<\/a>.<\/p>\nas long as the thickness is small enough, the volume of the shell can be approximated<\/em> by the formula:<\/p>\nv<\/em> = 2πrhw<\/em><\/p>\nnote that the volume is simply the circumference (2πr<\/em>) times the height (h<\/em>) times the thickness (w<\/em>). in fact, you can think of cutting the shell along its height and “unrolling” it to produce a thin rectangular slab. then the volume is simply length<\/em> × height<\/em> × width<\/em> as in any rectangular solid.<\/p>\nnow suppose we have a solid of revolution with generating region being the area under a function y<\/em> = f<\/em>(x<\/em>) between x<\/em> = a<\/em> and x<\/em> = b<\/em>. and suppose that the y<\/em>-axis as its axis of symmetry. (this is the easiest case).<\/p>\nf<\/em>(x<\/em>) = x<\/em>2<\/sup> + 1, between 2 and 6, revolved around the y<\/em>-axis, generating a solid of revolution using shell method” width=”300″ height=”300″ class=”size-full wp-image-12021″ \/> <\/p>\nimagine what happens to a thin vertical strip of the region as it revolves around the y<\/em>-axis. because the axis is also vertical, the strip will sweep out a cylindrical shell. furthermore, we have the following info about each shell:<\/p>\n\n- its radius is r<\/em> = x<\/em> (distance from a typical point to the y<\/em>axis).<\/li>\n
- the height is h<\/em> = y<\/em> = f<\/em>(x<\/em>).<\/li>\n
- its thickness is a small change in x<\/em>, which we label as δx<\/em> or dx<\/em>.<\/li>\n<\/ul>\n
therefore, the approximate volume of a typical shell is:<\/p>\n
v<\/em> = 2πx<\/em> × f<\/em>(x<\/em>) × dx<\/em><\/p>\nintegration<\/h3>\n
but remember, that’s only a single shell. the solid consists of shells that were sliced at various positions xk<\/sub><\/em> along the x<\/em>-axis. so we should add them up to get the approximate volume of the entire solid.<\/p>\n<\/p>\n
finally, after taking the limit as n<\/em> → ∞ (so that we have infinitely many shells to fill out the solid), we get the exact formula.<\/p>\n<\/p>\n
example 1<\/h3>\n
find the volume of the solid generated by revolving the region under f<\/em>(x<\/em>) = x<\/em>2<\/sup> + 1, where 2 ≤ x<\/em> ≤ 6, around the y<\/em>-axis.<\/p>\nsolution<\/h4>\n
it might help to sketch a figure. fortunately, this is exactly what’s pictured in the figure above.<\/p>\n
first identify the dimensions of a typical shell.<\/p>\n
\n- r<\/em> = x<\/em><\/li>\n
- h<\/em> = f<\/em>(x<\/em>) = x<\/em>2<\/sup> + 1<\/li>\n
- thickness = dx<\/em><\/li>\n<\/ul>\n
in addition, we use a<\/em> = 2 and b<\/em> = 6 because we have 2 ≤ x<\/em> ≤ 6.<\/p>\nnow set up the shell method integral and evaluate to find the volume.<\/p>\n
<\/p>\n
thus the volume is equal to 672π cubic units.<\/p>\n
more than one function<\/h2>\n
when the generating region is defined as the area between two functions, then we have to modify the formula somewhat. <\/p>\n
consider the region between two curves, y<\/em> = f<\/em>(x<\/em>) on top and y<\/em> = g<\/em>(x<\/em>) on bottom, between x<\/em> = a<\/em> and x<\/em> = b<\/em>.<\/p>\n<\/p>\n
the height at a typical sample x<\/em>-value is equal to the difference of the two function values. that is,<\/p>\nh<\/em> = y<\/em>2<\/sub> – y<\/em>1<\/sub> = f<\/em>(x<\/em>) – g<\/em>(x<\/em>)<\/p>\nthis observation leads directly to the following version of the shell method formula:<\/p>\n
<\/p>\n
revolving around different axes<\/h3>\n
there are also variations of the formula to cover cases in which the axis of revolution is not the y<\/em>-axis. in all cases though, the axis must not be interior to the region itself. <\/p>\nbe careful!<\/em><\/strong> the horizontal axis cases require that the functions be solved for x<\/em> rather than y<\/em>.<\/p>\n\n- vertical axis x<\/em> = h<\/em> entirely to the left of the region bounded above by y<\/em> = f<\/em>(x<\/em>) and below by y<\/em> = g<\/em>(x<\/em>):\n
<\/p>\n<\/li>\n
- horizontal axis y<\/em> = k<\/em> entirely below the region bounded on the right by x<\/em> = f<\/em>(y<\/em>) and on the left by x<\/em> = g<\/em>(y<\/em>):\n
<\/p>\n<\/li>\n<\/ul>\n
example 2<\/h3>\n
let r<\/em> be the region bounded by y<\/em> = x<\/em>4<\/sup> and y<\/em> = 3x<\/em>3<\/sup>. find the volume of the solid generated by revolving r<\/em> around the line x<\/em> = -2.<\/p>\nsolution<\/h4>\n
let’s make a sketch. there is only a tiny sliver of area between the two curves.<\/p>\n
<\/p>\n
to find the bounds of integration, we need to set the two functions equal.<\/p>\n
<\/p>\n
so we get x<\/em> = 0 and x<\/em> = 3. these two values will be our a<\/em> and b<\/em> in the integral.<\/p>\nthen identify the radius, height, and thickness of the typical shell.<\/p>\n
\n- r<\/em> = x<\/em> – (-2) = x<\/em> + 2<\/li>\n
- h<\/em> = f<\/em>(x<\/em>) – g<\/em>(x<\/em>) = 3x<\/em>3<\/sup> – x<\/em>4<\/sup><\/li>\n
- thickness = dx<\/em><\/li>\n<\/ul>\n
finally, put it all together and evaluate the definite integral.<\/p>\n
<\/p>\n","protected":false},"excerpt":{"rendered":"
the shell method is a technique for finding the volume of a solid of revolution. just as in the disk\/washer method (see ap calculus review: disk and washer methods), the exact answer results from a certain integral. in this article, we’ll review the shell method and show how it solves volume problems on the ap […]<\/p>\n","protected":false},"author":223,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[240],"tags":[241],"ppma_author":[24932],"acf":[],"yoast_head":"\n
ap calculus review: shell method - magoosh blog | high school<\/title>\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\t\n\t\n\t\n