imported>CaptainAngus: Cleaned up instances of raw HTML tags

2025-09-28T12:57:30Z

Cleaned up instances of raw HTML tags

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	{\| role="presentation" class="wikitable mw-collapsible mw-collapsed"		{\| role="presentation" class="wikitable mw-collapsible mw-collapsed"
	\| ~~<strong>~~A way to obtain the formula~~</strong>~~		\| '''A way to obtain the formula'''
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	\| The method's formula can be derived as follows:		\| The method's formula can be derived as follows:

imported>B.Frassek: /* A factor of 8 was missing in the function term; function term, integral, result and function term in text of "Comparison With Disc Integration" has been corrected.*/

2024-12-04T11:00:25Z

A factor of 8 was missing in the function term; function term, integral, result and function term in text of "Comparison With Disc Integration" has been corrected.

New page

{{Short description|Method for calculating the volume of a solid of revolution}}
[[File:Shell integral undergraph - around y-axis.png|thumb|right|300px|A volume is approximated by a collection of hollow cylinders. As the cylinder walls get thinner the approximation gets better. The limit of this approximation is the shell integral.]]

{{Calculus |Integral}}

'''Shell integration''' (the '''shell method''' in [[integral calculus]]) is a method for [[calculation|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.

==Definition==

The shell method goes as follows: Consider a volume in three dimensions obtained by rotating a cross-section in the {{mvar|xy}}-plane around the {{mvar|y}}-axis. Suppose the cross-section is defined by the graph of the positive function {{math|''f''(''x'')}} on the interval {{math|[''a'', ''b'']}}. Then the formula for the volume will be:

:<math>2 \pi \int_a^b x f(x)\, dx</math>

If the function is of the {{mvar|y}} coordinate and the axis of rotation is the {{mvar|x}}-axis then the formula becomes:

:<math>2 \pi \int_a^b y f(y)\, dy</math>

If the function is rotating around the line {{math|''x'' {{=}} ''h''}} then the formula becomes:<ref>{{Cite web|url=https://math.la.asu.edu/~dheckman/11%20-%20Volume%20-%20Shell%20Method.pdf|title=Volume – Shell Method|last=Heckman|first=Dave|date=2014|access-date=2016-09-28}}</ref>

:<math>\begin{cases}
\displaystyle 2 \pi \int_a^b (x-h) f(x)\,dx, & \text{if}\ h \le a < b\\
\displaystyle 2 \pi \int_a^b (h-x) f(x)\,dx, & \text{if}\ a < b \le h,
\end{cases}</math>
and for rotations around {{math|''y'' {{=}} ''k''}} it becomes
:<math>\begin{cases}
\displaystyle 2 \pi \int_a^b (y-k) f(y)\,dy, & \text{if}\ k \le a < b\\
\displaystyle 2 \pi \int_a^b (k-y) f(y)\,dy, & \text{if}\ a < b \le k.
\end{cases}</math>

The formula is derived by computing the [[double integral]] in [[polar coordinates]].

=== Derivation of the formula ===

{| role="presentation" class="wikitable mw-collapsible mw-collapsed"
| <strong>A way to obtain the formula</strong>
|-
| The method's formula can be derived as follows:
Consider the function <math>f( x)</math> which describes our cross-section of the solid, now the integral of the function can be described as a Riemann integral:
<math>\int\limits_{a}^{b} f(x) dx = \lim_{n \to \infty} \sum_{i=1}^n f(a + i \Delta x) \Delta x</math>

Where <math> \Delta x = \frac{b-a}{n} </math> is a small difference in <math> x </math>

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:

<math> f(a + k\Delta x) \Delta x</math>

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 <math>a + (k+1)\Delta x </math> with height <math>f(a + k \Delta x)</math> , and substracting it another smaller cylinder of radius <math>a + k\Delta x </math>, with the same height of <math>f(a + k\Delta x)</math> , this difference of cylinder volumes is:

<math> \pi (a+(k+1) \Delta x)^2 f(a+k \Delta x) - \pi (a + k\Delta x)^2 f(a+ k \Delta x) </math>

<math>= \pi f(a + k \Delta x) ( (a + (k+1)\Delta x )^2 - (a + k\Delta x )^2)</math>

By difference of squares , the last factor can be reduced as:

<math> \pi f(a + k \Delta x) (2a + 2k\Delta x + \Delta x) \Delta x</math>

The third factor can be factored out by two, ending up as:

<math> 2\pi f(a + k\Delta x) (a + k\Delta x + \frac{\Delta x}{ 2 }) \Delta x</math>

This same thing happens with all terms, so our total sum becomes:

<math> \lim_{n \to \infty}
2\pi \sum_{i=1}^n f(a + i\Delta x) (a + i\Delta x + \frac{\Delta x}{ 2 }) \Delta x</math>

In the limit of <math> n \rightarrow \infin</math>, we can clearly identify that:

* <math> f(a + i\Delta x)</math> as <math> \Delta x</math> tends to 0 ends up becoming <math> f(x)</math>
* <math> (a + i \Delta x + \frac{\Delta x}{2}) </math> becomes <math> x</math> itself, going from a to b (ignoring the last term which vanishes)
* <math> \Delta x</math> becomes the infinitesimal <math> dx </math>

Thus, at the limit of infinity, the sum becomes the integral:

<math> 2\pi \int\limits_{a}^{b} x f(x) dx</math>

QED <math> \square</math>.
|}

==Example==

Consider the volume, depicted below, whose cross section on the interval [1, 2] is defined by:

:<math>y = 8(x-1)^2(x-2)^2</math>

{{Multiple image
| align = none
| total_width = 600
| image1 = Shell_2d_example.png
| width1 = 481 | height1 = 307
| caption1 = Cross-section
| image2 = Shell_3D_example.png
| width2 = 632 | height2 = 463
| caption2 = 3D volume
| caption_align = center
}}

With the shell method we simply use the following formula:

:<math>V = 16 \pi \int_1^2 x ((x-1)^2(x-2)^2) \,dx </math>

By expanding the polynomial, the integration is easily done giving {{sfrac|8|10}}<math>\pi</math> cubic units.

=== Comparison With Disc Integration ===

Much more work is needed to find the volume if we use [[disc integration]]. First, we would need to solve <math>y = 8(x-1)^2(x-2)^2</math> for {{mvar|x}}. Next, because the volume is hollow in the middle, we would need two functions: one that defined an outer solid and one that defined the inner hollow. After integrating each of these two functions, we would subtract them to yield the desired volume.

==See also==
*[[Solid of revolution]]
*[[Disc integration]]

==References==
{{Reflist}}
*{{MathWorld|title=Method of Shells|urlname=MethodofShells}}
*[[Frank J. Ayres|Frank Ayres]], [[Elliott Mendelson]]. ''[[Schaum's Outlines]]: Calculus''. McGraw-Hill Professional 2008, {{isbn|978-0-07-150861-2}}. pp. 244–248 ({{Google books|Ag26M8TII6oC|online copy|page=244}})

{{Calculus topics}}

[[Category:Integral calculus]]

Shell integration - Revision history

imported>CaptainAngus: Cleaned up instances of raw HTML tags

imported>B.Frassek: /* A factor of 8 was missing in the function term; function term, integral, result and function term in text of "Comparison With Disc Integration" has been corrected.*/