Editing Shell integration (section)

==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>.
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