course Phy 201 002. Volumes*********************************************
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Given Solution: `aIf we orient this object so that its 3 cm dimension is its 'height', then it will be 'resting' on a rectangular base whose dimension are 5 cm by 7 cm. This base can be divided into 5 rows each consisting of 7 squares, each 1 meter by 1 meter. There will therefore be 5 * 7 = 35 such squares, showing us that the area of the base is 35 m^2. Above each of these base squares the object rises to a distance of 3 meters, forming a small rectangular tower. Each such tower can be divided into 3 cubical blocks, each having dimension 1 meter by 1 meter by 1 meter. The volume of each 1-meter cube is 1 m * 1 m * 1 m = 1 m^3, also expressed as 1 cubic meter. So each small 'tower' has volume 3 m^3. The object can be divided into 35 such 'towers'. So the total volume is 35 * 3 m^3 = 105 m^3. This construction shows us why the volume of a rectangular solid is equal to the area of the base (in this example the 35 m^2 of the base) and the altitude (in this case 3 meters). The volume of any rectangular solid is therefore V = A * h, where A is the area of the base and h the altitude. This is sometimes expressed as V = L * W * h, where L and W are the length and width of the base. However the relationship V = A * h applies to a much broader class of objects than just rectangular solids, and V = A * h is a more powerful idea than V = L * W * h. Remember both, but remember also that V = A * h is the more important. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique Rating: OK ********************************************* Question: `q002. What is the volume of a rectangular solid whose base area is 48 square meters and whose altitude is 2 meters? ********************************************* Your solution: V = l * w * h = A * h V = 48 * 2 =96 m^3 Confidence Assessment: 3
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Given Solution: `aUsing the idea that V = A * h we find that the volume of this solid is V = A * h = 48 m^2 * 2 m = 96 m^3. Note that m * m^2 means m * (m * m) = m * m * m = m^2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique Rating: OK ********************************************* Question: `q003. What is the volume of a uniform cylinder whose base area is 20 square meters and whose altitude is 40 meters? ********************************************* Your solution: V = A * h =20 * 40 =800 m^3 Confidence Assessment: 3
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Given Solution: `aV = A * h applies to uniform cylinders as well as to rectangular solids. We are given the altitude h and the base area A so we conclude that V = A * h = 20 m^2 * 40 m = 800 m^3. The relationship V = A * h applies to any solid object whose cross-sectional area A is constant. This is the case for uniform cylinders and uniform prisms. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique Rating: OK ********************************************* Question: `q004. What is the volume of a uniform cylinder whose base has radius 5 cm and whose altitude is 30 cm? ********************************************* Your solution: Area of circle = π * r^2 =π(5)62 =25π V = 25π(30) = 750π cm^3 = 2356.19 cm^3 Confidence Assessment: 3
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Given Solution: `aThe cylinder is uniform, which means that its cross-sectional area is constant. So the relationship V = A * h applies. The cross-sectional area A is the area of a circle of radius 5 cm, so we see that A = pi r^2 = pi ( 5 cm)^2 = 25 pi cm^2. Since the altitude is 30 cm the volume is therefore V = A * h = 25 pi cm^2 * 30 cm = 750 pi cm^3. Note that the common formula for the volume of a uniform cylinder is V = pi r^2 h. However this is just an instance of the formula V = A * h, since the cross-sectional area A of the uniform cylinder is pi r^2. Rather than having to carry around the formula V = pi r^2 h, it's more efficient to remember V = A * h and to apply the well-known formula A = pi r^2 for the area of a circle. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique Rating: OK ********************************************* Question: `q005. Estimate the dimensions of a metal can containing food. What is its volume, as indicated by your estimates? ********************************************* Your solution: Diameter = approx 4 inches Altitude = approx 7 inches Area of a circle = π * 2^2 = 4π V = 4π(7) = 28π inches^3 = 87.96 inches^3 Confidence Assessment: 3
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Given Solution: `aPeople will commonly estimate the dimensions of a can of food in centimeters or in inches, though other units of measure are possible (e.g., millimeters, feet, meters, miles, km). Different cans have different dimensions, and your estimate will depend a lot on what can you are using. A typical can might have a circular cross-section with diameter 3 inches and altitude 5 inches. This can would have volume V = A * h, where A is the area of the cross-section. The diameter of the cross-section is 3 inches so its radius will be 3/2 in.. The cross-sectional area is therefore A = pi r^2 = pi * (3/2 in)^2 = 9 pi / 4 in^2 and its volume is V = A * h = (9 pi / 4) in^2 * 5 in = 45 pi / 4 in^3. Approximating, this comes out to around 35 in^3. Another can around the same size might have diameter 8 cm and height 14 cm, giving it cross-sectional area A = pi ( 4 cm)^2 = 16 pi cm^2 and volume V = A * h = 16 pi cm^2 * 14 cm = 224 pi cm^2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique Rating: OK ********************************************* Question: `q006. What is the volume of a pyramid whose base area is 50 square cm and whose altitude is 60 cm? ********************************************* Your solution: V = 1/3*A*h = 1/3(50)(60) = 1000 cm^3 Confidence Assessment: 3
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Given Solution: `aWe can't use the V = A * h idea for a pyramid because the thing doesn't have a constant cross-sectional area--from base to apex the cross-sections get smaller and smaller. It turns out that there is a way to cut up and reassemble a pyramid to show that its volume is exactly 1/3 that of a rectangular solid with base area A and altitude h. Think of putting the pyramid in a box having the same altitude as the pyramid, with the base of the pyramid just covering the bottom of the box. The apex (the point) of the pyramid will just touch the top of the box. The pyramid occupies exactly 1/3 the volume of that box. So the volume of the pyramid is V = 1/3 * A * h. The base area A is 30 cm^2 and the altitude is 60 cm so we have V = 1/3 * 50 cm^2 * 60 cm = 1000 cm^3. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique Rating: OK ********************************************* Question: `q007. What is the volume of a cone whose base area is 20 square meters and whose altitude is 9 meters? ********************************************* Your solution: V = 1/3 * A * h =1/3(20)(9) = 60m^3 Confidence Assessment: 3
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Given Solution: `aJust as the volume of a pyramid is 1/3 the volume of the 'box' that contains it, the volume of a cone is 1/3 the volume of the cylinder that contains it. Specifically, the cylinder that contains the cone has the base of the cone as its base and matches the altitude of the cone. So the volume of the cone is 1/3 A * h, where A is the area of the base and h is the altitude of the cone. In this case the base area and altitude are given, so the volume of the cone is V = 1/3 A * h = 1/3 * 20 m^2 * 9 m = 60 m^3. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique Rating: OK ********************************************* Question: `q008. What is a volume of a sphere whose radius is 4 meters? ********************************************* Your solution: V = (4/3)π * r^3 =(4/3)π * (4)^3 =(256/3)π m^3 = 268.08m^3 Confidence Assessment: 3
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Given Solution: `aThe volume of a sphere is V = 4/3 pi r^3, where r is the radius of the sphere. In this case r = 4 m so V = 4/3 pi * (4 m)^3 = 4/3 pi * 4^3 m^3 = 256/3 pi m^3. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique Rating: OK ********************************************* Question: `q009. What is the volume of a planet whose diameter is 14,000 km? ********************************************* Your solution: Radius = 7,000 km V = (4/3)π * (7000)^3 =(4.57 * 10^11)π m^3 Confidence Assessment: 3
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Given Solution: `aThe planet is presumably a sphere, so to the extent that this is so the volume of this planet is V = 4/3 pi r^3, where r is the radius of the planet. The diameter of the planet is 14,000 km so the radius is half this, or 7,000 km. It follows that the volume of the planet is V = 4/3 pi r^3 = 4/3 pi * (7,000 km)^3 = 4/3 pi * 343,000,000,000 km^3 = 1,372,000,000,000 / 3 * pi km^3. This result can be approximated to an appropriate number of significant figures. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique Rating: OK ********************************************* Question: `q010. Summary Question 1: What basic principle do we apply to find the volume of a uniform cylinder of known dimensions? ********************************************* Your solution: Volume of a uniform cylinder is equal to the area of a cross-section multiplied by the height. Confidence Assessment: 3
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Given Solution: `aThe principle is that when the cross-section of an object is constant, its volume is V = A * h, where A is the cross-sectional area and h the altitude. Altitude is measure perpendicular to the cross-section. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique Rating: OK ********************************************* Question: `q011. Summary Question 2: What basic principle do we apply to find the volume of a pyramid or a cone? ********************************************* Your solution: The basic principle that is applied to the volume of a pyramid or cone is that it’s 1/3 the volume of a rectangle. Therefore, the volume is equal to 1/3 multiplied by the height multiplied by the area of the base. Confidence Assessment: 3
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Given Solution: `aThe volumes of these solids are each 1/3 the volume of the enclosing figure. Each volume can be expressed as V = 1/3 A * h, where A is the area of the base and h the altitude as measured perpendicular to the base. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique Rating: OK ********************************************* Question: `q012. Summary Question 3: What is the formula for the volume of a sphere? ********************************************* Your solution: V of a sphere = (4/3)π * r^3 Confidence Assessment: 3
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Given Solution: `aThe volume of a sphere is V = 4/3 pi r^3, where r is the radius of the sphere. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique Rating: OK ********************************************* Question: `q013. Explain how you have organized your knowledge of the principles illustrated by the exercises in this assignment. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Again, I feel like this exercise helped me to bring back the volume formulas that I may have forgotten or that may have been a bit rusty. It was good to work the problems and then summarize them again in the end. This helps to remember the formulas. Self-critique Rating: OK