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The volume and surface of the sphere
We have reached an important solid in our study of three-dimensional objects, which holds a unique position among solids in more than one sense. This object is nothing else than the sphere, commonly considered as the most perfectly shaped solid.
This object has a distinguished role in nature as well, as astronomical objects look like spheres and and this is the form of a drop of water falling from an icicle.
In everyday use, the solid itself is called sphere just as the surface bounding it. Usually it is easy to decide upon hearing a sentence which one of the two different concepts is needed.
The sphere is the set of points with equal distance from a fixed point, the centre of the sphere. This object is what we get by rotating a circle around one of its diameters.
The ball is the set of points which are no further away from a given point, the centre of the ball, than a given distance \latex{ r }.
Any planar section of a sphere is a circle. If this planar section contains the centre of the sphere, then it is called a great circle of the sphere.
This time we are talking about an object which is not bounded by plane figures, and it cannot be laid into the plane, creating much harder problems during the determination of the volume and the surface area than in the previous lessons.
The volume of the ball
The previously formulated principle of Cavalieri is a great asset in determining the volume of the ball. Without it, we could not manage but with the help of advanced mathematical tools.
First, divide the ball of radius \latex{ r } into two half-balls, and determine the volume of these.
Place a half-ball on the plane of its great circle. Inspect its planar sections parallel to this plane: we have to determine the area of a planar section at height \latex{ d }. (Figure 62)
\latex{ d }
\latex{ r }
\latex{ r }
\latex{ d }
\latex{ d }
\latex{ d }
\latex{ r }
\latex{ r }
Figure 62
The planar section is a disc with radius \latex{\sqrt{r^2-d^2} }, thus its area is
\latex{t=\left(\sqrt{r^2-d^2} \right)^2 \times \pi =(r^2-d^2)\times \pi =r^2\times \pi -d^2\times \pi}.
Now take a cylinder with radius and altitude both being \latex{ r }. If we remove an upside down cone, with radius and altitude being \latex{ r }, from this cylinder, then the remaining solid will satisfy the conditions of Cavalieri's principle. We will show that the area of any planar section of this solid and the half-ball are equal.
In case of the cylinder, this area is the area of an annulus, that is,
\latex{t=r^2\times \pi -d^2\times \pi}.
Thus the volume of the two solids are indeed equal.
Vhalf-ball = Vcylinder - Vcone= \latex{r^2\times \pi\times r-\frac{r^2\times\pi \times r}{3}=\frac{2\times r^3\times \pi }{3}}.
Therefore the volume of the whole ball is
\latex{V=\frac{4\times r^3\times \pi }{3} }
Example 1
Prove the following proposition first stated by Archimedes: the volume of the ball is equal to the two third of the volume of its circumscribed cylinder.
Solution
Let \latex{ r } denote the radius of the ball. It is easy to see that the radius of the base disc of its circumscribed cylinder will also be \latex{ r }, and its altitude is \latex{2 \times r}. This type of cylinder is called equilateral cylinder. (Figure 63)
The volume of the cylinder is
\latex{V_c=r^2\times \pi \times 2\times r=2\times r^3\times \pi}.
\latex{ 2r }
\latex{ r }
Figure 63
If we take two third of this, then we end up with the volume of a ball with radius\latex{ r }:
\latex{\frac{2}{3}\times V_c=\frac{2}{3}\times 2\times r^3\times \pi=\frac{4}{3}\times r^3\times \pi}.
Thus the proposition of Archimedes is proved.
The surface of the sphere
Determining the surface area is an even harder task, as we do not have any help similar to the principle of Cavalieri. We will just say intuitively that the surface area can be approximated as the sum of the areas of the lateral surfaces of appropriately chosen conical frustums, but actually computing this sum would be above our current knowledge of mathematics.
In absence of these skills we will show a method, which is not a proper proof, yet the described steps lead to the correct formula for the surface area.
\latex{ r }
\latex{ h }
Figure 64
Take a spherical shell with inner radius \latex{ r } and width \latex{ h }. (Figure 64)
It follows easily using the formula for the volume of the ball that the volume of this shell equals to the difference of the larger and smaller balls:
\latex{V=\frac{4\times (r+h)^3\times \pi}{3}-\frac{4\times r^3\times \pi}{3}=}
\latex{=\frac{4\times \pi}{3}\times (r^3+3\times r^2\times h+3\times r\times h^2+h^3-r^3)=} \latex{=\frac{4\times \pi}{3}\times (3\times r^2\times h+3\times r\times h^2+h^3)}.
On the other hand, if we choose \latex{ h } to be very small, then the volume can be viewed as the volume of a solid with area of base being the surface of a sphere with radius \latex{ r }, and altitude \latex{ h }:
\latex{V\approx A_{\text{sphere}}\times h}.
Thus the surface area of the sphere is
\latex{A_{\text{sphere}}\approx \frac{V}{h}=\frac{4\times \pi\times (3\times r^2\times h+3\times r\times h^2+h^3)}{3\times h}=4\times \pi\times \left(r^2+r\times h+\frac{h^2}{3} \right)}.
It can be shown that this equality gives exactly the surface area of the sphere if we are choosing h to be smaller and smaller, approaching zero. In this case the last two terms in the bracket will also approach zero. Thus we get the following formula for the surface area of the sphere:
\latex{A_{\text{sphere}}=4\times r^2\times \pi}.
This result shows that the other statement of Archimedes, that is, that the surface area of a sphere equals to four times the area of one of its great discs.
Example 2
Determine the size of the hole inside an iron ball with a radius of \latex{ 10\, cm } which floats when thrown into water.
Solution
According to the principle of Archimedes, an upward buoyant force is exerted on a body immersed in a fluid, which is equal to the weight of the fluid displaced by the body. The volume of the water displaced by the iron ball is:
\latex{V=\frac{4\times r^3\times \pi}{3} \approx 4188.8 \;\text{cm}^3}.
The size of the buoyant force is (\latex{\rho_{\text{water}}=1\frac{\text{kg}}{\text{dm}^3},g\approx 10\frac{\text{m}}{\text{s}^2}} acceleration due to gravity):
\latex{F=\rho \times V\times g\approx 41.88N}.
If the ball floats, then its weight equals to this buoyant force. Using this, the volume of the iron ball can be expressed:
\latex{V_{\text{iron}}=\frac{F}{\rho_{\text{iron}}\times g } \approx 537\;\text{cm}^2\;\left(\rho_{\text{iron}}=7.8\frac{\text{kg}}{\text{dm}^3}\right)}.
It follows that the volume of the hole in the ball is
\latex{V_{\text{hole}}=V-V_{\text{iron}}=3651.8\;\text{cm}^3}.
This is the \latex{ 87.2 }% of the volume of the ball.
Exercises
{{exercise_number}}. What is the volume and surface area of the ball whose
- radius is \latex{ 110\, cm };
- diameter is \latex{ 40\, cm }?
{{exercise_number}}. The surface area of a ball is \latex{ 1,000\, m^2 }. What is its volume?
{{exercise_number}}. The volume of a ball is \latex{ 100\, cm^3 }. What is its surface area?
{{exercise_number}}. What is the area of the disc which is the intersection of a ball with radius \latex{ r } and a plane at a distance of from the centre of the ball? What is the ratio of the area of the disc and the surface area of the ball?
{{exercise_number}}. How far is the plane from the centre of a ball with radius \latex{ r } for which the area of the intersection is one tenth of the surface area of the ball?
{{exercise_number}}. Prove that the ratio of the solids seen in the Figure satisfy the following:
\latex{V_{\text{cylinder}}:V_{\text{sphere}}:V_{\text{cone}}=3:2:1}
\latex{V_{\text{cylinder}}:V_{\text{sphere}}:V_{\text{cone}}=3:2:1}
{{exercise_number}}. How much does a ninepin bowling ball of radius \latex{ 12\, cm } weigh, if its density is \latex{ 3\, kg/dm^3 }?
{{exercise_number}}. Determine the volume and surface area of a ball whose density is \latex{ 5\, kg/dm^3 } and weight is \latex{ 8\, kg }.
{{exercise_number}}. A spherical cap of height \latex{ h } is being cut off a ball of radius \latex{ r }. Determine the volume of the resulting solid. (Use Cavalieri's principle.)
{{exercise_number}}. There are three equal sized balls of radius \latex{ R } on a table such that they pairwise touch each other. Find the volume of the small sphere which can be placed between the three large spheres on the table such that it touches all three.
{{exercise_number}}. The sum of the lengths of all edges of a cuboid is \latex{ 400\, cm } and its surface area is \latex{ 3,600\, cm^2 }. Determine the volume and surface area of its circumscribed sphere.