Volumes by cylindrical shells

Sometimes it is easier to calculate the volume of a solid (formed by the revolution of a surface about an axis of revolution) by dividing it up in a different way than into thin cross-sections. We will still use cylinders, but this time they will be cylindrical shells, where the height of the cylindrical shell mimics the function. This figure illustrates how this works for a particular function (in this case f(x)=\sqrt{1-x^2} between x=0 and x=1.

cylshell

  • Top Left image: the function \sqrt{1-x^2}.
  • Top right image: The volume formed by revolving this about the y axis.
  • Bottom left image: Breaking this shape up into cylindrical shells (think of taking a wire loop of varying diameters and slicing vertically through the function.
  • Bottom right image: The cylindrical shells when pulled apart (this is just for illustration purposes). In this case the central shell is taller than the outer shells because in the centre of the shape the function is highest and towards the edges it is lowest.

Let’s first think about the volume of a single shell which has inner radius r_1, outer radius r_2 and height $h$ as shown here

cylex

The volume of such a cylindrical shell is just going to be the height times the area of the cylindrical face (which in this case is an annulus). The area of the annulus is:

 

Area_{annulus}=\pi( r_2^2-r_1^2)

 

so the volume of this object is h\pi (r_2^2-r_1^2). We can rewrite this in a way which will become much more useful shortly:

 

h\pi (r_2^2-r_1^2)=h\pi 2\frac{r_2+r_1}{2}(r_2-r_1)

 

We can see then that \frac{r_2+r_1}{2} is the radius up to the middle of the shell’s edge and r_2-r_1 is the thickness of the shell. Eventually we are going to want to take thin shells and add them together, so we can think of r_2-r_1 as \Delta x. Also, when the shells are thin, we can just call the distance to the middle of the shell’s edge x. If we want to mimic the shape made by rotating a given function around the y-axis, the height of the cylindrical shell will then just be f(x). Thus we can write the volume of a given shell which is of radius x as:

 

V_{\text{cylindrical shell}}=\Delta x\pi 2x f(x)

 

If we add all of the shells which are different radii x then we will end up with a Riemann sum and we can take the limit of the Riemann sum. This gives the exact answer to the volume as:

 

\int_a^b 2\pi x f(x) dx

 

where a and b are the inner and outer radii of the shape of interest (in the case of the above figure this will be x=0 and x=1 but you have to be careful with this as we will see in the next example.

The important thing to note about this expression is that when we used the method of cylindrical cross-sections we needed to know f^{-1}(y) which is not always possible. Now we just need to know how to integrate xf(x) which may be easier.

Let’s look at an example:

Calculating the volume of a torus using the method of cylindrical shells
 
A torus is created by taking a circular cross section in the (x,y) plane and revolving it around the y axis, as shown here:

torus

It will actually be easier to work with half of this torus (the top half or the bottom half) because the function is then single valued. If we choose the top half, then the equation that defines the half circle is y=\sqrt{r_2^2-(x-r_1)^2}. We want to slice this function into cylindrical pieces which will look like figure the figure below if we take just five pieces. Of course we will take the limit that we have an infinite number of pieces, but for now this is an illustrative approximation.

torpieces
The torus starts at x=r_1-r_2 and ends at x=r_1+r_2. We can thus write down the formula for the integration by the method of singular shells as (remembering that we are just looking at the half torus, so to get the full torus volume we multiply by two):

 

2\int_{r_1-r_2}^{r_1+r_2}2\pi x \sqrt{r_2^2-(x-r_1)^2}dx

 

Now we do a u-substitution of the form x-r_1=ur_2. This gives the integration limits in the u variable as \pm 1, so the integration is then:

 

2\int_{-1}^12\pi (u r_2+r_1)r_2 \sqrt{1-u^2}r_2du

 

We can split this up into two integrals:

 

2\int_{-1}^12\pi (u r_2)r_2 \sqrt{1-u^2}r_2du+2\int_{-1}^12\pi (r_1)r_2 \sqrt{1-u^2}r_2du

 

The first integral is an odd function which, when integrated from -1 to 1 necessarily gives 0. The second term is an even function so we can write:

 

Volume_{torus}=4\int_{0}^12\pi (r_1)r_2^2 \sqrt{1-u^2}du

 

and the integral of \int\sqrt{1-u^2}du is \frac{1}{2} \left(\sqrt{1-u^2} u+\sin ^{-1}(u)\right). The first term vanishes at both ends of the limits (both at 0 and at 1) and the second term has a contribution only at x=1, (\sin^{-1}1=\frac{\pi}{2}). Thus the final result for the integration gives:

 

Volume_{torus}=2\pi^2 r_1 r_2^2

 

This in itself is a rather lovely result because it can be written as 2\pi r_1\times \pi r_2^2 which is the perimeter of a circle of diameter r_1 times the area of a circle of radius r_2, which is roughly (but not quite) what a torus is. If you image a cylinder that is 2\pi r_1 in length and r_2 in radius then this is the volume that it represents. You can imagine then curling this round to make a torus (take a tube and join the ends). There is a rather lovely explanation here: http://whistleralley.com/torus/torus.htm as to why this is the case. It might seem obvious at first glance, then it seems surprising on further inspection, and finally it should seem very elegant!

Let’s now make sure that we get the same answers that we got for some of our previous calculations using this new method. One of the questions we looked at before was this figure:

yrot

 

If we now use the method of cylindrical shells we will get:

 

Volume=\int_0^1 2\pi x f(x)dx=\int_0^1 2\pi x\sqrt{x} dx=2\pi \frac{2}{5}x^\frac{5}{2}|_0^1=\pi\frac{4}{5}

 

which is exactly what we got when we did it with the method of cylindrical cross-sections.

Let’s take an example which we wouldn’t have been able to do with the previous method. Let’s look at the function f(x)=2x^2-x^3 revolved around the y-axis between 0 and 2. With this function we wouldn’t have been able to invert it easily. In fact it doesn’t even invert to a single function. We would need to define multiple functions and it would all be very messy. Now, using the method of cylindrical shells, it’s easy:

V=\int_0^2 2\pi x \left(2x^2-x^3\right)dx=\frac{16 \pi }{5}

 

What if we would like to calculate something a bit trickier, like this figure (where we have rotated the figure on the left around the line x=-1: ?

sqrtxm1

Now we have again a series of shells, but the radius and height of the shells is something that we have to be very careful of. At a position x, the height is going to be x-x^2 but the radius isn’t going to be x any more but x+1, so now we have:

 

Volume=\int_0^1 2\pi (x+1) \left(f(x)-g(x)\right)dx=\int_0^1 2\pi (x+1)\left(x-x^2\right)dx=\int_0^1 2\pi\left(x-x^3\right)=\frac{\pi}{2}

 

which is precisely what we got last time, but this time we didn’t have to do anything complicated with the inverse function.

How clear is this post?