Wikipedia:Reference desk/Archives/Mathematics/2007 March 15

= March 15 =

Solids of revolution
To begin, I apologize for asking a rather specific calculus homework question. But I'm a little stumped and I have about five problems left to finish, all of which involve the same twist in the normal solid of revolution process...the regions we are rotating contain the axis of revolution. For instance, one problem asks us to find the volume of the region bounded by $$y = -x+3$$ and $$y = x^2-3x$$ when it is rotated about the x-axis. The curves intersect at (&minus;1, 4) and (3, 0), but I'm not sure where to move from here. None of the typical methods seems to yield the correct answer, 20.367π. If anyone has a suggestion or a prod in the right direction, I'd be grateful. Thanks! 12.226.88.48 22:16, 15 March 2007 (UTC)

Disk_integration may be of help. See the part about "hollow". -- Ķĩřβȳ ♥  Ťįɱé  Ø  22:26, 15 March 2007 (UTC)


 * If you rotate by 180°, the effect is that of replacing x by −x. This means the region can also be described as bounded by y = x + 3 and y = x2 + 3x. If you graph all four functions (which you need to do the understand the following), you see a few more points of intersection, including at (0, 0), (0, 3), and (−2 + √7, 5 − √7) – the latter where y = −x + 3 intersects y = x2 + 3x. In rotating full circle, the triangle-like figure having these three points as corners sweeps out a volume that is also sweeped out by the part having positive x-values of the "blade" bounded by the original two functions. This means that standard methods for determining the volume will count this twice. You have to compensate for this, which you can do by separately determining the volume sweeped out by the triangle-like figure and subtracting it from the original answer, or by first splitting up the area into the "positive" part of the blade, and the smaller triangle-like figure with corners (0, 3), (−2 + √7, 5 − √7), and (1, 4), next determining for each the volume of the region sweeped out by a rotation, and finally adding the results.  --Lambiam Talk  23:40, 15 March 2007 (UTC)
 * Perhaps I'm missing something, but it seems to me that Lambiam is describing a rotation around the y axis. As the original question was about a rotation about the x axis, the specific details of this solution may not be relevant --- though the overall tactic is an excellent one. Tesseran 07:08, 16 March 2007 (UTC)
 * Right, my specific dissection was for a rotation about the y-axis, but the same approach with the roles of x and y swapped should work for the y-axis. A new point of intersection is then (1, 2). --Lambiam Talk  08:58, 16 March 2007 (UTC)

Hey guys, are you sure you talk about same rotation? Original poster says the area is rotated about the x-axis, Labmiam talks about replacing x with -x, wwich suggests rotation about y-axis... —CiaPan 07:23, 16 March 2007 (UTC)
 * See above. --Lambiam Talk  08:58, 16 March 2007 (UTC)

Nevertheless Lambian has identified the key point of this problem. The region bounded by the curves has two parts, one below the x-axis and one above the x-axis. Each of these parts sweeps out a solid of revolution and these two solids have a common part. The volume of this common part needs to be subtracted from the sum of the volumes of the solids of revolution in order to get the answer. Geoffcobra 23:19, 16 March 2007 (UTC)