In this video,
we will derive the volume formula of a sphere
by integrating the surface area,
where the volume of a sphere equals 4/3 pi r-cubed
and the surface area equals four pi r-squared.
In an earlier video, we discovered
the derivative of the volume equals the surface area,
where dV/dr equals
4/3 pi times a derivative of r-cubed
with respect to r, which is three r-squared.
Simplifying this does give us four pi r-squared,
which is the surface area.
And therefore, it should make sense,
if we integrate the surface area,
we should be able to recover the volume.
Let's explain why this makes sense.
Again, we already know dV/dr equals four pi r-squared.
And now if we write this equation in differential form,
we have dV, which represents the change in the volume,
equals four pi r-squared, which is the surface area,
times differential r,
which represents the change in the radius.
So four pi r-squared times differential r
gives us the change in the volume,
which is the volume of one layer or one shell of the sphere
given a small change in the radius.
And therefore, the sum of the volumes
of the layers of the sphere will give us the total volume,
where, again, each thickness or layer of the sphere
is equal to differential r or delta r.
To better understand this,
if we take a look at the picture of the jawbreaker here,
it should make sense that if we determine the volume
of each layer of the jawbreaker,
then the sum of the volumes of the layers
will give us the total volume of the jawbreaker.
Well, if we look at this graphically,
if we have a sphere and cut in half,
this graph shows five layers of the sphere,
where, again, if we determine the volume of each layer,
the sum of the volumes of the layers
will give us the total volume of the sphere.
And now let's take a look at this on the coordinate plane.
A sphere is three-dimensional,
but here we'll take a look at it
only using the x- and y-axes.
Let's visualize this yellow part as one layer of the sphere,
with a thickness of the layer,
because we're on the coordinate plane,
is going to be differential x or delta x.
And the radius would be some x value
in some subinterval of x,
based upon delta x or the change of x.
And now let's introduce some calculus notation.
Let's first focus on the sum here,
where we have the sum from i equals one to n of four pi
times the square of c sub i,
where c sub i is some x value
in some small subinterval of x times delta x.
Well, four pi times the square of c sub i
is the surface area,
and delta x would be the thickness of the layer.
And therefore, this product represents the volume
of one thin layer of the sphere.
And therefore, the sum is equal to the sum
of the volume of n layers of the sphere.
And now if we let the number of layers approach infinity
or, in this case, n approach infinity,
we have the limit as n approaches infinity
of the sum from i equals one to n
of four pi times the square of c sub i times delta x,
which indicates, as the number of layers approach infinity,
the sum of the volumes of the layers
approach the total volume of the sphere.
And we should recognize the limit of this sum
is equal to the def integral from zero to r
of four pi times x-squared, differential x.
So this integral is equal to the volume of the sphere.
The next step, we factor out four pi
and then integrate with respect to x,
which gives us x-cubed divided by three.
And now we find big F of r minus big F of zero.
If we substitute r for x, we get r-cubed divided by three.
If we substitute zero for x,
we get zero-cubed divided by three.
Of course, this is equal to zero,
giving us four pi times r-cubed divided by three,
which equals 4/3 pi r-cubed,
which is the volume of a sphere.
Thank you for watching.