 Lesson Video: Relating Volumes and Surface Areas | Nagwa Lesson Video: Relating Volumes and Surface Areas | Nagwa

# Lesson Video: Relating Volumes and Surface Areas Mathematics • 8th Grade

In this video, we will learn how to relate volumes of different shapes, surface areas of different shapes, and relate a shape’s volume and surface area.

17:10

### Video Transcript

In this video, we will learn how to relate volumes of different shapes, surface areas of different shapes, and a shape’s volume and surface area. Since we are interested in relating the volumes and surface areas of shapes, we will begin by recalling the formulas that exist for calculating these quantities.

We will begin by considering the volume and surface area of a sphere. If we let 𝑟 be the radius of a sphere, then its volume 𝑉 is given by 𝑉 is equal to four-thirds 𝜋𝑟 cubed. And its surface area 𝑆 is given by four 𝜋𝑟 squared. We note that both formulas have only one variable, the radius 𝑟. This means that if we are given the radius, we can calculate both the volume and surface area. Alternatively, if we are given the surface area 𝑆, we can rearrange this formula to make 𝑟 the subject. Dividing through by four 𝜋 and then square rooting both sides, we have 𝑟 is equal to the square root of 𝑆 over four 𝜋. We could then substitute this value of the radius into our first formula in order to calculate the volume. This process also applies the other way round. If we are given the volume, we can find the radius and by extension the surface area.

We will now consider an example of this type.

Given that the volume of a sphere is 562.5𝜋 cubic centimeters, find its surface area in terms of 𝜋.

We begin by recalling that the volume 𝑉 of a sphere can be calculated using the formula four-thirds 𝜋𝑟 cubed, where 𝑟 is the sphere’s radius. We can also calculate the surface area 𝑆 of a sphere using the formula 𝑆 is equal to four 𝜋𝑟 squared. In this question, we are given the volume in cubic centimeters. And we can use this to calculate the radius of the sphere and then its surface area.

Substituting 𝑉 is equal to 562.5𝜋, we have the following equation. We can solve this for 𝑟 by firstly dividing through by 𝜋. We can then divide both sides by four-thirds, which is the same as multiplying both sides by three-quarters. This gives us 421.875 is equal to 𝑟 cubed. Finally, we can cube root both sides of this equation to calculate the value of 𝑟. This is equal to 7.5. The radius of the sphere is therefore equal to 7.5 centimeters.

We can now substitute this value into our formula for the surface area. We have 𝑆 is equal to four 𝜋 multiplied by 7.5 squared. Squaring 7.5 gives us 56.25. And multiplying this by four 𝜋 gives us 𝑆 is equal to 225𝜋. Since the radius was measured in centimeters, we can conclude that the surface area of the sphere is 225𝜋 square centimeters.

In our next example, we will consider a cylinder. So we will first recall the formulas for the volume and surface area of a cylinder. The cylinder drawn has a base radius 𝑟 and a height ℎ. We know that its volume 𝑉 is equal to 𝜋𝑟 squared multiplied by ℎ. The total surface area 𝑆 is given by a more complicated formula. This is equal to two 𝜋𝑟ℎ plus two 𝜋𝑟 squared. This comes from the fact that the curved surface of a cylinder is a rectangle and the two ends are circles. We know that the area of a circle is equal to 𝜋𝑟 squared, and we have two of these. We know that the area of a rectangle is equal to its length multiplied by its width. And in this case, the length of the rectangle is equal to the circumference of the circle at either end.

It is clear from the two formulae that if we know the radius and height of the cylinder, we can calculate its volume and surface area. Additionally, if we are given one of the radius or height and one of the volume or surface area, it is possible to find the other two measurements. We do this by rearranging the formulas to calculate the missing variables. In the example that follows, we are given the curved surface area and the diameter of a cylinder and are asked to find its volume.

A cylinder has a curved surface area of 656 square centimeters and a diameter of 10.2 centimeters. Find its volume, giving your answer to the nearest cubic centimeter.

In this question, we will begin by picking out the key information from the question. Firstly, we are told that the curved surface area of a cylinder is equal to 656 square centimeters. We recall that this curved surface is equal to the area of a rectangle with base equal to two 𝜋𝑟, the circumference of the circle of the cylinder, and height equal to ℎ. The curved surface area is therefore equal to two 𝜋𝑟ℎ. And as such, we can set this equal to 656.

We are also told that the diameter of the cylinder is 10.2 centimeters. As the radius is half of this, this is equal to 5.1 centimeters. Substituting this value into our equation, we have two 𝜋 multiplied by 5.1 multiplied by ℎ is equal to 656. The left-hand side simplifies to 10.2𝜋 multiplied by ℎ. And dividing through by 10.2𝜋, we have the following expression for ℎ. Dividing 656 by 10.2, we see that ℎ is equal to 3280 over 51𝜋.

We now have an expression for the height of the cylinder in centimeters. The question asks us to calculate the volume of the cylinder. In order to do this, we recall that the volume of any cylinder is equal to 𝜋𝑟 squared ℎ. Letting the volume of this cylinder be 𝑉, we have 𝑉 is equal to 𝜋 multiplied by 5.1 squared multiplied by 3280 over 51𝜋. We can cancel a factor of 𝜋 from the numerator and denominator. And typing the rest of the right-hand side into our calculator, we have 𝑉 is equal to 1672.8.

We are asked to give our answer to the nearest cubic centimeter. And this means that a cylinder with a curved surface area of 656 square centimeters and a diameter of 10.2 centimeters has a volume of 1673 cubic centimeters to the nearest cubic centimeter.

Before looking at our next example, we will recall the formulas we use to calculate the volume and surface area of cuboids and cubes. If a cuboid has length 𝑙, width 𝑤, and height ℎ as shown, then its volume 𝑉 is given by 𝑉 is equal to 𝑙 multiplied by 𝑤 multiplied by ℎ. The surface area 𝑆 of the cuboid is given by 𝑆 is equal to two multiplied by 𝑙𝑤 plus 𝑤ℎ plus ℎ𝑙. If we have a cube on the other hand, where the length, width, and height are equal, the volume 𝑉 is equal to 𝑙 cubed and the surface area 𝑆 is equal to six 𝑙 squared. In our next example, we will need to relate the volume of two different shapes.

If a sphere is inscribed in a cube of volume eight cubic centimeters, what is the volume of the sphere?

The key to this question is being able to relate the dimensions of the cube to the sphere. Since the sphere is inscribed in the cube, this means that the sphere is touching each face of the cube without any gaps. As such, the diameter 𝑑 of the sphere is equal to the length 𝑙 of the cube. Alternatively, the radius 𝑟 is half of the length of the cube.

Next, we recall that the volume of any cube is equal to its side length cubed. And in this question, we’re told that this volume is eight cubic centimeters. This means that 𝑙 cubed is equal to eight. We can then cube root both sides such that 𝑙 is equal to two. The side length of the cube is therefore equal to two centimeters.

We have already established that the radius of the sphere is half of this. And this is therefore equal to one centimeter. We can calculate the volume of any sphere when we know its radius. The volume is equal to four-thirds 𝜋𝑟 cubed. If we let the volume of our sphere be 𝑉, we have 𝑉 is equal to four-thirds 𝜋 multiplied by one cubed. And this is simply equal to four-thirds 𝜋. We can therefore conclude that if the volume of the cube is eight cubic centimeters and the sphere is inscribed in the cube, then its volume is four-thirds 𝜋 cubic centimeters.

We will now consider one final example where we have multiple spheres combined together and we have to consider their relation to a cylinder.

Three spheres are inscribed in a cylinder, as shown in the figure. If the volume of a sphere is 36𝜋 cubic centimeters, find the total surface area of the cylinder.

In this question, we need to consider how a single sphere relates to the dimensions of the cylinder that contains the three spheres. To do this, we make use of the fact that the spheres are inscribed, which means they fit exactly into the cylinder with no gaps in the top, bottom, or sides. In particular, it means that the radii of the spheres in the cylinder must be the same. It also means that the height of the cylinder must be equal to three times the diameter of a single sphere, or six times the radius of one. If we let 𝑟 be the radius, then the height of the cylinder is six 𝑟.

In order to find the total surface area of the cylinder, we firstly need to find the value of 𝑟, as this will give us the radius and height of the cylinder. We recall that the volume of a sphere is equal to four-thirds 𝜋𝑟 cubed. And in this question, we are told that the volume is equal to 36𝜋. Setting these equal to one another, we have four-thirds 𝜋𝑟 cubed is equal to 36𝜋.

We can solve this equation for 𝑟 by firstly dividing through by 𝜋. We then divide both sides of this equation by four-thirds, which is the same as multiplying by three-quarters, giving us 𝑟 cubed is equal to 27. Cube rooting both sides of this equation, we have 𝑟 is equal to three. The radius of each of the spheres and, hence, the cylinder is equal to three centimeters. Multiplying this by six, we see that the height of the cylinder is 18 centimeters.

Next, we recall that the total surface area of a cylinder is equal to two 𝜋𝑟ℎ plus two 𝜋𝑟 squared. This is made up of the curved surface, which is a rectangle, and the two circles at either end. Letting this total surface area be 𝑆 and substituting in our values of 𝑟 and ℎ, we have 𝑆 is equal to two 𝜋 multiplied by three multiplied by 18 plus two 𝜋 multiplied by three squared. This is equal to 108𝜋 plus 18𝜋, which simplifies to 126𝜋. The total surface area of the cylinder in the figure is 126𝜋 square centimeters.

We will now finish this video by summarizing the key points. In this video, we saw that for a sphere, if we have any one of the volume 𝑉, radius 𝑟, or surface area 𝑆, we can find the other two measurements using the formulae 𝑉 is equal to four-thirds 𝜋𝑟 cubed and 𝑆 is equal to four 𝜋𝑟 squared. When dealing with a cylinder, if we have the radius 𝑟 and the height ℎ, then we can find the volume 𝑉 and surface area 𝑆 using the formulae 𝑉 is equal to 𝜋𝑟 squared ℎ and 𝑆 is equal to two 𝜋𝑟ℎ plus two 𝜋𝑟 squared.

Furthermore, if we are given either the volume or surface area and either the radius or height, we can calculate the other two measurements by rearrangement. For a cuboid or cube, if we have the length 𝑙, width 𝑤, and height ℎ, then we can find the volume 𝑉 and surface area 𝑆 using the formulae 𝑉 is equal to 𝑙𝑤ℎ and 𝑆 is equal to two multiplied by 𝑙𝑤 plus 𝑤ℎ plus ℎ𝑙. Additionally, if we are given two of the dimensions of a cuboid and either the volume or surface area, we can once again find the remaining measurements by rearrangement. Finally, in the last example, we saw that in problems where we are given different shapes, we can relate them to each other by using the information given to us in the problem and rearrangement of the above formulae.