Review 5: Solids of Revolution - Volume of Revolution: Example 2
For this example, let us consider the region bounded by the curve given by the function f(x)=1−x2 with
0≤x≤1 along with both the
x and
y-axes.
We choose to rotate this region around the y-axis which creates a volume similar to an upside-down bowl.
Again, our question is, "What is the volume of the solid?" In order to answer this we will again choose to partition the x-axis. However, this time let's do the partition before the rotation. The following picture shows the discretization of
the
x-axis and highlights one particular inscribed rectangle.
When the region is rotated, the highlighted rectangle will make a thin cylinder like shape with thickness Δx. We will call such a region a "shell."
If we imagine doing this for every part of the partition, we can think of the volume made up of a bunch of these shells, one inside the other, like nested Russian dolls.
As a result, the volume of the solid will be approximately the sum of the volume of the shells. So the question is, what is the volume of one of the shells? If we can discover this, then we can construct the Riemann sum and the associated definite integral.
The key to the volume of a single shell is the warm up exercise on the label of the can. In the following picture, you can see how "cutting" the shell and "unfolding it" gives a very easy to compute volume of a thin rectangular block.
So a generic shell will have volume ChΔx where
C is the circumference of the shell,
H is the height, and
Δx is the thickness. If you refer back to figure 4, you will see that the height will be
f(x∗i)=1−(x∗i)2 where
x∗i is a representative value for the
i-th part of the partition. Moreover, the circumference can be computed by using
x∗i as the value for the radius. So we have the volume of the
i-th shell is
Vi=2πx∗i(1−(x∗i)2)Δx
and the Riemann sum approximation:
∑ni=12πx∗i(1−(x∗i)2)Δx.