16 Volume of a Sphere
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To the left we have a picture of our beautiful planet, Earth. Although the picture itself is two dimensional, we know that Earth is three dimensional. This implies that Earth has a volume. What is also true is that Earth is an example of a sphere.
So far we’ve calculated the volume of cubes, rectangular tanks, and cylinders. Using that information, how do you think we would calculate the volume of a sphere? What variables do you think you would use? Take a few moments to think about it before we go through the explanation below.
Just in case you were wondering, Earth has a volume of…
[latex]\Large 1,083,206,916,846 \text{ cubic kilometres}[/latex]
There are two things to note here:
- The answer is once again in cubic units.
- That’s a lot of volume.
We should start this explanation off by revisiting the formula for a circle. Remember that a circle had a radius, diameter, and circumference. Also remember that to find the formula for a circle, we could have used one of two formulas.
[latex]\Large \begin{array}{ll} \text{Formula one:} & \text{ area} = {\text{d}}^{2} \times 0.7854 \times \text{h} \\ \text{Formula two:} & \text{ area} = \pi {\text{r}}^{2} \end{array}[/latex]
When dealing with the formula for a sphere, we can use formula two as a starting point. The formula for a sphere has a similar feel to it but with a little twist. The similarities include having both pi and radius in the formula, but that is where the similarities end.
Here is the formula:
[latex]\Large \text{volume} = \dfrac{4}{3} \pi {\text{r}}^{3}[/latex]
The question becomes, “Where does the 4/3 come from?” Well the explanation for that is actually quite a long one, and I’ll leave that for another day. Quickly stated, it comes from the fact that if you took two cones with similar measurements to the sphere, it would end up that the volume of those two cones would equal the volume of the sphere. Using a bit of mathematical wizardry the 4/3 ends up being derived from this fact.
The pi in the formula is the constant that we use when finding the circumference of a circle, and the radius, as you might remember, is half the length of the diameter. The last thing is that the radius is cubed. This relates to the fact that in the end we are solving for volume, which has three dimensions.
Now I didn’t expect you to get the formula as it’s quite a bizarre one, but understanding where it comes from helps to conceptualize the formula. As we talked about previously, the purpose of this exercise is to rely less on memorization and more on an understanding of how stuff is derived.
Example
Find the volume of a sphere with a radius of 7 inches.
Step 1: As usual write down the formula.
[latex]\Large \text{volume} = \dfrac{4}{3} \pi {\text{r}}^{3}[/latex]
Step 2: Plug in the variables and solve for volume
[latex]\Large \begin{array}{c}\text{volume} = \dfrac{4}{3} \pi {\text{r}}^{3} \\ \text{vo