 𝑡ℎ𝑒 𝑠𝑖𝑙𝑜𝑠 𝑤𝑖𝑙𝑙 𝑏𝑒 𝑒𝑚𝑝𝑡𝑦 𝑖𝑛 183 𝑑𝑎𝑦𝑠.
 SOLID GEOMETRY Example #6

Find the shaded area in the figure shown if the length of the outer radius is 36.0 ft. and the length of the
inner radius is 20.0 ft.
Solution:
𝑅 = 36′ 𝐴 = 𝐴𝑠𝑒𝑐𝑡𝑜𝑟1 − 𝐴𝑠𝑒𝑐𝑡𝑜𝑟2
60°
1 2 1 2
𝐴= 𝑅 𝜃− 𝑟 𝜃
𝑟 = 20′
2 2

1
𝐴 = 𝜃(𝑅2 − 𝑟 2 )
2

1 5
𝐴= × 𝜋 362 − 202
2 3
60°
𝜽 = 360 − 60 = 𝟑𝟎𝟎°
𝜃 𝜋 𝟓 𝑨 = 𝟐𝟑𝟒𝟓. 𝟕𝟐𝟑 𝒇𝒕𝟐
𝜽 = 300° × = 𝝅 𝒓𝒂𝒅
180° 𝟑
 SOLID GEOMETRY Example #6

Find the shaded area in the figure shown if the length of the outer radius is 36.0 ft. and the length of the
inner radius is 20.0 ft.
Solution:
Set Calculator to Radian Mode:
60° 𝑑𝐴 = 𝑑𝐴𝑠𝑒𝑐𝑡𝑜𝑟1 − 𝑑𝐴𝑠𝑒𝑐𝑡𝑜𝑟2

𝜃 1 2 1 2
𝑑𝐴 = 𝑅 𝑑𝜃 − 𝑟 𝑑𝜃
2 2
1 2 1 2
𝐴 = න 𝑅 𝑑𝜃 − න 𝑟 𝑑𝜃
2 2
5
60° 1 2 3𝜋
𝑑𝜃 𝐴= (𝑅 − 𝑟 2 ) න 𝑑𝜃
𝜽 = 360 − 60 = 𝟑𝟎𝟎° 2 0
𝜋 𝟓
𝜽 = 300° × = 𝝅
180° 𝟑 𝑨 = 𝟐𝟑𝟒𝟓. 𝟕𝟐𝟑 𝒇𝒕𝟐
 SOLID GEOMETRY Example #7

A rectangular lead sleeve is shown in the figure below. (a) How many cubic inches of lead are contained
in this sleeve? (b) Find its weight if lead weighs 708 𝑙𝑏/𝑓𝑡 3 .

Solution:
A:
y Volumes of containers (or solid) with uniform cross-
𝟏. 𝟓′′ sectional area:
𝑉 = 𝐴𝑏 ℎ
𝟏. 𝟓′′ 12′′

𝟐𝟒′′ 𝑉 = 𝑊𝐿 − 𝑤𝑙 ℎ

𝟗′′ 𝑉 = ( 9 × 12 − 6 × 9 ) × 24

𝑽 = 𝟏𝟐𝟗𝟔 𝒄𝒖. 𝒊𝒏
 SOLID GEOMETRY Example #7

A rectangular lead sleeve is shown in the figure below. (a) How many cubic inches of lead are contained
in this sleeve? (b) Find its weight if lead weighs 708 𝑙𝑏/𝑓𝑡 3 .

Solution:
B:
y 𝑊 = 𝛾𝑉
𝟏. 𝟓′′
3
𝑙𝑏 3
1𝑓𝑡
𝑊 = 708 3 1296 𝑖𝑛 ×
𝟏. 𝟓′′ 12′′
𝑓𝑡 12 𝑖𝑛
𝟐𝟒′′

𝑾 = 𝟓𝟑𝟏 𝒍𝒃
𝟗′′

𝑉 = 1296 𝑐𝑢. 𝑖𝑛
SOLID GEOMETRY Example #8

A cylindrical buoy, 0.60 m outside diameter and 2 m high, weighs 2000 N. It is moored in salt water to
a 12 m chain length weighing 120 N/m. At high tide, the height of the buoy protruding above water is
0.85 m. What would be the total volume of the chain? Density of steel is 77 𝑘𝑁/𝑚3

Solution:
𝑤 = 120𝑁/𝑚

𝑊 = 𝑤𝑙 𝑊 = 𝛾𝑉

𝑁
𝑊 = 120(12) 1440𝑁 = 77 × 103 ×𝑉
𝑚3

𝑊 = 1440 𝑁 𝑽 = 𝟎. 𝟎𝟏𝟖𝟕 𝒎𝟑
 SOLID GEOMETRY Example #9

Frank’s fish tank is in the shape of a hexagonal prism, as shown. Use the dimensions shown in the figure:
1. Determine the volume of the fishtank in cubic inches.
2. Determine the volume of the fishtank in gallons (round your answer to the nearest gallon) .

Solution:
Uniform Cross Section: A:
𝑉 = 𝐴𝑏 ℎ

𝐴𝑏 = 2𝐴 𝑇𝑟𝑎𝑝

1
=2 × 8 + 16 × 8
2

𝐴𝑏 = 192 𝑖𝑛2

𝑛𝑎2 𝑉 = 192 𝑖𝑛2 × 24 = 𝟒𝟔𝟎𝟖 𝐢𝐧𝟑

𝐴𝑏 =
180°
4 tan
𝑛
 SOLID GEOMETRY Example #9

Frank’s fish tank is in the shape of a hexagonal prism, as shown. Use the dimensions shown in the figure:
1. Determine the volume of the fishtank in cubic inches.
2. Determine the volume of the fishtank in gallons (round your answer to the nearest gallon) .

Solution:
3 3
B: 3
𝑉 = 4608 in ×
2.54 𝑐𝑚
×
1𝑚
1 𝑔𝑎𝑙 1 𝑖𝑛 100 𝑐𝑚
3
𝑉 = 4608 in ×
231 𝑖𝑛3
𝑉 = 0.075512 𝑚3
𝑉 = 𝟏𝟗. 𝟗𝟒𝟖 𝒈𝒂𝒍
1000𝐿
𝑉 = 0.075512 𝑚3× = 75.512 𝑙𝑖𝑡
1𝑚3
Convert Function (CONVT)

𝑉 = 𝟒𝟔𝟎𝟖 𝐢𝐧𝟑
 SOLID GEOMETRY Example #9

Frank’s fish tank is in the shape of a hexagonal prism, as shown. Use the dimensions shown in the figure:
1. Determine the volume of the fishtank in cubic inches.
2. Determine the volume of the fishtank in gallons (round your answer to the nearest gallon) .

Solution:
B:
𝑉 = 𝟒𝟔𝟎𝟖 𝐢𝐧𝟑

Convert Function (CONVT)

𝑉 = 𝟒𝟔𝟎𝟖 𝐢𝐧𝟑
 SOLID GEOMETRY Example #10

The solid represented in the figure has square bases of edge a in parallel planes and every section of
the solid parallel to the base is a square of edge a. If its altitude is h, find its volume.

Solution:
Volumes of containers (or solid) with uniform cross-
sectional area:

ℎ 𝑉 = 𝐴𝑏 ℎ

𝑽 = 𝒂𝟐 𝒉
SOLID GEOMETRY Example #11

Find the volume of the rectangular pyramid:

Solution:
Volume of pyramids and cones:
24.6 𝑖𝑛
1
𝑉 = 𝐴𝑏 ℎ
3

1
𝑉 = 18.6 𝑖𝑛 × 10.1 𝑖𝑛 × 24.6 𝑖𝑛
3
10.1 𝑖𝑛
𝑽 = 𝟏𝟓𝟒𝟎. 𝟒𝟓𝟐 𝒊𝒏𝟑
18.6 𝑖𝑛
 SOLID GEOMETRY Example #12

Find (a) the volume and (b) the lateral surface area of the right circular cone feed storage bin as shown
below.
Solution:
2.50 𝑚
A:
Volume of pyramids and cones:
1
𝑉 = 𝐴𝑏 ℎ
3
2.10 𝑚

1 2
𝑉 = 𝜋𝑟 ℎ
3
1
𝑉 = 𝜋 1.25 2 (2.10)
3
2.50
𝑟=
2
= 1.25 𝑚 𝑽 = 𝟑. 𝟒𝟒 𝒎𝟑
 SOLID GEOMETRY Example #12

Find (a) the volume and (b) the lateral surface area of the right circular cone feed storage bin as shown
below.
Solution:
2.50 𝑚
B:
Getting the slant height, l first:

𝑙 2 = ℎ2 + 𝑟 2
2.10 𝑚
𝑙= 2.10 2 + 1.25 2 = 2.44 𝑚
𝑙
Lateral Surface Area for Cones and Pyramids:

1 𝐴𝑙 = 𝜋𝑟𝑙
𝐴𝑙 = 𝑝𝑏 𝑙
2
= 𝜋(1.25)(2.44)
2.50 1
𝑟= = 1.25 𝑚 𝐴𝑙 = 2𝜋𝑟 𝑙 𝑨𝒍 = 𝟗. 𝟓𝟖 𝒎𝟐
2 2

𝐴𝑙 = 𝜋𝑟𝑙 →Derived for Cones

SOLID GEOMETRY Example #13

Find (a) the volume and (b) the lateral surface area of the storage tank with hemispherical one end:

Solution:
A:
10.3 𝑐𝑚 𝑉 = 𝑉𝑐𝑦𝑙𝑖𝑛𝑑𝑒𝑟 + 𝑉ℎ𝑒𝑚𝑖𝑠𝑝ℎ𝑒𝑟𝑒
1 4 3
= 𝜋𝑟 2 ℎ + 𝜋𝑟
2 3
28.5 𝑐𝑚
2 (28.5 𝑐𝑚)
2 3
= 𝜋 10.3 𝑐𝑚 + 𝜋 10.3
3
𝑉𝑐𝑦𝑙𝑖𝑛𝑑𝑒𝑟 = 𝜋𝑟 2 ℎ
𝑽 = 𝟏𝟏, 𝟖𝟎𝟎 𝒄𝒎𝟑

1 4 3
𝑉ℎ𝑒𝑚𝑖𝑠𝑝ℎ𝑒𝑟𝑒 = 𝜋𝑟
2 3
SOLID GEOMETRY Example #13

Find (a) the volume and (b) the lateral surface area of the storage tank with hemispherical one end:

Solution:
B:
10.3 𝑐𝑚
𝐴 = 𝐴𝑐𝑖𝑟𝑐𝑙𝑒 + 𝐴𝑙𝑎𝑡𝑐𝑦𝑙𝑖𝑛𝑑𝑒𝑟 + 𝐴ℎ𝑒𝑚𝑖𝑠𝑝ℎ𝑒𝑟𝑒
2
1
= 𝜋𝑟 + 2𝜋𝑟ℎ + (4𝜋𝑟 2 )
2
1
28.5 𝑐𝑚 = 𝜋(10.3𝑐𝑚) +2𝜋(10.3𝑐𝑚)(28.5𝑐𝑚) + × 4𝜋(10.5𝑐𝑚)2
2
2

𝑨 = 𝟐𝟖𝟒𝟎𝟒. 𝟑𝟎𝟒 𝒄𝒎𝟐

𝐴𝑙𝑎𝑡𝑐𝑦𝑙𝑖𝑛𝑑𝑒𝑟 = 2𝜋𝑟ℎ

𝐴𝑠ℎ𝑒𝑟𝑒 = 4𝜋𝑟 2
SOLID GEOMETRY Example #14

The circular building in the figure shown is made of metal weighing 17.8 𝑙𝑏/𝑓𝑡 2 . (a) What is the
weight of the top? (b) What is the weight of the top and sides? The bottom consists of concrete. (c)
Find the total volume capacity of the building.
Solution:
𝑙 10 𝑓𝑡 A:
Lateral Area for Cones and Pyramids:
1
𝐴𝑙 = 𝑝𝑏 𝑙
2

30 𝑓𝑡 𝑙𝑏
1 𝑤 = 17.8 2
𝐴𝑙 = 2𝜋𝑟 𝑙 𝑓𝑡
2

𝐴𝑙 = 𝜋𝑟𝑙 𝑊 = 𝑤𝐴
30 𝑓𝑡
𝑙𝑏
30 = 𝜋 × 15 × 102 + 152 𝑊 = 17.8 2 × 849.538 𝑓𝑡 2
𝑟= = 15𝑓𝑡 𝑓𝑡
2
𝐴𝑙 = 849.538 𝑓𝑡 2
𝑙= 102 + 152 𝑊 = 𝟏𝟓𝟏𝟐𝟏. 𝟕𝟖 𝒍𝒃
 SOLID GEOMETRY Example #14

The circular building in the figure shown is made of metal weighing 17.8 𝑙𝑏/𝑓𝑡 2 . (a) What is the
weight of the top? (b) What is the weight of the top and sides? The bottom consists of concrete. (c)
Find the total volume capacity of the building.
Solution:
𝑙 10 𝑓𝑡 B:
𝐴𝐿𝑡𝑜𝑡𝑎𝑙 = 𝐴𝑡𝑜𝑝 + 𝐴𝑠𝑖𝑑𝑒𝑠 𝑊 = 𝑤𝐴
𝑙𝑏
𝐴𝑠𝑖𝑑𝑒𝑠 = 2𝜋𝑟ℎ 𝑊 = 17.8 2 × 3676.971 𝑓𝑡 2
30 𝑓𝑡 𝑓𝑡
= 2𝜋 × 15 × 30
𝑾 = 𝟔𝟓𝟒𝟓𝟎. 𝟎𝟗 𝒍𝒃
𝐴𝑠𝑖𝑑𝑒𝑠 = 2827.433 𝑓𝑡 2
30 𝑓𝑡

𝑟=
30
= 15𝑓𝑡
𝐴𝐿𝑡𝑜𝑡𝑎𝑙 = 849.538 + 2827.433
2
𝐴𝑡𝑜𝑝 = 849.538 𝑓𝑡 2 𝐴𝐿𝑡𝑜𝑡𝑎𝑙 = 3676.971 𝑓𝑡 2
𝑙𝑏
𝑤 = 17.8 2
𝑓𝑡
SOLID GEOMETRY Example #14

The circular building in the figure shown is made of metal weighing 17.8 𝑙𝑏/𝑓𝑡 2 . (a) What is the
weight of the top? (b) What is the weight of the top and sides? The bottom consists of concrete. (c)
Find the total volume capacity of the building.
Solution:
𝑙 10 𝑓𝑡 C:
𝑉𝑡𝑜𝑡𝑎𝑙 = 𝑉𝑐𝑜𝑛𝑒 + 𝑉𝑐𝑦𝑙𝑖𝑛𝑑𝑒𝑟

1
𝑉𝑡𝑜𝑡𝑎𝑙 = 𝐴𝑏 ℎ1 + 𝜋𝑟 2 ℎ2
3
30 𝑓𝑡
1
𝑉𝑡𝑜𝑡𝑎𝑙 = (𝜋𝑟 2 )ℎ1 + 𝜋𝑟 2 ℎ2
3

1
30 𝑓𝑡 𝑉𝑡𝑜𝑡𝑎𝑙 = 𝜋(152 × 10) + 𝜋(152 × 30)
𝑟 = 15𝑓𝑡 3
1
𝑉𝑐𝑜𝑛𝑒 = 𝐴𝑏 ℎ
3 𝑽𝒕𝒐𝒕𝒂𝒍 = 𝟐𝟑𝟓𝟔𝟏. 𝟗𝟒𝟓 𝒇𝒕𝟑

𝑉𝑐𝑦𝑙𝑖𝑛𝑑𝑒𝑟 = 𝜋𝑟 2 ℎ
 SOLID GEOMETRY Example #15
Try
Determine the volume of the remaining solid after the cylinder, triangular prism, and square
prism have been cut from the solid.
 SOLID GEOMETRY Example #15
Try
Determine the volume of the remaining solid after the cylinder, triangular prism, and square
prism have been cut from the solid.

Answer: 𝑽 = 𝟑𝟓𝟖. 𝟑𝟎 𝒊𝒏𝟑

 SOLID GEOMETRY Example #16

A diving bell has the shape of a frustrum of a right circular cone with closed top (3m in diameter), open
bottom (4.2 m in diameter) and is 3m high. It is lowered in an upright position until it rests on an ocean
floor (unit weight = 10.05 𝑘𝑁/𝑚3 ). In this position, the compressed air inside the diving belle has an
altitude of 1.2 m. Assuming atmospheric pressure to be 101.325 kPa, determine the depth of the ocean
floor.
3𝑚

3𝑚

ℎ 𝑑
4.2 𝑚
 SOLID GEOMETRY Example #16

A diving bell has the shape of a frustrum of a right circular cone with closed top (3m in diameter), open
bottom (4.2 m in diameter) and is 3m high. It is lowered in an upright position until it rests on an ocean
floor (unit weight = 10.05 𝑘𝑁/𝑚3 ). In this position, the compressed air inside the diving belle has an
altitude of 1.2 m. Assuming atmospheric pressure to be 101.325 kPa, determine the depth of the ocean
floor.
3𝑚
𝑉1

3𝑚

ℎ 𝑑
4.2 𝑚

𝑉2
 SOLID GEOMETRY Example #16
A diving bell has the shape of a frustrum of a right circular cone with closed top (3m in diameter), open
bottom (4.2 m in diameter) and is 3m high. It is lowered in an upright position until it rests on an ocean
floor (unit weight = 10.05 𝑘𝑁/𝑚3 ). In this position, the compressed air inside the diving belle has an
altitude of 1.2 m. Assuming atmospheric pressure to be 101.325 kPa, determine the depth of the ocean
floor. 3𝑚 Solution:
Determine the total volume of frustum and the volume of air
when it rests on an ocean floor.
3𝑚
Volume for a frustrum of a solid:
ℎ
4.2 𝑚 𝑉1 = (𝐴 + 𝐴2 + 𝐴1 𝐴2 )
3 1
𝑉1
𝜋ℎ 2
𝑉1 = (𝑟 + 𝑅2 + 𝑟𝑅) → Derived volume for a frustrum of a cone
3
3𝑚
3
ℎ 𝑑 𝜋×3 𝑟 = = 1.5 𝑚
𝑉1 = (1.52 + 2.12 + 1.5 × 2.1) 2
3 𝑉1 3𝑚
4.2
𝑅= = 2.1 𝑚
𝑽𝟏 = 𝟑𝟎. 𝟖𝟏𝟗 𝒎𝟑 2
𝑉2 4.2 𝑚
 SOLID GEOMETRY Example #16
A diving bell has the shape of a frustrum of a right circular cone with closed top (3m in diameter), open
bottom (4.2 m in diameter) and is 3m high. It is lowered in an upright position until it rests on an ocean
floor (unit weight = 10.05 𝑘𝑁/𝑚3 ). In this position, the compressed air inside the diving belle has an
altitude of 1.2 m. Assuming atmospheric pressure to be 101.325 kPa, determine the depth of the ocean
floor. 3𝑚 Solution:
1.5𝑚

𝑧 1.2
= 𝑉2 1.2 𝑚
0.6 3.0 3𝑚 𝒛
3𝑚
𝑧 = 0.24

4.2 𝑚 𝑅2 = 1.5 + 0.24 = 1.74 𝑚 2.1𝑚

𝑉1
𝜋ℎ 2 0.6 𝑚
𝑉2 = (𝑟 + 𝑅2 + 𝑅𝑟)
3
𝑽𝟏 = 𝟑𝟎. 𝟖𝟏𝟗 𝒎𝟑 𝜋 × 1.2
𝑉2 = (1.52 + 1.742 + 1.5 × 1.74)
ℎ 𝑑 𝑽𝟐 = 𝟗. 𝟗𝟏𝟐 𝒎𝟑 3

𝑉2 = 9.912 𝑚3

𝑉2
 SOLID GEOMETRY Example #16
A diving bell has the shape of a frustrum of a right circular cone with closed top (3m in diameter), open
bottom (4.2 m in diameter) and is 3m high. It is lowered in an upright position until it rests on an ocean
floor (unit weight = 10.05 𝑘𝑁/𝑚3 ). In this position, the compressed air inside the diving belle has an
altitude of 1.2 m. Assuming atmospheric pressure to be 101.325 kPa, determine the depth of the ocean
floor. 3𝑚 Solution: 1.5 𝑚

Another way to solve 𝑉 by integral: 𝑥

3𝑚 𝑧
3𝑚 𝑧 𝑥 𝑑𝑥
= 𝑟
0.6 3.0
4.2 𝑚 𝑧 = 0.2 𝑥
𝑉1 0.6 𝑚
𝑟 = 1.5 + 0.2𝑥
3
𝑑𝑉 = 𝜋𝑟 2 𝑑𝑥 𝑉1 = න 𝜋 1.5 + 0.2𝑥 2 𝑑𝑥 = 𝟑𝟎. 𝟖𝟏𝟗 𝒎𝟑
0
ℎ 𝑑 𝑑𝑉 = 𝜋𝑟 2 𝑑𝑥 1.2
𝑥2 𝑉2 = න 𝜋 1.5 + 0.2𝑥 2 𝑑𝑥 = 𝟗. 𝟗𝟏𝟐 𝒎𝟑
0
𝑉 = න 𝜋𝑟 2 𝑑𝑥
𝑉2 𝑥1
 SOLID GEOMETRY Example #16
A diving bell has the shape of a frustrum of a right circular cone with closed top (3m in diameter), open
bottom (4.2 m in diameter) and is 3m high. It is lowered in an upright position until it rests on an ocean
floor (unit weight = 10.05 𝑘𝑁/𝑚3 ). In this position, the compressed air inside the diving belle has an
altitude of 1.2 m. Assuming atmospheric pressure to be 101.325 kPa, determine the depth of the ocean
floor. 3𝑚 Solution:
𝑉1 = 30.819 𝑚3 𝑃1 = 101.325 𝑘𝑃𝑎
3𝑚
𝑉2 = 9.912 𝑚3 𝑃2 = 101.325 𝑘𝑃𝑎 + 10.05ℎ

4.2 𝑚
𝑃1 𝑉1 = 𝑃2 𝑉2
𝑉1
101.325 × 30.819 = (101.325 𝑘𝑃𝑎 + 10.05ℎ) × 9.912 𝑚3

ℎ = 21.2658

ℎ 𝑑
𝑑 = ℎ + (3 − 1.2)
𝑑 = 21.2658 + (3 − 1.2)
𝑉2 𝒅 = 𝟐𝟑. 𝟎𝟔𝟔 𝒎
 SOLID GEOMETRY Example #17

A sphere with a diameter of 1.5cm is partially submerged in liquid with a draft of 1.2 m. Find the volume
displaced.

Solution:
Volume of a Spherical Segment of One Base:

𝜋ℎ2
𝑉𝐷 = (3𝑅 − ℎ)
3

𝜋 1.2 2
𝑉𝐷 = (3(0.75) − 1.2)
3

𝑽𝑫 = 𝟏. 𝟓𝟖𝟑 𝒎𝟑
 SOLID GEOMETRY Example #18

A spherical tank having a radius of 2.5 m is filled with water to a depth of 1.8 m. If a solid sphere having a
diameter of 1.80 m is placed on the bowl, determine how high will the water rise on the bowl.

∆ℎ
ℎ1
1.8 𝑚

Solution:
Volume of water: Volume the solid sphere placed:
𝜋ℎ2 1
𝑉𝑤 = (3𝑅 − ℎ) 𝑟𝑠 = 1.80 = 0.90
3 2
𝜋 1.8 2 4 3
𝑉𝑤 = (3(2.5) − 1.8) 𝑉𝑠 = 𝜋𝑟
3 3
4
𝑉𝑤 = 19.34 𝑚3 𝑉𝑠 = 𝜋(0.90)3
3
𝑉𝑠 = 3.054
 SOLID GEOMETRY Example #18

A spherical tank having a radius of 2.5 m is filled with water to a depth of 1.8 m. If a solid sphere having a
diameter of 1.80 m is placed on the bowl, determine how high will the water rise on the bowl.

∆ℎ
ℎ1
1.8 𝑚

Solution:
Total Volume: Rise in depth:
𝑉𝑤 = 19.34 𝑚3 𝑉𝑇 = 19.34 + 3.054 = 22.394 𝑚3 ∆ℎ = 1.966 − 1.80
𝑉𝑠 = 3.054 𝑚3 𝜋ℎ12
𝑉𝑇 = (3𝑅 − ℎ1 ) ∆ℎ = 0.166 𝑚
3
𝜋ℎ12 ∆𝒉 = 𝟏𝟔𝟔 𝒎𝒎
22.394 = (3 × 2.5 − ℎ1 )
3

ℎ1 = 1.966 𝑚
 SOLID GEOMETRY Example #19

The famous pyramid of Egypt has an altitude of 15 m and weight of 4000 kg. It has a square base. At
what distance from its base must it be cut by a plane parallel to its base so that the two solids of equal
weight will be formed?
Solution:
15 𝑚
By property of similar figures:
15 𝑚 3
𝑉1 𝑥1 𝑥1
𝑥 𝑉𝑡𝑜𝑝 𝑉2
=
𝑥2

𝑉𝑏𝑜𝑡 𝑉1 𝑥1 3
=
𝑧 2𝑉1 15
𝑉𝑡𝑜𝑝 = 𝑉1 𝑉2 = 2𝑉1
𝑠 𝑠
1 𝑥1 3
𝑠 𝑠 =
2 15
𝑉𝑡𝑜𝑝 = 𝑉𝑏𝑜𝑡
𝑥1 = 11.906 𝑥1
𝑥2
𝑉𝑤ℎ𝑜𝑙𝑒 = 𝑉𝑡𝑜𝑝 + 𝑉𝑏𝑜𝑡
= 𝑉𝑡𝑜𝑝 + 𝑉𝑡𝑜𝑝 𝑧 = 15 − 𝑥1

𝑽𝒘𝒉𝒐𝒍𝒆 = 𝟐𝑽𝒕𝒐𝒑 𝒛 = 𝟑. 𝟏 𝒎
 SOLID GEOMETRY Example #20
INTRODUCTION
An open rectangular tank mounted on a truck 5 m long, 2.5 m wide, and 2.5 m high is filled with water to a
depth of 2m. The truck accelerated and it causes the tank to spill out water. Determine the following:
a. Find the volume spill, refer to figure 1.
b. Find the volume spill, refer to figure 2.

𝒂
𝒂

2.5 𝑚 2.5 𝑚
2𝑚 𝜃 = 15° 2𝑚
𝜃 = 35°

5𝑚 5𝑚
 SOLID GEOMETRY Example #20
INTRODUCTION
An open rectangular tank mounted on a truck 5 m long, 2.5 m wide, and 2.5 m high is filled with water to a
depth of 2m. The truck accelerated and it causes the tank to spill out water. Determine the following:
a. Find the volume spill, refer to figure 1.
b. Find the volume spill, refer to figure 2.
𝒂

2.5 𝑚
2𝑚 𝜃 = 15°

5𝑚

a. 𝑉𝑠𝑝𝑖𝑙𝑙 = 2.125 𝑚3
 SOLID GEOMETRY Example #20
INTRODUCTION
An open rectangular tank mounted on a truck 5 m long, 2.5 m wide, and 2.5 m high is filled with water to a
depth of 2m. The truck accelerated and it causes the tank to spill out water. Determine the following:
a. Find the volume spill, refer to figure 1.
b. Find the volume spill, refer to figure 2.

2.5 𝑚
2𝑚
𝜃 = 35°

5𝑚

b. 𝑉𝑠𝑝𝑖𝑙𝑙 = 13.84 𝑚3
 SOLID GEOMETRY Example #21

A trough whose ends are isosceles right triangle with vertical axis, is 6 m long. If it contains 800 liters of
water, how deep is the water?

𝑑 90°

𝑑 90°

𝑑 = 0.365 𝑚
 SOLID GEOMETRY Example #22
Nov 2023 CELE
Dale has a water trough whose ends are trapezoids and whose sides are rectangles as illustrated. He
is afraid that the base it is sitting on it will not support the weight of the trough when it is filled with
water. He knows that the base will support 4800 lb.
A. If the trough is filled with water, determine the number of cubic feet of water contained in the
trough.
B. Determine the total weight, assuming that the trough weighs 375 lb and the water weighs
62.4 lb per cubic foot. Is the base strong enough to support the troughfilled with water?
C. If 1 gal of water weighs 8.3 lb, how many gallons of water will the trough hold?
 SOLID GEOMETRY Example #22
Nov 2023 CELE
Dale has a water trough whose ends are trapezoids and whose sides are rectangles as illustrated. He
is afraid that the base it is sitting on it will not support the weight of the trough when it is filled with
water. He knows that the base will support 4800 lb.
A. If the trough is filled with water, determine the number of cubic feet of water contained in the
trough.

𝟔𝟕. 𝟖𝟖𝟐 𝒇𝒕𝟑

 SOLID GEOMETRY Example #22
Nov 2023 CELE
Dale has a water trough whose ends are trapezoids and whose sides are rectangles as illustrated. He
is afraid that the base it is sitting on it will not support the weight of the trough when it is filled with
water. He knows that the base will support 4800 lb.
B. Determine the total weight, assuming that the trough weighs 375 lb and the water weighs
62.4 lb per cubic foot. Is the base strong enough to support the troughfilled with water?

𝟒𝟔𝟏𝟎. 𝟖𝟓𝟐 𝒍𝒃, 𝒚𝒆𝒔

 SOLID GEOMETRY Example #22
Nov 2023 CELE
Dale has a water trough whose ends are trapezoids and whose sides are rectangles as illustrated. He
is afraid that the base it is sitting on it will not support the weight of the trough when it is filled with
water. He knows that the base will support 4800 lb.
C. If 1 gal of water weighs 8.3 lb, how many gallons of water will the trough hold?

1𝑔𝑎𝑙
4235.852 𝑙𝑏 ×
8.3 𝑙𝑏
𝟓𝟏𝟎. 𝟑𝟒 𝒈𝒂𝒍
 SOLID GEOMETRY Example #23

Find the (a) volume and (b) the total surface area of the sphere:

41.8 𝑓𝑡

(a) 38,240.84 𝑓𝑡 3
(b) 5489.12 𝑓𝑡 2
 SOLID GEOMETRY Example #24

Find (a) how many gallons of water can be stored in a spherical water tank 60 ft in diameter (7.48
gal are contained in 1 𝑓𝑡 3 ), (b) the surface area of the water tank (c) how many gallons of paint
would be necessary to paint the tank if 1 gal covers 125 𝑓𝑡 2 .

(a) 846,000 𝑔𝑎𝑙 (b) 11309.73 𝑓𝑡 2 (c) 90.46 𝑔𝑎𝑙

SOLID GEOMETRY Example #24

4 3
𝑉 = 𝜋𝑟 𝐴 = 4𝜋𝑟 2
3
4 𝐴 = 4𝜋 × 302
𝑉 = 𝜋 × 303 = 113097.3355 𝑓𝑡 3
3
7.48𝑔𝑎𝑙 𝐴 = 11309.734 𝑓𝑡 2
3
𝑉 = 113097.3355 𝑓𝑡 × = 845,968.07 𝑔𝑎𝑙
1𝑓𝑡 3
𝑽 ≈ 𝟖𝟒𝟔, 𝟎𝟎𝟎 𝒈𝒂𝒍

2
1𝑔𝑎𝑙
𝑃 = 11309.734 𝑓𝑡 ×
125 𝑓𝑡 2

𝑷 = 𝟗𝟎. 𝟒𝟕𝟖 𝒈𝒂𝒍

 SOLID GEOMETRY Example #25

A closed cylindrical tank is 9 m long and 3 m in diameter. When lying in a horizontal position, the water is
2 m deep. Compute the following:
a. Volume of water
b. The wetted lateral area of the tank
c. The depth of water in the tank if it is upright position

A. 𝑉 = 𝐴𝑏 ℎ θ1 = 360 − 141.058°
𝐴𝑏 = 𝐴𝑠𝑒𝑐𝑡𝑜𝑟 + 𝐴𝑡𝑟𝑖𝑎𝑛𝑔𝑙𝑒 θ1 = 218.942°

1 1 𝐴𝑏
𝐴𝑏 = 𝑟 2 𝜃1 + 𝑟 2 sin 2𝜃2 1 𝜋
2 2 2
= × 1.5 218.942° ×
2 180°
1
0.5 + × 1.52 sin 141.058°
cos 𝜃2 = 2
1.5
θ2 = 70.529° 𝐴𝑏 = 5.006𝑚2
2θ2 = 141.058°
 SOLID GEOMETRY Example #2

A closed cylindrical tank is 9 m long and 3 m in diameter. When lying in a horizontal position, the water is
2 m deep. Compute the following:
a. Volume of water
θ1 = 218.942°
b. The wetted lateral area of the tank
c. The depth of water in the tank if it is upright position 2θ2 = 141.058°
𝐴𝑏 = 5.006𝑚2
𝑉 = 𝐴𝑏 ℎ

𝑉 = 5.006𝑚2 × 9𝑚

𝑽 = 𝟒𝟓. 𝟎𝟓𝟒 𝒎𝟑
 SOLID GEOMETRY Example #2

A closed cylindrical tank is 9 m long and 3 m in diameter. When lying in a horizontal position, the water is
2 m deep. Compute the following:
a. Volume of water
θ1 = 218.942°
b. The wetted lateral area of the tank
c. The depth of water in the tank if it is upright position 2θ2 = 141.058°
𝐴𝑏 = 5.006𝑚2
𝑉 = 𝐴𝑏 ℎ
𝑏
𝐴𝑏 = න 2 𝑅2 − 𝑥 2 𝑑𝑥 𝑉 = 5.006 𝑚2 × 9 𝑚
𝑎
0.5 𝑽 = 𝟒𝟓. 𝟎𝟓 𝒎𝟑
𝐴𝑏 = න 2 1.52 − 𝑥 2 𝑑𝑥
−1.5

0.4999
𝐴𝑏 = න 2 1.52 − 𝑥 2 𝑑𝑥 = 5.006 𝑚2
−1.4999
 SOLID GEOMETRY Example #2
INTRODUCTION

A closed cylindrical tank is 9 m long and 3 m in diameter. When lying in a horizontal position, the water is
2 m deep. Compute the following: θ1 = 218.942°
a. Volume of water
b. The wetted lateral area of the tank 2θ2 = 141.058°
c. The depth of water in the tank if it is upright position 𝐴 = 5.006𝑚2
𝑏
𝑉 = 45.05 𝑚3
B. C.
𝐴𝑙𝑤𝑒𝑡𝑡𝑒𝑑 = 𝑝𝑏 ℎ 𝑉 = 𝐴𝑏 ℎ

𝐴𝑙𝑤𝑒𝑡𝑡𝑒𝑑 = (𝑟𝜃1 )(ℎ) 𝑉 = 𝜋𝑟 2 ℎ

𝜋 45.05 = 𝜋(1.5)2 ℎ
𝐴𝑙𝑤𝑒𝑡𝑡𝑒𝑑 = 1.5 × 218.942° × (9)
180°
𝒉 = 𝟔. 𝟑𝟕𝟑 𝒎
𝑨𝒍𝒘𝒆𝒕𝒕𝒆𝒅 = 𝟓𝟏. 𝟓𝟖𝟕 𝒎𝟐
DAILY REMINDER

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END