Ontario 09 Academic (MPM1D)
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Volume of spheres
Lesson

Volumes of Spheres

So who cares what the volume of a sphere is. Well...

  • A toy manufacturer needs to figure out how much plastic they need to make super bouncy rubber balls would need to know the volume of the ball. 
  • An astronomer who wants to calculate how much the sun weighs would need to know its volume. 
  • A pharmaceutical company making round pills needs to know the dosage of medicine in each pill, which is found by its volume.
  • A party balloon gas hire company needs to know how much gas fits inside each of their balloons, which is found by its volume.

So apart from having to complete questions at school on volumes of random shapes, volume calculations have a wide variety of applications.  

The Formula

The Volume of a sphere with radius $r$r can be calculated using the following formula:

Volume of Sphere

$\text{Volume of sphere }=\frac{4}{3}\pi r^3$Volume of sphere =43πr3

 

Proof of the formula

Whilst the proof is not typically included as part of the needs of the curriculum, this particular one is a nice introduction to thinking about proofs and abstract proofs. So you don't have to read this if you don't want to but aren't you curious to know where this funny formula came from?

The proof

If four points on the surface of a sphere are joined to the centre of the sphere, then a pyramid of perpendicular height r is formed, as shown in the diagram. Consider the solid sphere to be built with a large number of these solid pyramids that have a very small base which represents a small portion of the surface area of a sphere.

If $A_1$A1, $A_2$A2, $A_3$A3, $A_4$A4, .... , $A_n$An represent the base areas (of all the pyramids) on the surface of a sphere (and these bases completely cover the surface area of the sphere), then,

$\text{Volume of sphere }$Volume of sphere $=$= $\text{Sum of all the volumes of all the pyramids }$Sum of all the volumes of all the pyramids
$V$V  $=$= $\frac{1}{3}A_1r+\frac{1}{3}A_2r+\frac{1}{3}A_3r+\frac{1}{3}A_4r$13A1r+13A2r+13A3r+13A4r   $\text{+ ... +}$+ ... + $\frac{1}{3}A_nr$13Anr
  $=$= $\frac{1}{3}$13  $($(   $A_1+A_2+A_3+A_4$A1+A2+A3+A4  $\text{+ ... +}$+ ... +  $A_n$An  $)$) $r$r 
  $=$= $\frac{1}{3}\left(\text{Surface area of the sphere }\right)r$13(Surface area of the sphere )r
  $=$= $\frac{1}{3}\times4\pi r^2\times r$13×4πr2×r
  $=$= $\frac{4}{3}\pi r^3$43πr3

where $r$r is the radius of the sphere.  

 

Worked Examples

Question 1

Find the volume of the sphere shown.

Round your answer to two decimal places.

Question 2

A sphere has a radius $r$r cm long and a volume of $\frac{343\pi}{3}$343π3 cm3. Find the radius of the sphere.

Round your answer to two decimal places.

Enter each line of working as an equation.

Outcomes

9D.MG2.06

Solve problems involving the surface areas and volumes of prisms, pyramids, cylinders, cones, and spheres, including composite figures

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