Shell integration (the shell method in integral calculus) is a method for calculating the volume of a solid of revolution, when integrating along an axis perpendicular to the axis of revolution. This is in contrast to disc integration which integrates along the axis parallel to the axis of revolution.
The shell method goes as follows: Consider a volume in three dimensions obtained by rotating a cross-section in the -plane around the -axis. Suppose the cross-section is defined by the graph of the positive function on the interval . Then the formula for the volume will be:
2\pi
| b | |
| \int | |
| a |
xf(x)dx
If the function is of the coordinate and the axis of rotation is the -axis then the formula becomes:
2\pi
| b | |
| \int | |
| a |
yf(y)dy
If the function is rotating around the line then the formula becomes:[1]
\begin{cases} \displaystyle2\pi
| b | |
| \int | |
| a |
(x-h)f(x)dx,&if h\lea<b\\ \displaystyle2\pi
| b | |
| \int | |
| a |
(h-x)f(x)dx,&if a<b\leh, \end{cases}
\begin{cases} \displaystyle2\pi
| b | |
| \int | |
| a |
(y-k)f(y)dy,&if k\lea<b\\ \displaystyle2\pi
| b | |
| \int | |
| a |
(k-y)f(y)dy,&if a<b\lek. \end{cases}
The formula is derived by computing the double integral in polar coordinates.
| A way to obtain the formula | |||||||||||||||||||||||||||||||||||||
| The method's formula can be derived as follows:Consider the function f(x)
f(x)dx=\limn
f(a+i\Deltax)\Deltax Where \Deltax=
x The Riemann sum can be thought up as a sum of a number n of rectangles with ever shrinking bases, we might focus on one of them: f(a+k\Deltax)\Deltax Now, when we rotate the function around the axis of revolution, it is equivalent to rotating all of these rectangles around said axis, these rectangles end up becoming a hollow cylinder, composed by the difference of two normal cylinders. For our chosen rectangle, its made by obtaining a cylinder of radius a+(k+1)\Deltax f(a+k\Deltax) a+k\Deltax f(a+k\Deltax) \pi(a+(k+1)\Deltax)2f(a+k\Deltax)-\pi(a+k\Deltax)2f(a+k\Deltax) =\pif(a+k\Deltax)((a+(k+1)\Deltax)2-(a+k\Deltax)2) By difference of squares, the last factor can be reduced as: \pif(a+k\Deltax)(2a+2k\Deltax+\Deltax)\Deltax The third factor can be factored out by two, ending up as: 2\pif(a+k\Deltax)(a+k\Deltax+
)\Deltax This same thing happens with all terms, so our total sum becomes: \limn2\pi
f(a+i\Deltax)(a+i\Delta+
)\Deltax In the limit of n → infin f(a+i\Deltax) \Deltax f(x) (a+i\Deltax+
) x \Deltax dx Thus, at the limit of infinity, the sum becomes the integral: 2\pi
xf(x)dx QED \square |
Consider the volume, depicted below, whose cross section on the interval [1, 2] is defined by:
y=(x-1)2(x-2)2
With the shell method we simply use the following formula:
V=2\pi
| 2 | |
| \int | |
| 1 |
x((x-1)2(x-2)2)dx
By expanding the polynomial, the integration is easily done giving cubic units.
Much more work is needed to find the volume if we use disc integration. First, we would need to solve
y=(x-1)2(x-2)2