Lesson Explainer: Surface Areas of Spheres | Nagwa Lesson Explainer: Surface Areas of Spheres | Nagwa

Lesson Explainer: Surface Areas of Spheres Mathematics • 8th Grade

In this explainer, we will learn how to use the formula for the surface area of a sphere in terms of its radius or diameter to find the sphere’s area, radius, or diameter.

Definition: Surface Area of a Sphere

A sphere is the three-dimensional analog of a circle. It can be defined as a perfectly rounded object that has no edges or vertices.

All points located on the surface of a sphere are at an equal distance from the center. This distance is called the radius, which is usually denoted by π‘Ÿ.

An interesting property of a sphere is that among all three-dimensional shapes with the same volume, it is the one with the minimum surface area. For this reason, spheres arise in a variety of physical systems where surface area is minimized, such as water droplets and soap bubbles!

The surface area of a sphere can be calculated by means of the following formula.

Formula: Surface Area of a Sphere

The surface area, 𝐴, of a sphere of radius π‘Ÿ is given by the formula 𝐴=4πœ‹π‘Ÿ.

Since πœ‹=3.14159… is just a number, this means that as long as we know the radius of a sphere, we can always apply this formula to find its surface area.

Let’s start with a simple example.

Example 1: Finding the Surface Area of a Sphere given Its Radius

Find the surface area of the given sphere to the nearest tenth.

Answer

Recall that the surface area, 𝐴, of a sphere of radius π‘Ÿ is given by the formula 𝐴=4πœ‹π‘Ÿ.

From the diagram, this sphere has radius π‘Ÿ=6, so we can substitute this value into the formula and rearrange to get 𝐴=4Γ—πœ‹Γ—π‘Ÿ=4Γ—πœ‹Γ—6=4Γ—πœ‹Γ—36=4Γ—36Γ—πœ‹=144Γ—πœ‹=452.389….

We were asked to round our answer to the nearest tenth. Remember that the tenths digit is the first digit after the decimal point, which in this case is a 3. The digit following this (the hundredths digit) is an 8, so the answer rounds up to 452.4 to the nearest tenth.

Since the radius of the sphere was given in centimetres, the surface area must be in square centimetres. The surface area of the sphere, rounded to the nearest tenth, is 452.4 cm2.

Next, we will work through an example where we are given the diameter of the sphere, not the radius. Our approach is very similar but with one additional step. Always make sure that you check whether you are given a radius or a diameter in the question.

Example 2: Finding the Surface Area of a Sphere given Its Diameter Using an Approximation of Pi

Find the surface area of a sphere whose diameter is 12.6 cm. Use πœ‹=227.

Answer

Recall that the surface area, 𝐴, of a sphere of radius π‘Ÿ is given by the formula 𝐴=4πœ‹π‘Ÿ.

Here, we are given the diameter of the sphere, 12.6 cm, which is twice its radius. To apply the formula to work out the surface area, we first need to calculate the radius, so we halve the diameter to get π‘Ÿ=12.6Γ·2=6.3. Then, substituting for π‘Ÿ in the formula, we have 𝐴=4Γ—πœ‹Γ—π‘Ÿ=4Γ—πœ‹Γ—(6.3)=4Γ—πœ‹Γ—39.69=4Γ—39.69Γ—πœ‹=158.76Γ—πœ‹.

Note that in the question, we are given the approximation 227 for πœ‹, so substituting this value gives 𝐴=158.76Γ—πœ‹=158.76Γ—227=158.76Γ—227=3492.727=498.96.

As the diameter was given in centimetres, the surface area must be in square centimetres; the surface area of the sphere is 498.96 cm2.

The formula for the surface area of a sphere contains only two variables, 𝐴 and π‘Ÿ. This means that if we are given the surface area of a sphere, then we can always work backward to find its radius. Once we have worked out the radius, if needed, we can double this value to obtain the diameter. The next example shows how to rearrange the formula to solve this type of problem.

Example 3: Finding the Diameter of a Sphere given Its Surface Area

What is the diameter of a sphere whose surface area is 36πœ‹ cm2?

Answer

First, we recall the formula to calculate the surface area, 𝐴, of a sphere of radius π‘Ÿ: 𝐴=4πœ‹π‘Ÿ.

We have been given the surface area, 𝐴=36πœ‹οŒ², so substituting into the formula, we have 36Γ—πœ‹=4Γ—πœ‹Γ—π‘Ÿ.

Conveniently, we have been given the surface area in terms of πœ‹, which allows us to neatly divide both sides of our equation by πœ‹ to get 36=4Γ—π‘Ÿ.

Then, we divide both sides by 4, giving 9=π‘Ÿ.

We can now find the radius of our sphere by taking the square roots of both sides of this equation: √9=βˆšπ‘Ÿ, so 3=π‘Ÿ, which is the same as π‘Ÿ=3. It is worth noting that there are actually two solutions to this equation. Although we could say that π‘Ÿ=Β±3, in this case, π‘Ÿ represents a length and we can therefore ignore the negative solution.

The units of the surface area are square centimetres, so the radius is in centimetres. Doubling the value of the radius, we find that the diameter of the sphere is 2Γ—3=6cm.

Notice that in the above example, we substituted the value for 𝐴, the surface area of the sphere, into the formula 𝐴=4πœ‹π‘ŸοŒ²οŠ¨ and then rearranged to find the value of the radius, π‘Ÿ. An alternative approach to this strategy is to rearrange the formula to make π‘Ÿ the subject and then substitute for 𝐴 directly, as follows.

Starting with the original formula 𝐴=4πœ‹π‘ŸοŒ²οŠ¨ and rewriting the right-hand side to include multiplication signs, we have 𝐴=4Γ—πœ‹Γ—π‘Ÿ.

Dividing both sides by 4 and then by πœ‹, we get 𝐴4πœ‹=π‘Ÿ.

Finally, we take the square roots of both sides: ο„žπ΄4πœ‹=βˆšπ‘Ÿ, so π‘Ÿ=ο„žπ΄4πœ‹.

Again, we can safely ignore the negative solution when defining this relationship, so we have obtained our formula for the radius. If we were to substitute 𝐴=36πœ‹οŒ² directly into this formula, we would get π‘Ÿ=3. As expected, this is the same value for the radius that we worked out in the previous example.

Formula: Radius of a Sphere given Its Surface Area

The radius, π‘Ÿ, of a sphere of surface area 𝐴 is given by the formula π‘Ÿ=ο„žπ΄4πœ‹.

Now, we examine a very important concept when studying spheres: the great circle.

Definition: Great Circle

A great circle is the largest circle that can be drawn on any given sphere. It can be formed on the surface of a sphere by the intersection of a plane that passes through the sphere’s center. Since a great circle shares its center with the parent sphere, it will also share its radius, π‘Ÿ.

A great circle will always cut a sphere into two equal hemispheres, as shown in Figure 1.

Figure 1

It is possible to draw other circles on the surface of a sphere that do not pass through the center of the sphere. These circles will not be great circles and will have a smaller radius than a great circle (and the parent sphere).

Figure 2 shows a great circle of radius π‘ŸοŠ§ and another circle that lies on the surface of the sphere and has a radius π‘ŸοŠ¨. The smaller circle divides the sphere into two unequal parts.

Figure 2

The centers of both circles share a common axis, which, by the definition of a great circle, also passes through the center of the sphere.

Figure 3 shows a top-down view of the sphere shown in Figure 2.

By taking this view, we can clearly see that the radius of a great circle, π‘ŸοŠ§, is the radius of the sphere. We can also see by comparison that the radius of any other circle on the surface of the sphere (represented by π‘ŸοŠ¨ in this example) will be smaller than the radius of the sphere, so π‘Ÿ<π‘ŸοŠ¨οŠ§.

Figure 3

Finally, we know that all circles on a given sphere that are classified as a great circle will have the same radius, π‘ŸοŠ§. We can therefore conclude that all great circles on a sphere will be identical to each other, even if they occupy a different set of points along the surface of the sphere.

Note that we discover another interesting fact by recalling the formula for the area of a circle, 𝐴, in terms of its radius: 𝐴=πœ‹π‘Ÿ.

By comparing this to the formula for the surface area of a sphere, we see that the sphere will have exactly four times the area of a circle with the same radius, which we now know is a great circle; that is, 𝐴=4πœ‹π‘Ÿ=4Γ—πœ‹π‘Ÿ=4𝐴.

Formula: Surface Area of a Sphere given Its Great Circle

The surface area of a sphere, 𝐴, is exactly four times the area of its great circle, 𝐴: 𝐴=4𝐴.

Since a great circle shares some of its properties with its parent sphere, you may be asked to solve questions using the relationship between these two shapes. Let’s look at an example.

Example 4: Finding the Surface Area of a Sphere given Information about Its Great Circle

Find the surface area of a sphere to the nearest tenth if the area of the great circle is 441πœ‹ in2.

Answer

Recall that the surface area, 𝐴, of a sphere of radius π‘Ÿ is given by the formula 𝐴=4πœ‹π‘Ÿ.

Our strategy will be to work out the value of 𝐴 from the area of the great circle given in the question.

Any great circle has the same radius as its parent sphere. Therefore, the area of this great circle, 𝐴, will be given by the formula 𝐴=πœ‹π‘Ÿ.

Using this last formula, we can substitute for πœ‹π‘ŸοŠ¨ in the surface area formula as follows: 𝐴=4πœ‹π‘Ÿ=4Γ—πœ‹π‘Ÿ=4𝐴.

Now that we have expressed the surface area of the sphere in terms of the area of its great circle, we can use the fact that 𝐴=441πœ‹οŒ’ to get 𝐴=4𝐴=4Γ—441πœ‹=4Γ—441Γ—πœ‹=1764Γ—πœ‹=5541.769….

From the question, we need to round our answer to the nearest tenth. The tenths digit is the first digit after the decimal point, which here is a 7. The digit following this (the hundredths digit) is a 6, so our answer must round up to 5β€Žβ€‰β€Ž541.8 to the nearest tenth.

The area of the great circle was given in square inches, so the surface area of the sphere will also be in square inches. We conclude that the surface area of the sphere is 5β€Žβ€‰β€Ž541.8 in2, rounded to the nearest tenth of a square inch.

Next, we have another example where we must use the properties of a great circle to work out the surface area of its parent sphere; this time we are given the circumference.

Example 5: Finding the Surface Area of a Sphere given the Great Circle’s Circumference

Find, to the nearest tenth, the surface area of a sphere whose great circle has a circumference of 140πœ‹ ft.

Answer

Recall the formula for the surface area, 𝐴, of a sphere of radius π‘Ÿ: 𝐴=4πœ‹π‘Ÿ.

Here, we are given the circumference of a great circle. Since we know that the radii of a great circle and its parent sphere are the same, then our first step will be to use this information to calculate the value of π‘Ÿ. We know that the relationship between circumference and radius is circumference=2πœ‹π‘Ÿ.

Now, we substitute the value 140πœ‹ for the circumference to get 140πœ‹=2πœ‹π‘Ÿ.

To solve this equation for π‘Ÿ, we divide both sides by 2πœ‹, so 140πœ‹2πœ‹=π‘Ÿ, which implies π‘Ÿ=70.

We now have a familiar situation, where the surface area of the sphere, 𝐴, can be found using our formula. Substituting for π‘Ÿ, we have 𝐴=4Γ—πœ‹Γ—π‘Ÿ=4Γ—πœ‹Γ—(70)=4Γ—πœ‹Γ—4900=4Γ—4900Γ—πœ‹=19600Γ—πœ‹=61575.216….

The question states that we should provide our answer to an accuracy of the nearest tenth. The tenths digit is the first digit after the decimal point, which here is a 2. The digit following this (the hundredths digit) is a 1, so our answer must round down to 61β€Žβ€‰β€Ž575.2 to the nearest tenth.

The circumference of the great circle was given in feet, so the surface area of the sphere will be in square feet. We conclude that the surface area of the sphere is 61β€Žβ€‰β€Ž575.2 ft2, rounded to the nearest tenth of a square foot.

Remember that a great circle will always cut a sphere into two equal hemispheres. Consequently, we can use information about great circles to help us calculate the surface area of corresponding hemispheres or other fractions of a sphere. Here is an example.

Example 6: Finding the Total Surface Area of a Hemisphere given Its Radius

Find the total surface area of the hemisphere. Round the answer to the nearest tenth.

Answer

The diagram includes a great circle with a radius of 18 cm. As we know that the radius of a great circle is also the radius, π‘Ÿ, of its parent sphere (or hemisphere), then π‘Ÿ=18.

Now, recall the formula for the surface area, 𝐴, of a sphere of radius π‘Ÿ: 𝐴=4πœ‹π‘Ÿ.

Notice that the total surface of a hemisphere is made up of a curved surface and a flat, circular surface. The area of the curved part, which we shall call 𝐴, is half the surface area of the corresponding sphere; that is, 𝐴=12×𝐴=12Γ—4πœ‹π‘Ÿ=2πœ‹π‘Ÿ.

Furthermore, the at surface is just a circle of radius π‘Ÿ, so writing 𝐴 for its area, we have 𝐴=πœ‹π‘Ÿ.

Therefore, the total surface area of the hemisphere, which we can write as 𝐴, must satisfy 𝐴=𝐴+𝐴.

Substituting for 𝐴 and 𝐴 from above, we get 𝐴=2πœ‹π‘Ÿ+πœ‹π‘Ÿ=3πœ‹π‘Ÿ.

Finally, we have π‘Ÿ=18, so making this substitution gives 𝐴=3Γ—πœ‹Γ—π‘Ÿ=3Γ—πœ‹Γ—(18)=3Γ—πœ‹Γ—324=3Γ—324Γ—πœ‹=972Γ—πœ‹=3053.628….

We were asked to give our answer to the nearest tenth. The tenths digit is the first digit after the decimal point, which here is a 6. The digit following this (the hundredths digit) is a 2, so our answer must round down to 3β€Žβ€‰β€Ž053.6 to the nearest tenth.

The radius of the great circle was given in centimetres, so the total surface area of the hemisphere will be in square centimetres.

The total surface area of the hemisphere is 3β€Žβ€‰β€Ž053.6 cm2, rounded to the nearest tenth of a square centimetre.

For our last example, we have a question with a real-world context that is given as a word problem and without a diagram. In cases like this, it is always important to read the question carefully and work out exactly what we are being asked to find.

Example 7: Solving a Word Problem Involving a Hemisphere

A water feature can be modeled as a hemisphere with its base set onto a square patio. If the diameter of the hemisphere is 4 feet and the patio has a side of length 6 feet, what would the visible area of the patio be? Give your answer accurate to two decimal places.

Answer

Recall that a great circle will always cut a sphere into two equal hemispheres. Therefore, the base of the hemisphere in the water feature will be a great circle. Moreover, a great circle and its parent hemisphere must have the same radius, π‘Ÿ. Our strategy will be to use information about the hemisphere to calculate the area of this great circle. We can then subtract it from the area of the whole patio to find the visible area of the patio.

We are told that the patio is square with a side of length 6 feet, so the area of the whole patio is 6Γ—6=36.

We also know that the diameter of the hemisphere is 4 feet, so, to find its radius π‘Ÿ, we divide by 2 to get π‘Ÿ=4Γ·2=2. Thus, its base will be a great circle with a radius of 2.

Moreover, the area of this great circle will be 𝐴=πœ‹Γ—π‘Ÿ=πœ‹Γ—2=πœ‹Γ—4=4πœ‹.

For now, we keep this answer in its exact form because we need to use it in a further calculation.

Our last step is to obtain the visible area of the patio by subtracting the area of the circle from the area of the square, giving 36βˆ’4πœ‹=36βˆ’12.566…=23.433…, which is 23.43, accurate to two decimal places.

The lengths in the question were given in feet, so this area must be in square feet. The visible area of the patio, accurate to two decimal places, is 23.43 ft2.

Let’s finish by recapping some key concepts from this explainer.

Key Points

  • The surface area, 𝐴, of a sphere of radius π‘Ÿ is given by the formula 𝐴=4πœ‹π‘Ÿ.
  • The formula can be rearranged in order to more easily find the radius (or diameter) of a sphere, given its surface area: π‘Ÿ=ο„žπ΄4πœ‹.
  • Always make sure that you check whether you are given the radius or the diameter of the sphere in the question.
  • A great circle is the intersection of a sphere with a plane that passes through the sphere’s center. It cuts the sphere exactly in half, forming two hemispheres.
  • A great circle is the largest circle that can be formed on the surface of its parent sphere, and both shapes share the same radius.
  • The area of a sphere, 𝐴, is exactly four times the area of its great circle, 𝐴.
  • We can use the formula to find the surface area of a hemisphere or other fractions of a sphere, in particular, in real-world questions presented as word problems.

Join Nagwa Classes

Attend live sessions on Nagwa Classes to boost your learning with guidance and advice from an expert teacher!

  • Interactive Sessions
  • Chat & Messaging
  • Realistic Exam Questions

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy