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Shell integration
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<h2 id="definition">Definition</h2> <p>The shell method goes as follows: Consider a volume in three dimensions obtained by rotating a cross-section in the xy-plane around the y-axis. Suppose the cross-section is defined by the graph of the positive function <i>f</i>(<i>x</i>) on the interval [<i>a</i>, <i>b</i>]. Then the formula for the volume will be: </p> 2 π ∫ a b x f ( x ) d x {\displaystyle 2\pi \int _{a}^{b}xf(x)\,dx} <p>If the function is of the y coordinate and the axis of rotation is the x-axis then the formula becomes: </p> 2 π ∫ a b y f ( y ) d y {\displaystyle 2\pi \int _{a}^{b}yf(y)\,dy} <p>If the function is rotating around the line <i>x</i> = <i>h</i> then the formula becomes:<a class="footnote-ref" id="fnref:1" href="#fn:1"><sup>1</sup></a> </p> { 2 π ∫ a b ( x − h ) f ( x ) d x , if h ≤ a < b 2 π ∫ a b ( h − x ) f ( x ) d x , if a < b ≤ h , {\displaystyle {\begin{cases}\displaystyle 2\pi \int _{a}^{b}(x-h)f(x)\,dx,&{\text{if}}\ h\leq a<b\\\displaystyle 2\pi \int _{a}^{b}(h-x)f(x)\,dx,&{\text{if}}\ a<b\leq h,\end{cases}}} <p>and for rotations around <i>y</i> = <i>k</i> it becomes </p> { 2 π ∫ a b ( y − k ) f ( y ) d y , if k ≤ a < b 2 π ∫ a b ( k − y ) f ( y ) d y , if a < b ≤ k . {\displaystyle {\begin{cases}\displaystyle 2\pi \int _{a}^{b}(y-k)f(y)\,dy,&{\text{if}}\ k\leq a<b\\\displaystyle 2\pi \int _{a}^{b}(k-y)f(y)\,dy,&{\text{if}}\ a<b\leq k.\end{cases}}} <p>The formula is derived by computing the <a href="/facts/Double_integral/Fvl3Bkam">double integral</a> in <a href="/facts/Polar_coordinates/HBko4YoN">polar coordinates</a>. </p> <h3>Derivation of the formula</h3> <table><tbody><tr><td>A way to obtain the formula</td></tr><tr><td>The method's formula can be derived as follows:<p>Consider the function f ( x ) {\displaystyle f(x)} which describes our cross-section of the solid, now the integral of the function can be described as a Riemann integral: ∫ a b f ( x ) d x = lim n → ∞ ∑ i = 1 n f ( a + i Δ x ) Δ x {\displaystyle \int \limits _{a}^{b}f(x)dx=\lim _{n\to \infty }\sum _{i=1}^{n}f(a+i\Delta x)\Delta x} </p><p>Where Δ x = b − a n {\displaystyle \Delta x={\frac {b-a}{n}}} is a small difference in x {\displaystyle x} </p><p>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:</p><p> f ( a + k Δ x ) Δ x {\displaystyle f(a+k\Delta x)\Delta x} </p><p>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 ) Δ x {\displaystyle a+(k+1)\Delta x} with height f ( a + k Δ x ) {\displaystyle f(a+k\Delta x)} , and substracting it another smaller cylinder of radius a + k Δ x {\displaystyle a+k\Delta x} , with the same height of f ( a + k Δ x ) {\displaystyle f(a+k\Delta x)} , this difference of cylinder volumes is:</p><p> π ( a + ( k + 1 ) Δ x ) 2 f ( a + k Δ x ) − π ( a + k Δ x ) 2 f ( a + k Δ x ) {\displaystyle \pi (a+(k+1)\Delta x)^{2}f(a+k\Delta x)-\pi (a+k\Delta x)^{2}f(a+k\Delta x)} </p><p> = π f ( a + k Δ x ) ( ( a + ( k + 1 ) Δ x ) 2 − ( a + k Δ x ) 2 ) {\displaystyle =\pi f(a+k\Delta x)((a+(k+1)\Delta x)^{2}-(a+k\Delta x)^{2})} </p><p>By difference of squares , the last factor can be reduced as:</p><p> π f ( a + k Δ x ) ( 2 a + 2 k Δ x + Δ x ) Δ x {\displaystyle \pi f(a+k\Delta x)(2a+2k\Delta x+\Delta x)\Delta x} </p><p>The third factor can be factored out by two, ending up as:</p><p> 2 π f ( a + k Δ x ) ( a + k Δ x + Δ x 2 ) Δ x {\displaystyle 2\pi f(a+k\Delta x)(a+k\Delta x+{\frac {\Delta x}{2}})\Delta x} </p><p>This same thing happens with all terms, so our total sum becomes:</p><p> lim n → ∞ 2 π ∑ i = 1 n f ( a + i Δ x ) ( a + i Δ x + Δ x 2 ) Δ x {\displaystyle \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} </p><p>In the limit of n → ∞ {\displaystyle n\rightarrow \infty } , we can clearly identify that:</p><ul><li> f ( a + i Δ x ) {\displaystyle f(a+i\Delta x)} as Δ x {\displaystyle \Delta x} tends to 0 ends up becoming f ( x ) {\displaystyle f(x)} </li><li> ( a + i Δ x + Δ x 2 ) {\displaystyle (a+i\Delta x+{\frac {\Delta x}{2}})} becomes x {\displaystyle x} itself, going from a to b (ignoring the last term which vanishes)</li><li> Δ x {\displaystyle \Delta x} becomes the infinitesimal d x {\displaystyle dx} </li></ul><p>Thus, at the limit of infinity, the sum becomes the integral:</p><p> 2 π ∫ a b x f ( x ) d x {\displaystyle 2\pi \int \limits _{a}^{b}xf(x)dx} </p><p>QED ◻ {\displaystyle \square } .</p></td></tr></tbody></table> <h2 id="example">Example</h2> <p>Consider the volume, depicted below, whose cross section on the interval [1, 2] is defined by: </p> y = 8 ( x − 1 ) 2 ( x − 2 ) 2 {\displaystyle y=8(x-1)^{2}(x-2)^{2}} Cross-section3D volume <p>With the shell method we simply use the following formula: </p> V = 16 π ∫ 1 2 x ( ( x − 1 ) 2 ( x − 2 ) 2 ) d x {\displaystyle V=16\pi \int _{1}^{2}x((x-1)^{2}(x-2)^{2})\,dx} <p>By expanding the polynomial, the integration is easily done giving 8/10 π {\displaystyle \pi } cubic units. </p> <h3>Comparison With Disc Integration</h3> <p>Much more work is needed to find the volume if we use <a href="/facts/Disc_integration/hGyQt19u">disc integration</a>. First, we would need to solve y = 8 ( x − 1 ) 2 ( x − 2 ) 2 {\displaystyle y=8(x-1)^{2}(x-2)^{2}} for 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. </p> <h2 id="see-also">See also</h2> <ul><li><a href="/facts/Solid_of_revolution/lDNv2P7p">Solid of revolution</a></li> <li><a href="/facts/Disc_integration/hGyQt19u">Disc integration</a></li></ul> <ul><li><a href="/facts/Eric_W._Weisstein/FbMuVDep">Weisstein, Eric W.</a> <a href="https://mathworld.wolfram.com/MethodofShells.html">"Method of Shells"</a>. <i><a href="/facts/MathWorld/yc1jodbq">MathWorld</a></i>.</li> <li><a href="/facts/Frank_J._Ayres/Co7SCcGv">Frank Ayres</a>, <a href="/facts/Elliott_Mendelson/S6PpLYGJ">Elliott Mendelson</a>. <i><a href="/facts/Schaum%2527s_Outlines/Dpskwhr9">Schaum's Outlines</a>: Calculus</i>. McGraw-Hill Professional 2008, <a href="/facts/ISBN_(identifier)/15AdSPa9">ISBN</a> 978-0-07-150861-2. pp. 244–248 (<i><a href="https://books.google.com/books?id=Ag26M8TII6oC&pg=PA244">online copy</a></i>, p. 244, at <a href="/facts/Google_Books/lnGRMWXs">Google Books</a>)</li></ul> <h2 id="references">References</h2> <ol> <li id="fn:1"><p>Heckman, Dave (2014). "Volume – Shell Method" (PDF). Retrieved 2016-09-28. <a href="https://math.la.asu.edu/~dheckman/11%20-%20Volume%20-%20Shell%20Method.pdf" target="_blank">https://math.la.asu.edu/~dheckman/11%20-%20Volume%20-%20Shell%20Method.pdf</a> <a href="#fnref:1" class="footnote-back-ref">↩</a></p></li> </ol>