002. Volumes 

Question: `q001. There are 14 questions in this document.

What is the volume of a rectangular solid whose dimensions are exactly 3 meters by 5 meters by 7 meters?

Your solution:

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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.

STUDENT QUESTION

I guess I am confused at what the length and the width are???? I drew a rectangle I made the top length 5
and the bottom lenghth 7 then the side 3. So the 7 and the 5 are both width and the 3 is the height??????

INSTRUCTOR RESPONSE

You can orient this object in any way you choose. The given solution orients it so that the base is 5 cm by 7 cm. The area of the base is then 35 cm^2. In this case the third dimension, 3 cm, is the height and we multiply the area of the base by the height to get 105 cm^3.

Had we oriented the object so that it rests on the 3 cm by 5 cm rectangle, the area of the base would be 15 cm^2. The height would be the remaining dimension, 7 cm. Multiplying the base by the height we would be 15 cm^2 * 7 cm = 105 cm^3.

We could also orient the object so its base is 3 cm by 7 cm, with area 21 cm^2. Multiplying by the 5 cm height we would again conclude that the volume is 105 cm^3.

All these results can be visualized in terms of 1-cm squares and 1-cm cubes, as explained in the given solution.

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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:

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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^3.

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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:

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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.

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Question: `q004. What is the volume of a uniform cylinder whose base has radius 5 cm and whose altitude is 30 cm?

Your solution:

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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.

STUDENT QUESTION

why do we not calculate the pi times the radius and then the height or calculate the pi after the height

why do we just leave the pi in the answer?

INSTRUCTOR RESPONSE

pi cannot be written exactly in decimal form; it's an irrational number and any decimal representation is going to have round-off error.

750 pi cm^3 is the exact volume of a cylinder with radius 5 cm and altitude 30 cm.

750 pi is approximately 2356.  However 2356 has two drawbacks:

• 2356 is a 4-significant-figure approximation of 750 pi.  It's not exact.  This might or might not be a disadvantage, but we're better off expressing the result as a multiple of pi, which we can then calculate to any desired degree of precision, than in using 2356, which already contains a roundoff error.

• It's hard to look at 2356 and see how it's related to 5 and 30.  You probably can't calculate that in your head.  However it's not difficult to see that 30 * 5^2 is 30 * 25 or 750.

When in doubt, we use the exact expression rather than the approximation.  It's fine to give an answer like the following:

The volume is 750 pi cm^3, which is approximately 2356 cm^3.

STUDENT QUESTION

I should have stated that my answer was an approximate. ???? When using pi, should I calculate this out or just leave pi in the solution?

INSTRUCTOR RESPONSE

I would say to do both when in doubt.

If the given dimensions are known to be approximate, and when the numbers aren't simple in the first place, it's appropriate to just multiply everything out and use an appropriate number of significant figures.

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Question: `q005. Estimate the dimensions of a metal can containing food. What is its volume, as indicated by your estimates?

Your solution:

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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^3.

STUDENT QUESTION

Should my in^3 come after the total solution even though it is associated with the 9? As in your example the in^3 is
associated with 224 but you have it at the end of the solution.

INSTRUCTOR RESPONSE

I wouldn't be picky at this point of the course, but the generally used order has the numbers first and the units last. 

This is what most readers will expect.  It's a lot like using good grammar, which makes everything easier to understand.

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Question: `q006. What is the volume of a pyramid whose base area is 50 square cm and whose altitude is 60 cm?

Your solution:

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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 50 cm^2 and the altitude is 60 cm so we have

V = 1/3 * 50 cm^2 * 60 cm = 1000 cm^3.

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Question: `q007. What is the volume of a cone whose base area is 20 square meters and whose altitude is 9 meters?

Your solution:

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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.

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Question: `q008. What is a volume of a sphere whose radius is 4 meters?

Your solution:

Confidence Assessment:

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.

STUDENT QUESTION:

How does a formula come up with multiplying by pi? I understand how to work a formula, but don’t know how to
calculate the formula. Does that make sense?

INSTRUCTOR RESPONSE:  It makes perfect sense to ask that question.

However the answer is beyond the scope of your course. 

(one answer, which will not make sense to anyone until at least the midway point of their third semester of a challenging calculus sequence, is that the volume of a sphere of radius R is the integral of rho^2 sin (phi) cos(theta) from rho = 0 to R, phi from 0 to pi and theta from 0 to 2 pi; also the surface area of a sphere of radius R is double the double integral of r / secant(theta), integrated in polar coordinates from r = 0 to R and theta from 0 to 2 pi) .

(there is another way of figuring this out using solid geometry, a topic with which few students are familiar).

In other words, at this point your best recourse is to just learn the formulas.

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Question: `q009. What is the volume of a planet whose diameter is 14,000 km?

Your solution:

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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.

STUDENT QUESTION

How did we go from 343,000,000,000 to 1,372,000,000,000?

INSTRUCTOR RESPONSE

We go from 4/3 pi * 343,000,000,000 to 1,372,000,000,000 / 3 * pi by multiplying 343 000 000 000 by 4. Like a lot of thing, this is fairly obvious once you see it, hard to see until you do.

Let me know if after thinking about it for a few minutes, then if necessary giving it a rest for awhile (say, a day) and coming back to it, you don't see it.

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Question: `q010. Summary Question 1: What basic principle do we apply to find the volume of a uniform cylinder of known dimensions?

Your solution:

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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.

STUDENT QUESTION

What does it mean “when the cross-section of an object is constant”? When would it not be
constant?

INSTRUCTOR RESPONSE  

For example the cross-sectional area of a cone, which tapers, is not constant; nor is the cross-sectional area of a sphere.

STUDENT QUESTION

And why is altitude measured perpendicular to the cross-section?

INSTRUCTOR RESPONSE

This is for essentially the same reason the altitude of a parallelogram is measured perpendicular to its base.

If you imagine nailing four sticks together to make a rectangle, then imagine partially 'collapsing' the rectangle into a parallelogram, you will see that the altitude of the resulting parallelogram is less than that of the original rectangle, and its area is correspondingly less.

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Question: `q011. Summary Question 2: What basic principle do we apply to find the volume of a pyramid or a cone?

Your solution:

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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.

STUDENT QUESTION

I thought I had the right idea but I got lost. I’m not sure how to handle the square roots,
even after reading the solution, I am confused about this one.

INSTRUCTOR RESPONSE

Think of a simple example, the equation x^2 = 25.

It should be clear that x = 5 is a solution to this equation, as is x = -5.

Now 5 is the square root of 25, since 25 is the square of 5. In notation, the same sentence would read

5 = sqrt(25) since 25 = 5^2.

So the solutions to this equation are x = sqrt(25) and x = -sqrt(25). We often write that as x = +- sqrt(25), where the '+-' means 'plus or minus'.

More generally, if c is any positive number, the equation x^2 = c has solutions x = +- sqrt(c).

Now sometimes only one of the two solutions makes sense.

In the present problem A radius is a distance, and a distance can't be negative. So after finding the two solutions, we discard the negative solution. However we always find both solutions before discarding everything, in order to make sure we don't throw out something important

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Question: `q012. Summary Question 3: What is the formula for the volume of a sphere?

Your solution:

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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.

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Question: `q013. Explain how you have organized your knowledge of the principles illustrated by the exercises in this assignment.

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Question: `q014. A cone has an altitude of 14 centimeters, above a circular base whose radius is 5 centimeters.  What is the volume of the cone?

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