MATH 113 Section 10.3: Surface Area and Volume
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1 MATH 113 Section 10.3: Surface Area and Volume Prof. Jonathan Duncan Walla Walla College Winter Quarter, 2007
2 Outline 1 Surface Area 2 Volume 3 Conclusion
3 Measuring Three Dimensional Shapes In the last section we looked at two measurements for two dimensional shapes. Two Dimensional Shapes Two dimensional shapes can be measured in two different ways: 1 Perimeter (one dimensional) 2 Area (two dimensional) In the same way, there are two important types of measurements for three dimensional shapes. Three Dimensional Shapes Three dimensional shapes have two important measurements: 1 Surface Area (two dimensional) 2 Volume (three dimensional)
4 What is Surface Area? Surface area is a term which is mainly used with three dimensional objects. It is like the perimeter of a two dimensional object. Surface Area The surface area of a three dimensional object is the amount of exposed area on the figures. For a two dimensional object, the surface area is simply the area of the object. With many three dimensional objects we can think of a net for the object, find the area of the net, and this will be the surface area of the object.
5 Surface Area of Right Prisms We begin by examining the surface area of right prisms. Find a general formula for the surface area of a right rectangular prism with a base of length l, width h, and a height of h. Is this formula unique to rectangular prisms? Find a general formula for the surface area of a right equilateral triangular prism with base side length s and height h. General Formula In general the surface area of a right prism is 2B + ph where B is the area of the base, h is the height, and p is the perimeter of the base.
6 Surface Area of Cylinders In many ways cylinders and prisms are alike. In fact, we can use a net to analyze the surface area. Find the surface area of a right circular cylinder with radius r and height h. Surface Area Formula Using the formula for prisms, 2B + ph, we can set B = πr 2 and p = 2πr to get 2πr 2 + 2πrh = 2πr(r + h).
7 Surface Area of Oblique Prisms In each of the previous examples, we dealt with right objects. That is, the sides made right angles with the base. How does surface area change in oblique prisms? Can we develop a surface area formula for an oblique prisms which is independent of the slant angle? Slant and Height Note that the length of the lateral faces in an oblique prism will vary depending on the height and the slant angle.
8 Surface Area of Pyramids By using nets and the formula for the area of a triangle we can develop a formula for the surface area of a pyramid. Develop a formula for the surface area of a right square pyramid with a base side length s and a slant height l. General Formula In general, a pyramid with an area of base B and perimeter p with slant height l has a surface area B pl.
9 Surface Area of Cones Just as we extended the formula for the surface area of a prism to one for a cylinder, we can extend the surface area for a pyramid to one for a cone. Find the surface area for a right circular cone with base radius r and slant height h. Surface Area Formula Using the formula for a pyramid with a base area of B = πr 2 and perimeter, or in this case circumference, of 2πr, the surface area is πr (2πr)l = πr(r + l).
10 Surface Area of a Sphere The final object which we will examine is the sphere. Surface Area of a Sphere The surface area of a sphere of radius r is 4πr 2. Radius of a Sphere? What is the radius of a sphere? It is the radius of a great circle or largest possible circle taken as a cross-section of the sphere. Unfortunately, to show that this formula works we would need to use calculus. This is beyond the scope of this class.
11 What is Volume? Just as area is measured in square units, so volume is measured in a different sort of unit called a cubic unit. Volume The volume of an object is the amount of space contained in the object. It can be thought of as the number of cubes of unit length which can be fit into the object leaving no empty space and with no overlap. Just as area can be found by multiplying the side length of a polygon by the height of the polygon, we can use multiplication to find the volume of certain objects.
12 Volume of Right Prisms The first, and simplest, object we will examine is a right rectangular prism. Find a formula for the volume of a right rectangular prism of base length l, width w and with height h. How many little one by one by one inch cubes could be packed into a box which is inches? Volume Formula In general the volume of a right prism is B h where B is the area of the base and h is the height.
13 Volume of Oblique Prisms What happens to the volume formula in an oblique prism? Find a formula for the volume of an oblique rectangular prism of base length l, width w, and with height h. Find a formula for the volume of a right triangular prisms with a triangular base with base length 3 and height 2 having prism height 5.
14 Volume of Cylinders and Pyramids How do these formulas change for cylinders and pyramids? Show that the volume of a circular cylinder is πr 2 h where r is the radius of the base and h is the height of the prism. In can be shown that three congruent pyramids each with square base B can be fit into a cube with sides B. Using this fact, find a formula for the volume of a square pyramid with base area B.
15 Volume of a Sphere As with surface area, the volume of a sphere is difficult to derive. Volume of a Sphere The volume of a sphere is 4 3 πr 3. Spheres and Cylinders From previous work, we know that: The volume of a right circular cylinder of radius r and height 2r is 2πr 3 The volume of a sphere of radius r is 4 3 πr 3. A sphere of radius r can be inscribed in a right circular cylinder of radius r and height 2r. Therefore, the volume of that sphere is 2 3 the volume of the cylinder.
16 Volume of Irregular Objects While the formulas above are necessary, their applicability in real live can be limited as we seldom have perfect prisms, cylinders, pyramids or spheres to work with. Archimedes was an ancient Greek mathematician. In one story Archimedes was asked to determine if a crown presented to his king was solid gold as the giver claimed or had sliver mixed into the crown. Archimedes knew that the mass of an object is given by mass = volume density and he knew that gold is about twice as dense as silver. Finding the volume and weight of the crown would let him solve the problem. But how could he determine the volume of the irregularly shaped crown? The solution came to Archimedes as he sat is his bath one day. He is said to have leapt out of the bath and run out into the street shouting Eureka!
17 Important Concepts Things to Remember from Section Surface Area and Volume for: 1 Right Prisms 2 Right Cylinders 3 Right Pyramids 4 Spheres 2 Strategies for finding the volume of irregular objects
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