# Lesson Video: Volumes of Spheres Mathematics • 8th Grade

In this video, we will learn how to find volumes of spheres and solve problems including real-life situations.

16:33

### Video Transcript

In this video, we will learn how to find volumes of spheres. We’ll first introduce the standard formula for calculating the volume of a sphere and then see how we can apply this to some examples. We’ll also see how we can work backwards from knowing the volume of a sphere to determining its radius or diameter, which will require some skill in rearranging formulae.

First of all, we remember that a sphere is a three-dimensional shape. Its size is completely determined by one measurement, its radius, which is the distance from the center of the sphere to any point on its surface. The formula we need for calculating the volume of any sphere is this. Four-thirds 𝜋𝑟 cubed, where 𝑟 is the radius of the sphere. We must remember that it is only the radius that is being cubed, not the factors of four-thirds and 𝜋.

Although we may not require them in this video, we can also recall some other key formulae relating to spheres and circles. Firstly, the surface area of a sphere is four 𝜋𝑟 squared. And in two dimensions, the area of a circle is simply 𝜋𝑟 squared. Notice that each of these formulae includes a factor of 𝜋 as we’re working with circles or objects related to them. We notice also that, in the formula for area and the formula for surface area, there’s a factor of 𝑟 squared, whereas in the formula for volume, there’s a factor of 𝑟 cubed.

This makes sense when we think about the dimensions of each of these. Area and surface area are measured in square units. So we’re multiplying two dimensions together, 𝑟 by 𝑟, which is 𝑟 squared, whereas volume is measured in cubic units. We’re multiplying three dimensions together, 𝑟 by 𝑟 by 𝑟, which is 𝑟 cubed.

Sometimes in a problem, we may be given the diameter rather than the radius of the sphere. That’s the total distance between two opposite points on the sphere’s surface passing through the center of the sphere. In this instance, we can calculate the radius using the relationship between the two. The diameter is equal to twice the radius, or equivalently the radius is equal to half the diameter.

Now that we’ve identified the key formulae we’re going to need, let’s consider some examples.

Work out the volume of the sphere, giving your answer accurate to two decimal places.

First, we remember the formula we need for calculating the volume of a sphere. It’s four-thirds 𝜋𝑟 cubed, where 𝑟 is the radius of the sphere. We can see from the diagram that the radius of this sphere is 6.3 centimeters. That’s the distance from any point on the sphere’s surface to the center of the sphere. So we can substitute this value of 𝑟 directly into our formula, giving that the volume of this sphere is equal to four-thirds 𝜋 multiplied by 6.3 cubed.

We must remember that it is only the radius that we are cubing, so only the value of 6.3, not the factors of four-thirds and 𝜋. As we’ve been asked to give our answer accurate to two decimal places, it’s reasonable to assume that we can use a calculator. So evaluating this on our calculators gives 1047.394424 continuing.

In order to round our answer to two decimal places, we need to consider the value in the third decimal place. It is a four. And as this is less than five, this tells us that we’re rounding down. So we have a value of 1047.39. The units of volume will be cubic units. And as the units given for the radius were centimeters, the units for the volume will be cubic centimeters. And so we have our answer to this problem. The volume of the sphere, given to two decimal places, is 1047.39 cubic centimeters.

Let’s now consider a problem in which we haven’t been given the radius of the sphere but instead we’ve been given its diameter.

What is the volume of a sphere whose diameter is 30?

We begin by recalling the key formula we need for calculating the volume of a sphere. It’s four-thirds 𝜋𝑟 cubed, where 𝑟 is the radius of the sphere. In this question, we haven’t been given the radius of the sphere. Instead, we’ve been told that its diameter is 30 units. That’s no great problem though because we know the relationship between the radius and diameter of a sphere. The diameter is twice the radius, or equivalently we can say that the radius is half the diameter.

So to find the radius of this sphere, we simply need to halve the measurement we were given for the diameter. The radius is 30 over two, which is 15 units.

Now that we know the radius, we can substitute directly into our formula for the volume of the sphere, giving four-thirds 𝜋 multiplied by 15 cubed. Now let’s consider how we could simplify this answer if we didn’t have access to a calculator.

We can begin by writing 15 cubed as 15 times 15 times 15. And then we know that three is a factor of 15. So we can cancel the three in the denominator with one factor of three in one of the 15s in the numerator, giving four 𝜋 multiplied by five multiplied by 15 multiplied by 15. We know that we can multiply in any order. So reordering the value slightly, this is equivalent to four multiplied by five multiplied by 𝜋 multiplied by 15 multiplied by 15.

Now four multiplied by five we know is equal to 20. And 15 multiplied by 15 we should be able to work out or we should know. It’s one of our square numbers. 15 squared is equal to 225. So our calculation becomes 20 multiplied by 𝜋 multiplied by 225. We can then think of 20 as 10 multiplied by two. So we have 10 multiplied by two multiplied by 𝜋 multiplied by 225. Multiplying two by 225 gives 450, and then multiplying by 10 gives 4,500. So we can simplify our answer to 4,500𝜋.

Now notice that we didn’t use a calculator at any point in this question. And this is called giving our answer as a multiple of 𝜋. Doing so means that we can answer questions about spheres and circles when we don’t have access to a calculator.

We weren’t given any units for the diameter in this question. So the units for our volume will just be general cubic units. We’ve found then that the volume of a sphere whose diameter is 30 is 4,500𝜋 cubic units.

In our next example, we’ll see how we can work backwards from knowing the volume of a sphere to determining its radius or diameter.

Find the radius of a sphere whose volume is nine over two 𝜋 cubic centimeters.

In this problem, we’ve been given the volume of a sphere and asked to work backwards to determine its radius. Let’s begin by recalling the formula we use for calculating the volume of a sphere. It’s this, four-thirds 𝜋𝑟 cubed, where 𝑟 represents the radius of the sphere.

Now as we’ve been given the volume and we know the general formula for working it out, we can set these two values or expressions equal to one another to give an equation. We have four-thirds 𝜋𝑟 cubed equals nine over two 𝜋. And in order to determine the radius of the sphere, we simply need to solve this equation.

First, we notice that there’s a factor of 𝜋 on each side of the equation. So we can cancel this. Or we can think of this as dividing through by 𝜋, to give four-thirds 𝑟 cubed equals nine over two. Next, we need to divide each side of the equation by four-thirds in order to leave 𝑟 cubed on its own on the left-hand side.

We recall that dividing by a fraction is equivalent to multiplying by the reciprocal of that fraction. So to divide by four-thirds, we can multiply each side of the equation by three-quarters. Doing so will eliminate the factor of four-thirds on the left-hand side, leaving just 𝑟 cubed. And on the right-hand side, we have nine over two multiplied by three over four. We multiply the numerators, giving 27, and multiply the denominators, giving eight. So we find that 𝑟 cubed is equal to 27 over eight. To find the value of 𝑟, we need to perform the inverse or opposite of cubing, which is cube rooting. So we find that 𝑟 is equal to the cubed root of 27 over eight.

Now at this point, we remember that, in order to find the cubed root of a fraction, we can find the cubed root of the numerator over the cubed root of the denominator. So we have that 𝑟 equals the cubed root of 27 over the cubed root of eight. And these are both integer values. The cubed root of 27 is three, and the cubed root of eight is two. So we find that the radius of this sphere is three over two or 1.5. And as the units for the volume were cubic centimeters, the units for the radius will be centimeters.

Now of course, we weren’t asked for it in this problem. But if we needed to calculate the diameter of the sphere, we just need to recall that the diameter is twice the radius. So if the radius is three over two centimeters, then the diameter is twice this. The diameter of the sphere is three centimeters. We’ve completed the problem there. The radius of the sphere whose volume is nine over two 𝜋 cubic centimeters is three over two centimeters.

Let’s now consider how we could find the volume of a hemisphere, which we recall is simply half of a sphere. We can therefore adapt the formula for the volume of a sphere to give a formula for finding the volume of a hemisphere. As the volume of a sphere is given by four-thirds 𝜋𝑟 cubed, the volume of the hemisphere with the same radius will be given by a half multiplied by four-thirds 𝜋𝑟 cubed. Of course, we can cross-cancel a factor of two in the denominator of the first fraction with the factor of two in the numerator of the second, to give a simplified fraction of two-thirds. And so we find that the volume of a hemisphere can be calculated using the formula two-thirds 𝜋𝑟 cubed.

Let’s consider one example of this.

A hemisphere has a radius of 15 inches. Work out its volume, giving your answer in terms of 𝜋.

We recall firstly that a hemisphere is simply half a sphere. So the formula for finding the volume of a hemisphere is just half the formula for finding the volume of a sphere. It’s two-thirds 𝜋𝑟 cubed. We’re told that the radius of this hemisphere is 15 inches. So we can substitute this value directly into our formula, giving two-thirds 𝜋 multiplied by 15 cubed.

Now we’re asked to give our answer in terms of 𝜋, which suggests that we don’t have access to a calculator for this problem. So we need to consider how to simplify the calculation without using a calculator. We can first write 15 cubed as 15 multiplied by 15 multiplied by 15. And then as three is a factor of 15, we can cancel the three in the denominator with a factor of three from one of the 15s in the numerator, giving two over one 𝜋 multiplied by five multiplied by 15 multiplied by 15.

We can perform this multiplication in any order. So perhaps the easiest is to think of it as two times five multiplied by 𝜋 multiplied by 15 multiplied by 15. Two times five is of course 10, and 15 multiplied by 15 we should know — it’s one of our square numbers — is 225. So we have 10 multiplied by 𝜋 multiplied by 225. 10 multiplied by 225 is 2,250. So our value in terms of 𝜋 is 2,250𝜋.

The units given for the radius were inches. And so the units given for the volume will be cubic inches. And so we have our answer to the problem. It’s 2,250𝜋 cubic inches.

Now we could’ve answered this problem a slightly different way. We could’ve simply calculated the volume of the full sphere using the formula four-thirds 𝜋𝑟 cubed and then divided our answer by two at the end to give the volume of a hemisphere. But of course, it would give the same result of 2,250𝜋 cubic inches.

Let’s now consider one final problem in which we introduce the definition of the great circle of a sphere.

Find, to the nearest tenth, the volume of a sphere given that the circumference of its great circle is 90𝜋 inches.

So first of all, what do we mean by the great circle of a sphere? Well, formally, it is the intersection of the sphere and any plane — that’s a two-dimensional slice — which passes through the center of the sphere. On our diagram, this is one such great circle. But in fact, there are infinitely many depending on the angle of the plane we draw.

Now we know that, in order to find the volume of any sphere, we use the formula four-thirds 𝜋𝑟 cubed. So in order to answer this question, we need to calculate the radius of our sphere. We can do this using the information given about the circumference of the great circle because we know that the circumference of any circle is found using the formula 𝜋𝑑 or two 𝜋𝑟.

We can therefore form an equation using the version of the circumference formula that involves 𝑟. Two 𝜋𝑟 is equal to 90𝜋. And we can solve this equation in order to determine the radius of the sphere. First, we cancel a factor of 𝜋 from each side, giving two 𝑟 equals 90. And then we divide by two to find that 𝑟 is equal to 45. The units for this will be inches. Finally, we can substitute this value for the radius into our formula for the volume of a sphere, giving four-thirds multiplied by 𝜋 multiplied by 45 cubed.

As we’re asked to give our answer to the nearest tenth, it’s reasonable to assume we can use a calculator to help with this question. So evaluating on a calculator gives 121,500𝜋. Or as a decimal, this is equivalent to 381,703.5074. As we’re rounding to the nearest tenth, our deciding digit is in the hundredths column. That’s a zero. So we’re rounding down.

So we find that the volume of the sphere whose great circle has a circumference of 90𝜋 inches, to the nearest tenth, is 381,703.5 cubic inches.

Let’s now summarize some of the key points that we’ve seen in this video. Firstly, the volume of a sphere whose radius is 𝑟 units is four-thirds 𝜋𝑟 cubed. And we must remember it is only the radius that is cubed. If instead of the radius we were given the diameter of a sphere, we need to halve it before substituting into our formula. So we must make sure we check carefully whether it’s the radius or diameter we’ve been given in each problem.

We also saw in the context of an example that we can work backwards from knowing the volume of a sphere to calculating its radius or diameter. And to do this, we need to form and then solve an equation. We also saw that the volume of a hemisphere is simply half the volume of the sphere with the same radius and can be found using the formula two-thirds 𝜋𝑟 cubed. We also saw that the great circle of a sphere is the intersection of the sphere with any plane that passes through the sphere’s center. And in fact, it divides the sphere up into two hemispheres.

Finally, we saw through our examples that we can give our answers as appropriately rounded decimals if we have access to a calculator. Or if we don’t have access to a calculator or simply if an exact answer is required, then we can give our answers as multiples of 𝜋.