4(pi)r^2 gives you the surface area of a sphere,
but where does the formula come from?
First we'll create a sphere and fill in its
surface area.
Next, we inscribe a polygon inside the sphere
and we rotate the polygon along its horizontal
axis to create a 3d model.
Now, if we increase the number of sides of
the inscribed polygon we can see how this
affects the shape of the model,
specifically notice how the surface area of
the model approximates the surface area of
the sphere.
So if we increase the sides of the inscribed
polygon to infinitely many sides, then the
surface area of the model becomes equal to
the surface area of the sphere.
So lets derive a formula for the surface area
of a model where the inscribed polygon has
an arbitrary number of sides.
We can see that the model is made up of two
cones at both ends and in the middle we have
shapes called frustums.
In order to find the surface area of the model
we need to find the surface area of the cones
and frustums without their bases.
First the surface area of a cone without its
base is equal to (pi) times the radius times
the lateral height.
And surface area of a frustum without its
top and bottom bases is equal to (pi) x the
radius of the top base + the radius of the
bottom base x the lateral height.
Next lets take a two dimensional view of our
shape, and we can see the equilateral polygon
that is inscribed inside the sphere.
We will label the vertices with capital letters.
And we label all the sides with the letter
s.
Next we need to point out some unique properties
related to an even sided equilateral polygon.
First the vertical lines that connect the
opposite vertices are parallel.
And the same is true if we connect points
CH and DG which are parallel with lines AB
and EF.
Another property is that the horizontal line
AE intersects the vertical lines at right
angles.
And there are multiple diagonals that form
right angles with the sides of the polygon.
One final property is that line AE is a diameter
which bisects the polygon.
Therefore if we label this line as r1 then
the line below also has length r1
We label the next line r2 and the last line
we label as r3.
Now as we can see, the polygon is made up
of different sized triangles. Before we can
find the surface area, we must first establish
an important relationship between these triangles.
First, we label the intersection points in
the middle and lets look at triangle ABI and
HIJ.
The vertical angles formed by both triangles
are equal.
Now since line AB is parallel to line CH,
the alternate interior angles are equal.
Since both triangles have side r1 in common
then by the Angle Side Angle theorem both
triangles are congruent.
Therefore all the corresponding angles are
equal and if we call this side of the triangle
d1 then the corresponding side of the other
triangle must also be d1.
Next lets look at these two triangles.
The vertical angles are equal and since line
BH is parallel to line CG then the alternate
interior angles are equal.
Notice that both triangles have all the same
angles.
If at least two angles of one triangle are
equal to two angles of another triangle, then
both triangles must be similar.
In the same way we can show that the following
triangles are also similar.
Next we need to point out that these pair
of triangles are congruent by the Angle side
Angle theorem.
So they have side d2 in common,
and the same is true with these pair of triangles
which have side d3 in common.
Finally we will look at this last triangle,
and it is also similar because it has two
of the same angles shared by all the other
similar triangles.
Since all these triangles are similar, we
can establish an important relationship between
them.
Specifically we will show that the corresponding
sides are all in the same proportion.
Therefore, r1 over d1 is equal to r2 over
d2 which is equal to r3 over d3 which is equal
to line AD over s.
Next, we will cross multiply each of these
terms and so we have
(r1)(s) is equal to (d1)(AD)
(r2)(s) is equal to (d2)(AD)
and
(r3)(s) is equal to (d3)(AD)
Next we will add these equations together
and we have,
r1s + r2s + r3s = d1 AD + d2 AD + d3 AD
On the left hand side we can factor out and
divide by s.
And on the right hand side we will factor
out line AD.
So lets take a look at what d1 + d2 + d3 actually
represents.
We can see that these are line segments that
form the diameter AE.
So lets arrange the segments around, and we
can see that d1 + d2 + d3 is equal to half
of line AE.
Therefore r1 + r2 + r3 = line AE times line
AD over 2s.
Finally this is the important equation we
need as a result of the similar triangles.
We can now begin to find the surface area
of the model by finding the surface area of
the cones and frustums without their bases.
So the surface area of the model is equal
to the surface area of the first cone
which is (pi) times (r1) times (s) + the surface
area of the first frustum which is (pi) times
(r1+r2) times (s) + the surface area of the
second frustum
which is (pi) times (r2+r3) times (s) + the
surface area of the 2nd cone which is (pi)
times (r3) times (s).
Next we will use algebra and factor out pi
and we will also factor out s.
We can then simplify further by combining
like terms,
And we will factor out the 2's.
Therefore the surface area of the model is
equal to
2(pi)s(r1+ r2+ r3)
Now notice that r1 + r2 + r3 is the equation
we derived as a result of the similar triangles.
Therefore the surface area of the model becomes
equal to
2(pi)s(line AE x line AD over 2s)
The two's will cancel out and the s's also
cancel out.
So the surface area of the model is now equal
to (pi) x line AE x line AD.
Next, lets focus on line AE and in terms of
our sphere, line AE is the diameter, and the
diameter is equal to twice the radius of the
sphere.
Therefore line AE is equal to 2r.
Next lets look at line AD.
So what we are going to do is increase the
number of sides of the inscribed polygon and
lets look at what happens to line AD.
And as we can see line AD begins to approach
the diameter.
So now we are going to increase the sides
of the inscribed polygon to infinitely many
sides, and the first thing we should point
out is the surface area of the model becomes
equal to the surface area of the sphere.
The second thing is that line AD eventually
becomes equal to the diameter which is equal
to 2r.
Finally the surface area of the sphere is
equal to 2(pi)r x 2r, which simplifies to
4(pi)r^2.
Therefore, 4(pi)r^2 gives you the surface
area of any sized sphere that exists.
