Mathematics
Volume of Cylinder
The volume of a cylinder is the amount of space it occupies and is calculated using the formula V = πr^2h, where V represents the volume, π is a constant (approximately 3.14159), r is the radius of the cylinder's base, and h is the height of the cylinder. This formula allows for the quick and accurate determination of a cylinder's volume in mathematical calculations.
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12 Key excerpts on "Volume of Cylinder"
eBook - PDFGeometry
A Self-Teaching Guide
- Steve Slavin, Ginny Crisonino(Authors)
- 2004(Publication Date)
- Wiley(Publisher)
A cylinder is a uniform solid whose base is a circle. The cylinder shown above has a radius of 2 and a height of 5. Volume of a cylinder formula V = πr 2 h where r is the radius and h is the height of the cylinder See if you can find the volume of this cylinder. V = πr 2 h = (3.14)(2) 2 (5) = 62.8 Example 22: Find the volume of a cylinder with a radius of 3.8 feet and a height of 4.6 feet. Solution: Volume = πr 2 h = (3.14)(3.8) 2 (4.6) = (3.14)(14.44)(4.6) = 208.571 cubic feet Example 23: Find the volume of a cylinder with a diameter of 10 inches and a height of 29 inches. Solution: First we’ll begin by finding the radius. r = = = 5 Volume = πr 2 h = (3.14)(5) 2 (29) = (3.14)(25)(29) = 2,276.5 cubic inches 10 2 d 2 5 2 Volume and Surface Area of Three-dimensional Polygons 181 Example 24: Find the surface area of a cylinder with a radius of 12 inches and a height of 5 inches. Solution: SA = 2πrh = 2(3.14)(12)(5) = 6.28(60) = 376.8 square inches Example 25: Find the surface area of a cylinder with a diameter of 9 feet and a height of 13.4 feet. Solution: Start by finding the radius. r = = = 4.5 SA = 2πrh = 2(3.14)(4.5)(13.4) = 6.28(60.3) = 378.684 square feet SELF-TEST 4 1. Find the volume of a cylinder with a a. radius of 12 feet and a height of 9 feet. b. radius of 4.5 yards and a height of 12 yards. c. diameter of 16 inches and a height of 23 inches. 2. Find the surface area of a cylinder with the same dimensions as those in problem 1. ANSWERS 1. a. Volume = πr 2 h = (3.14)(12) 2 (9) = (3.14)(144)(9) = 4,069.44 cubic feet b. Volume = πr 2 h = (3.14)(4.5) 2 (12) = (3.14)(20.25)(12) = 763.02 cubic yards c. Volume = πr 2 h = (3.14)(8) 2 (23) = (3.14)(64)(23) = 4,622.08 cubic inches 2. a. SA = 2πrh = 2(3.14)(12)(9) = (6.28)(108) = 678.24 square feet b. SA = 2πrh = 2(3.14)(4.5)(12) = (6.28)(54) = 339.12 square yards c. SA = 2πrh = 2(3.14)(8)(23) = (6.28)(184) = 1,155.52 cubic inches 9 2 d 2 182 GEOMETRY
No longer available |Learn more- Andre Castagna, Andre, Ph.D. Castagna, Barron's Educational Series(Authors)
- 2021(Publication Date)
- Barrons Educational Services(Publisher)
V = Bh to calculate the volume of the prism.Circular Cylinders
A circular cylinder is a solid figure with two parallel and congruent circular bases and a curved lateral surface. The height is the perpendicular distance between the bases. As with the prisms, cylinders can be right or oblique. Figure 12.3 shows each type of cylinder.Figure 12.3 CylinderVolume of a Circular Cylinder
Volume (V) = πr2 hwhere r is the radius of the base and h is the height.Example 1
A right circular cylinder has a diameter of 10 cm and a height of 5 cm. Find the volume. Express your answers in terms of π.Solution:We can work backward to find the height or the radius given the volume of a cylinder.Example 2
A right circular cylinder has a volume of 200 in.3 If the height of the cylinder is 6 in., what is the radius of the cylinder? Round your answer to the nearest hundredth.Solution: Set up the formula for volume. Substitute the known values. Then solve for the radius.
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12.2 Cones, Pyramids, and Spheres
Key Ideas
A circular cone is a solid with one circular base and an apex. A pyramid is a solid with one polygonal base and an apex. A sphere is the solid that is the set of all points a fixed distance from a center point.Circular Cone Pyramid Sphere Volume (V) B is the area of the base, h is the height Circular Cones
A circular cone is a solid with one circular base and a curved lateral surface that comes to a point at the apex. The height of the cone is the length of the segment from the apex perpendicular to the base. In a right circular cone, the height will intersect at the center of the circular base.Figure 12.4 Right circular coneVolume of a Circular Cone
where r is the radius of the base and h is the height.These formulas are applied in a manner similar to those for the cylinder. The Pythagorean theorem can be used to relate slant height, height, and radius. So if you know any two, you can find the third.
eBook - PDF- John Peterson, Robert Smith(Authors)
- 2019(Publication Date)
- Cengage Learning EMEA(Publisher)
The formula for computing volumes of right circular cylinders is the same as that of prisms. The volume of a right circular cylinder is equal to the product of the base area and height. V 5 A B h where V 5 volume A B 5 area of base, A B 5 p r 2 , or 0.7854 d 2 (rounded) h 5 height Example 1 Find the volume of a cylinder with a base area of 30.0 square centimeters and a height of 6.0 centimeters. V 5 30.0 cm 2 3 6.0 cm 5 180 cm 3 Ans UNIT 62 VOLUMES OF PRISMS AND CYLINDERS 467 Example 2 A length of pipe is shown in Figure 62-5. Find the volume of metal in the pipe. Find the area of the outside circle. A B < 0.7854 d 2 A B < (0.7854)(4.00 in.) 2 < 12.566371 sq in. Find the area of the hole. A B < (0.7854)(3.40 in.) 2 < 9.079224 sq in. Find the cross-sectional area. 12.566371 sq in. 2 9.079224 sq in. < 3.487147 sq in. Find the volume. V < 3.487147 sq in. 3 50.0 in. < 174 cu in. Ans COMPUTING HEIGHTS AND BASES OF PRISMS AND CYLINDERS The height of a prism or a right circular cylinder can be determined if the base area and volume are known. Also, the base area can be found if the height and volume are known. Substitute the known values in the volume formula and solve for the unknown value. Example 1 Find the height of a spacer block in the shape of a right rectangular prism that has a base area of 68.40 square inches and a volume of 366.60 cubic inches. Substitute values in the formula and solve for h . 366.6 cu in. 5 68.40 sq in. 3 h h < 5.36 in. Ans Example 2 Calculate the diameter of a piston designed to have a volume of 450.0 cubic centimeters and a length (height) of 10.800 centimeters. Find the base area. 450.0 cm 3 5 A B (10.800 cm) A B < 41.6667 cm 2 Find the piston diameter. 41.6667 cm 2 < 0.7854 d 2 d 2 < 53.051566 cm 2 d < 7.284 cm Ans (rounded) 50.0 in. 3.40 in. Inside Diameter 4.00 in. Outside Diameter FIGURE 62-5 APPLICATION Tooling Up 1. Find the volume of a steel shaft that is 18.64 cm long and has a radius of 1.75 cm.
eBook - PDFSpellman's Standard Handbook for Wastewater Operators
Volume I, Fundamental Level, Second Edition
- Frank R. Spellman(Author)
- 2010(Publication Date)
- CRC Press(Publisher)
Pi (π)—A number used in calculations involving circles, spheres, or cones (π = 3.14). Radius— The distance, measured in linear units, from the center of a circle to the edge. 3 Basic Mathematics 45 Sphere— A container shaped like a ball. Square units— Measurements used to express area (e.g., square feet, square meters, acres). Volume— The capacity of a unit (how much it will hold), measured in cubic units (e.g., cubic feet, cubic meters) or in liquid volume units (e.g., gallons, liters, million gallons). Width— The distance from one from one side of the tank to the other, measured in linear units. 3.10.2 Relevant Geometric Equations Circumference C of a circle: C = π d = 2π r Perimeter P of a square with side a : P = 4 a Perimeter P of a rectangle with sides a and b : P = 2 a + 2 b Perimeter P of a triangle with sides a , b , and c : P = a + b + c Area A of a circle with radius r ( d = 2 r ): A = π d 2 /4 = π r 2 Area A of a square with sides a : A = a 2 Area A of a rectangle with sides a and b : A = ab Area A of a triangle with base b and height h : A = 0.5 bh Area A of an ellipse with major axis a and minor axis b : A = π ab Area A of a trapezoid with parallel sides a and b A = 0.5( a + b ) h and height h : Area A of a duct (in ft 2 ) when d is in inches: A = π d 2 /576 = 0.005454 d 2 Volume V of a sphere with a radius r ( d = 2 r ): V = 1.33π r 3 = 0.1667π d 3 Volume V of a cube with sides a : V = a 3 Volume V of a rectangular solid (sides a and b and height c ): V = abc Volume V of a cylinder with a radius r and height h : V = π r 2 h = π d 2 h /4 Volume V of a pyramid with base area B and height h : V = 0.33 Bh 46 Spellman’s Standard Handbook for Wastewater Operators: Volume I, Fundamental Level 3.10.3 Geometrical Calculations 3.10.3.1 Perimeter and Circumference On occasion, it may be necessary to determine the distance around grounds or landscapes.
No longer available |Learn more- Daniel C. Alexander, Geralyn M. Koeberlein, , , Daniel C. Alexander, Geralyn M. Koeberlein(Authors)
- 2014(Publication Date)
- Cengage Learning EMEA(Publisher)
EXS. 8–10 Figure 9.43 THEOREM 9.4.3 The volume V of a sphere with a radius of length r is given by . V 4 3 p r 3 EXAMPLE 4 Find the exact volume of a sphere whose length of radius is 1.5 in. SOLUTION This calculation can be done more easily if we replace 1.5 by . 9 p 2 in 3 4 3 p 3 2 3 2 3 2 V 4 3 p r 3 3 2 D iscover A farmer’s silo is a composite shape. That is, it is actually composed of two solids. What are they? ANSWER Cylinder and hemisphere © James Horning/Shutterstock.com Copyright 2013 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. EXAMPLE 5 A spherical propane gas storage tank has a volume of . Using , find the radius of the sphere. SOLUTION becomes . Then . In turn, The radius of the tank is 3 ft. Just as two concentric circles have the same center but different lengths of radii, two spheres can also be concentric. This fact is the basis for the solution of the problem in the following example. EXAMPLE 6 A child’s hollow plastic ball has an inside diameter length of 10 in. and is approxi-mately thick (see the cross-section of the ball in Figure 9.44). Approximately how many cubic inches of plastic were needed to construct the ball? SOLUTION The volume of plastic used is the difference between the outside volume and the inside volume. Where R denotes the length of the outside radius and r denotes the length of the inside radius, and . Then The volume of plastic used was approximately . Like circles, spheres may have tangent lines and secant lines as illustrated in Figure 9.45(a). However, spheres also have tangent planes as shown in Figure 9.45(b).
- Frank R. Spellman, Nancy E. Whiting(Authors)
- 2013(Publication Date)
- CRC Press(Publisher)
Sphere— A container shaped like a ball. Square units— Measurements used to express area (e.g., square feet, square meters, acres). Volume— The capacity of a unit (how much it will hold), measured in cubic units (e.g., cubic feet, cubic meters) or in liquid volume units (e.g., gallons, liters, million gallons). Width— The distance from one side of the tank to the other, measured in linear units. 61 Basic Math Operations 2.11.2 R ELEVANT G EOMETRIC E QUATIONS Circumference C of a circle: C = π d = 2 π r Perimeter P of a square with side a : P = 4 a Perimeter P of a rectangle with sides a and b : P = 2 a + 2 b Perimeter P of a triangle with sides a , b , and c : P = a + b + c Area A of a circle with radius r ( d = 2 r ): A = π d 2 /4 = π r 2 Area A of duct in square feet when d is in inches: A = 0.005454 d 2 Area A of a triangle with base b and height h : A = 0.5 bh Area A of a square with sides a : A = a 2 Area A of a rectangle with sides a and b : A = ab Area A of an ellipse with major axis a and minor axis b : A = π ab Area A of a trapezoid with parallel sides a and b and height h : A = 0.5( a + b ) h Area A of a duct in square feet when d is in inches: A = π d 2 /576 = 0.005454 d 2 Volume V of a sphere with a radius r ( d = 2 r ): V = 1.33 π r 3 = 0.1667 π d 3 Volume V of a cube with sides a : V = a 3 Volume V of a rectangular solid (sides a and b and height c ): V = abc Volume V of a cylinder with a radius r and height H : V = π r 2 h = π d 2 h /4 Volume V of a pyramid: V = 0.33 2.11.3 G EOMETRICAL C ALCULATIONS 2.11.3.1 Perimeter and Circumference On occasion, it may be necessary to determine the distance around grounds or landscapes. To mea-sure the distance around property, buildings, and basin-like structures, it is necessary to determine either perimeter or circumference. The perimeter is the distance around an object; a border or outer boundary. Circumference is the distance around a circle or circular object, such as a clarifier.
eBook - PDFMathematics for Elementary Teachers
A Contemporary Approach
- Gary L. Musser, Blake E. Peterson, William F. Burger(Authors)
- 2013(Publication Date)
- Wiley(Publisher)
This property holds for right and oblique cones. Volume of a Cone The volume V of a cone whose base has area A and whose height is h is V Ah = 1 3 . h r A Common Core – Grade 7 Know the formulas for the vol- umes of cones, cylinders, and spheres and use them to solve real-world and mathematical problems. T H E O R E M 1 3 . 1 7 For a cone with a circular base of radius r , the volume of the cone is 1 3 2 p r h. Check for Understanding: Exercise/Problem Set A #8–9 ✔ r r 2r Figure 13.74 Volume of a Sphere The volume V of a sphere with radius r is V r = 4 3 3 p . r T H E O R E M 1 3 . 1 8 Section 13.4 Volume 703 Table 13.11 summarizes the volume and surface area formulas for right prisms, right circular cylinders, right regular pyramids, right circular cones, and spheres. The indicated dimensions are the area of the base, A; the height, h; the perimeter or cir- cumference of the base, P or C ; and the slant height, l . By observing similarities, one can minimize the amount of memorization. TABLE 13.11 GEOMETRIC SHAPE SURFACE AREA VOLUME Right prism S A Ph = + 2 V Ah = Right circular cylinder S A Ch = + 2 V Ah = Right regular pyramid S A Pl = + 1 2 V Ah = 1 3 Right circular cone S A Cl = + 1 2 V Ah = 1 3 Sphere S r = 4 2 π V r = 4 3 3 π The remainder of this section presents a more formal derivation of the volume and surface area of a sphere. First, to find the volume of a sphere, we use Cavalieri’s prin- ciple, which compares solids where cross-sections have equal areas. As an aid to recall and distinguish between the formulas for the surface area and volume of a sphere, observe that the r in 4 2 p r is squared, an area unit, whereas the r in 4 3 3 p r is cubed, a volume unit. An ice cream shop has sugar cones with a slant height of 13 cm and the diameter of the base is 10 cm (see Figure 13.75).
- James Stewart, Daniel K. Clegg, Saleem Watson, , James Stewart, James Stewart, Daniel K. Clegg, Saleem Watson(Authors)
- 2020(Publication Date)
- Cengage Learning EMEA(Publisher)
Definition of Volume Let S be a solid that lies between x - a and x - b. If the cross-sectional area of S in the plane P x , through x and perpendicular to the x-axis, is As xd, where A is a continuous function, then the volume of S is V - lim n l` o n i-1 As x i * d Dx - y b a As xd dx It can be proved that this definition is independent of how S is situated with respect to the x-axis. In other words, no matter how we slice S with parallel planes, we always get the same answer for V. When we use the volume formula V - y b a As xd dx , it is important to remember that As xd is the area of a moving cross-section obtained by slicing through x perpendicular to the x-axis. Notice that, for a cylinder, the cross-sectional area is constant: As xd - A for all x. So our definition of volume gives V - y b a A dx - Asb 2 ad; this agrees with the formula V - Ah. EXAMPLE 1 Show that the volume of a sphere of radius r is V - 4 3 r 3 . SOLUTION If we place the sphere so that its center is at the origin, then the plane P x intersects the sphere in a circle whose radius (from the Pythagorean Theorem) is FIGURE 3 Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
eBook - PDF- Frank R. Spellman(Author)
- 2015(Publication Date)
- CRC Press(Publisher)
77 Math Operations Review Volume V of a rectangular solid (sides a and b and height c ): V = abc Volume V of a cylinder with a radius r and height H : V = π r 2 h = π d 2 h /4 Volume V of a pyramid: V = 0.33 G EOMETRICAL C ALCULATIONS Perimeter and Circumference On occasion, it may be necessary to determine the distance around grounds or landscapes. To measure the distance around property, buildings, and basin-like structures, either the perimeter or circumfer-ence must be determined. The perimeter is the distance around an object; a border or outer boundary. Circumference is the distance around a circle or circular object, such as a clarifier. Distance is a linear measurement that defines the distance (or length) along a line. Standard units of measurement such as inches, feet, yards, and miles and metric units such as centimeters, meters, and kilometers are used. The perimeter ( P ) of a rectangle (a four-sided figure with four right angles) is obtained by adding the lengths ( L i ) of the four sides (see Figure 3.1): Perimeter = L 1 + L 2 + L 3 + L 4 (3.4) ■ EXAMPLE 3.40 Problem: Find the perimeter of the rectangle shown in Figure 3.2. Solution: P = 35 ft + 8 ft + 35 ft + 8 ft = 86 ft ■ EXAMPLE 3.41 Problem: What is the perimeter of a rectangular field if its length is 100 ft and its width is 50 ft? Solution: P = (2 × length) + (2 × width) = (2 × 100 ft) + (2 × 50 ft) = 200 ft + 100 ft = 300 ft L 1 L 4 L 2 L 3 FIGURE 3.1 Perimeter. 35' 8' 8' 35' FIGURE 3.2 Perimeter of a rectangle for Example 3.40. 78 Handbook of Environmental Engineering ■ EXAMPLE 3.42 Problem: What is the perimeter of a square with 8-in. sides? Solution: P = (2 × length) + (2 × width) = (2 × 8 in.) + (2 × 8 in.) = 16 in. + 16 in. = 32 in. The circumference is the length of the outer border of a circle.
eBook - ePub- John Bird(Author)
- 2019(Publication Date)
- Routledge(Publisher)
Chapter 14 Volumes of common solidsWhy it is important to understand: Volumes of common solidsThere are many practical applications where volumes and surface areas of common solids are required. Examples include determining capacities of oil, water, petrol and fish tanks, ventilation shafts and cooling towers, determining volumes of blocks of metal, ball-bearings, boilers and buoys, and calculating the cubic metres of concrete needed for a path. Finding the surface areas of loudspeaker diaphragms and lampshades provide further practical examples. Understanding these calculations is essential for the many practical applications in engineering, construction, architecture and science.At the end of this chapter, you should be able to:- state the SI unit of volume
- calculate the volumes and surface areas of cuboids, cylinders, prisms, pyramids, cones and spheres
- appreciate that volumes of similar bodies are proportional to the cubes of the corresponding linear dimensions
14.1 Introduction
The volume of any solid is a measure of the space occupied by the solid. Volume is measured in cubic units such as mm3 , cm3 and m3 .This chapter deals with finding volumes of common solids; in engineering it is often important to be able to calculate volume or capacity, to estimate, say, the amount of liquid, such as water, oil or petrol, in differently shaped containers.A prism is a solid with a constant cross-section and with two ends parallel. The shape of the end is used to describe the prism. For example, there are rectangular prisms (called cuboids), triangular prisms and circular prisms (called cylinders).On completing this chapter you will be able to calculate the volumes and surface areas of rectangular and other prisms, cylinders, pyramids, cones and spheres. Volumes of similar shapes are also considered.14.2 Calculating volumes and surface areas of common solids
Science and Mathematics for Engineering. 978-0-367-20475-4, © John Bird. Published by Taylor & Francis. All rights reserved.14.2.1 Cuboid or rectangular prism
eBook - PDF- John Peterson, Robert Smith(Authors)
- 2018(Publication Date)
- Cengage Learning EMEA(Publisher)
If the volumes of the prism and pyramid are measured, the volume of the pyramid will be one-third the volume of the prism. Also, if the volumes of a cylinder and a cone with identical bases and heights are measured, the volume of the cone will be one-third the volume of the cylinder. The formulas for computing volumes of prisms and right circular cylinders are the same. Therefore, the formulas for computing volumes of regular pyramids and right circular cones are the same. The volume of a regular pyramid or a right circular cone equals one-third the product of the area of the base and height. V 5 1 3 A B h where V 5 volume A B 5 area of the base h 5 height Copyright 2019 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. Unit 29 PYRAMIDS AND CONES: VOLUMES, SURFACE AREAS, AND WEIGHTS 719 These examples illustrate the method of computing volumes of regular pyramids and right circular cones. EXAMPLES 1. Compute the volume of a pyramid that has a base area of 24.0 square feet and a height of 6.0 feet. V 5 24.0 sq ft 3 6.0 ft 3 5 48 cu ft Ans 2. A bronze casting in the shape of a right circular cone is 14.52 inches high and has a base diameter of 10.86 inches. a. Find the volume of bronze required for the casting. b. Find the weight of the casting to the nearest pound. Bronze weighs 547.9 pounds per cubic foot. a. V < s 3.1416 ds 5.430 in. d 2 s 14.52 in. d 3 < 448.3 cu in. Ans b. 448.3 cu in. 4 1728 cu in. y cu ft < 0.2594 cu ft 0.2594 cu ft 3 547.9 lb y cu ft < 142 lb Ans Calculator Method a. V 5 p 3 5.43 x 2 3 14.52 4 3 5 448.3269989, 448.3 cu in.
eBook - PDFMathematical Practices, Mathematics for Teachers
Activities, Models, and Real-Life Examples
- Ron Larson, Robyn Silbey(Authors)
- 2014(Publication Date)
- Cengage Learning EMEA(Publisher)
Round your 4 mm 3 mm answer to the nearest tenth. SOLUTION Begin by drawing a net of the cylinder. S = 2πr 2 + 2πrh 3 mm 4 mm 4 mm = 2π(4) 2 + 2π(4)(3) = 32π + 24π = 56π ≈ 175.84 The surface area is about 175.8 square millimeters. There are many types of cylinders. Their bases can be ellipses, and their lateral surfaces can meet the base at angles other than right angles. The formula for the surface area of a right circular cylinder does not apply to elliptical or oblique cylinders. In this book, only right cylinders that have circular bases are discussed. h h Right cylinder Oblique cylinder Classroom Tip Copyright 2014 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 470 Chapter 12 Circles and Circular Solids EXAMPLE 2 Identifying the Dimensions of a Cylinder A student is constructing a cylinder, as shown. What is the height of the cylinder that the student is constructing? Explain your reasoning. 3.5 cm SOLUTION The radius of each circle is 3.5 centimeters. Looking at the dimensions of the rectangle, one dimension is the height of the cylinder, and the other is the circumference of each base. Use the formula for circumference C = 2πr to determine which is which. 10 = ? 2π(3.5) 22 = ? 2π(3.5) 10 ≈ 21.98 ✗ 22 ≈ 21.98 ✓ So, it follows that 22 centimeters is the circumference. This implies that the height of the cylinder is 10 centimeters. EXAMPLE 3 Finding a Surface Area in Real Life The tank of a truck is a stainless steel cylinder. a. Find the surface area of the tank. Round your answer to the nearest tenth.
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