Discovering Solids: Volumes of Pyramids, Cones and Spheres (1959)

[Music] because we live in a three dimensional world we are concerned with objects having depth or height as well as length and width if a given object has a closed solid form often times we become primarily concerned with the space it will occupy if however the object is a container for liquids or other materials then we are more often concerned with its capacity such a property of an object is referred to as its volume the volume of a geometric solid may be further defined as the number of cubic units of a given kind which a solid contains the question of course is how many cubic units does it take to make up the solid this is true regardless of the shape of the solid under consideration when we studied prisms and cylinders we found that the volume could be obtained by multiplying the number of area units in the base by the number of linear units in the height provided of course we use the same linear unit in all our measurements in a prism the number of cubic units is computed with the formula V equal [Music] BH the volume of a cylinder is obtained with the formula v = p r² h in each formula the same principle is applied the volume of either the prism or the cylinder is equal to the product of the number of area units in its base times the number of linear units in its height now let's see if there's any relationship between these geometric solids the prism And the cylinder and these the pyramid and The Cone for this observation let's compare a pyramid with a prism having the same measur for base and height first of all we need to remember that a prism which has a rectangle for a base is a special kind of prism a rectangular prism similarly a pyramid whose base is a rectangle is called a rectangular pyramid the shape of such a pyramid is largely determined by the shape of its base the faces of a pyram are triangles all of which meet at a given point called the vertex the dimension we call height is measured from the vertex along a line perpendicular to the base once we know the number of linear units in the height and the number of area units in the base we then can apply the formula V equals BH but obviously the capacity of the and that of the pyramid are not the same the volume of the pyramid is less than that of the prism we can see this but now we need to know how much perhaps a formula for finding the volume can be developed how many pyramids of sand will it take to fill the prism one [Music] two [Music] three the volume of a prism is three times that of a pyramid having an equal base and height we have now established a meaningful relationship between these two geometric solids because we know the formula for finding the volume of the prism we can now determine a formula for finding the volume of a pyramid V equal 13 the product of the base time the height a simple illustration of how such a formula could be used in modern industry is that of a box designer who was given the job of creating a new popcorn box one that would have one3 the volume of the Box ordinarily used he could design another prism shape actually 13 of the original box but the shape and form of such a change might not be pleasing to the eye of the customer it looks too much like what it is 1/3 the original amount if however a pamal box is used the effect is that of an optical illusion the customer thinks he's getting more than he actually does there's that same familiar opening at the top and the same measurement in height but we know as did the designer that according to the formula V = 13 BH there are however many objects called pyramids whose bases are not rectangles for example this pyramid is called a triangular pyramid because its base is a triangle a square pyramid would have for its base a square the base of a pentagonal pyramid is five sided a pentagon a six-sided pyramid a hexagonal pyramid is soal because its base is a hexagon the volume of any of these pyramids can be computed if we know the area of the base and the dimension we call height by observing once more a rectangular pyramid whose volume is obtained with the formula V = 13 BH we find that if we BCT the rectangular base we form two triangular bases hence two triangular pyramids the height Remains the Same but the each triangular base is half that of the rectangular base the same formula applies in finding the volume of the halves as in finding the volume of the whole V equal 13 BH even though the bases to which we refer now are the Triangular bases B1 and B2 now we have a key for the volume of any pyramid regardless of its shape or size for those pyramids having four or more sides we simply divide them into triangular pyramids we use the same formula in finding the volume of any triangular pyramid regardless of its size V equal 13 BH so to find the total volume of the original pyramid we simply add the volumes of the parts for example if we take a hexagonal pyramid a pyramid with six lateral faces and divide the base into four triangles we then can determine the number of area units in each base by using the formula for finding the area of a triangle then we multiply the area units in each base by the height we divide these results by three add them together and we have the volume of the hole the hexagonal pyramid with which we started the one thing we need to remember is this the volume of any pyramid is equal to 13 the product of its base and height V = 13 H as the number of lateral faces of a pyramid increases notice what seems to be happening to the figure more and more our pyramid begins to look like a cone doesn't it if our figure actually becomes a cone then the question arises how do we find its volume does the same relationship exist between a cone and a cylinder as existed between a prism and a pyramid let's conduct another experiment as we did before with the pyramid and the prism let's compare the volume of a cone with the volume of a cylinder having the same size base and the same linear measurement for height by filling the cone with colored sand and pouring this into the cylinder shaped container we quickly determine how many cones it will take to fill it three cones are necessary to fill the cylinder shaped container therefore we can make the following deduction since the formula for finding the volume of a cylinder is stated as vun r^2 H then the formula for finding the volume of a cone can be stated as V = 13 R 2 h this formula can be applied to a cone of any size and is used many times in our present scientific world for example it's very important to know the exact volume of the nose cone of a rocket a certain capacity is needed to carry instruments which record scientific data during the flight through space and yet size itself is always critical weight must be kept to a minimum to determine capacity the engineers who work on a nose cone use the same formula the student uses in the classroom V = 13 R 2 H we should remember that Pi can be stated as 3.14 or 3 and 17 or 227 to make use of our formula let's find the volume of a nose cone whose diameter is 3T or whose radius is 1 and 1/2 ft or simply 3es of a foot the height of the cone is 7t using our formula we find that the volume of the cone is 13 * 22 7 time 3es * 3times 7B feet or 33 hves or 16 and5 cubic feet a nose cone may be used to help launch another man-made geometric solid a satellite in the shape of a sphere the type of solid whose surface at every point is equidistant from its Center the line from the center to the surface is known as the radius such a line might extend from the center to any point on the surface area a straight line extending from a point on the surface through the center to another point on the surface is a diameter of the sphere a circular line having the same diameter and radius as the sphere is called a great circle to develop a formula for finding the volume of a sphere let's remove one part and divide that part into a number of small pyramid shaped pieces the vertex of each pyramid is of course the center of the sphere the base of each pyramid is a minute portion of the surface even though part of a curved surface the small base areas in each of these pyramids may be thought of as polygons or squares in this case the height of each pyramid or H is the radius of the sphere or R the base of each pyramid designated here as B is actually a part of the surface of our sphere or S the pyramid volume formula could could be applied to each of these small sections and when the sum of the volumes of all the pyramids for the entire sphere were added together we would have an amount equal to the volume of the sphere this then is the formula the volume of the sphere is equal to 13 radius times the surface area of the sphere or more simply V = 13 RS s representing the total surface area we also know that the surface area of a sphere is four times the area of its great circle and can be expressed as 4 pi r 2 simplified the formula for finding the volume of a sphere now reads V = 13 R * 4un r or 43 pi R cubed when men work with geometric solids not only must measurements be taken but calculations must be made carefully and accurately with the proper use of proven formulas in the future find new uses for the shapes themselves but the mathematical laws for finding their volumes will still apply [Music]