Surface Area of Solid of RevolutionDate: 05/21/2001 at 09:32:13 From: Stan Winston Subject: Alternate Surface Area of Solid of Revolution Formula My teacher mentioned that we were going to be looking at surface areas of solids of revolution shortly after we did volumes of rotation. So, before checking it out in the book, I tried to derive it myself. I figured that area is just height times width, so why not take the integral of the circumference of the solid of revolution? When the multiple circumferences are accumulated by integration, it's the same as adding the areas of multiple cylindrical strips with infinitely small widths, which should give me the surface area. For example, the surface area of the cone found by rotating y = x around the x-axis from x = 0 to x = 4 should be: 2*pi*(integral of x dx) from 0 to 4 Unfortunately this doesn't give me the answer I know to be correct, as you have to use the length of a curve formula within the integral for it to work. In fact, the only solid that I've found my formula to work for is the cylinder. What's wrong with my analysis? Why doesn't it work? Date: 05/21/2001 at 13:07:48 From: Doctor Rick Subject: Re: Alternate Surface Area of Solid of Revolution Formula Hi, Stan. You've done a pretty good analysis of your own problem. All the pieces are there for you to see why your formula doesn't work. Consider the strips or sections of surface whose areas you want to sum - before they are reduced to infinitesimal width. These strips are not cylindrical in shape (as your formula assumes), but conical. (That is, they are shaped like sections, or frustums, of cones.) This explains your observations: in the case of a cylinder, the strips are indeed cylindrical, and so your formula works, but for any other shape - and most obviously for cones - the strips are conical. The situation is analogous to the curve-length formula: following the same approach in that case, you would just integrate the width of each infinitesimal segment, giving the very uninteresting integral[dx] = x. The length along the curve of the sloped segment, sqrt(dx^2 + dy^2), does NOT approach dx as dx goes to 0; rather it approaches sqrt(1+(dy/dx)^2)*dx. Therefore this factor must be retained in the curve length formula. In the same way, the surface area of a conical section does not approach the surface area of a cylindrical section as dx goes to zero. The difference in the radii at the two ends of the conical section is second-order in dx, so it does go away as dx goes to zero. The length perpendicular to the circumference, however, is not dx but the length along the curve; thus the same factor sqrt(1+(dy/dx)^2) must be retained in the formula. The product of the (mean) circumference of the segment and the length along the curve gives the area of the conical section, as dx goes to zero. You might take a look at the answer in the Dr. Math archives concerning the derivation of the surface area of a sphere. This answer does not use calculus, but it shows the conical section I am talking about, and compares it with a cylindrical section. Volume of a Sphere http://mathforum.org/dr.math/problems/banijamal05.28.99.html - Doctor Rick, The Math Forum http://mathforum.org/dr.math/ Date: 05/21/2001 at 13:23:30 From: Doctor Rob Subject: Re: Alternate Surface Area of Solid of Revolution Formula Thanks for writing to Ask Dr. Math, Stan! Don't feel upset with this. It is a common error. In fact, I fell into the same trap when I first saw this stuff. Let delta-x and delta-y be small, so that the curve is very well approximated by a straight line near (x,y). Then you have the following picture, where the diagonal line "is" the curve: (x+delta-x,y+delta-y) _,o _,-' | delta-s _,-' | _,-' |delta-y _,-' | _,-' t | (x,y) o-----------------------o delta-x (x+delta-x,y) The problem is that the ratio of the length delta-s of a small segment of the curve being revolved to the length delta-x of its projection on the x-axis does not approach 1 as those lengths approach zero. In fact, it approaches the secant of the angle of inclination t of the tangent line there. That also means that the limit of the ratio of the lateral area of the frustum of a cone generated by revolving the segment delta-s about the x-axis to the area of the cylinder generated by revolving the segment delta-x about the x-axis is also not 1, but in fact also equals sec(t). Now dy/dx is the slope of the tangent line, which is the tangent of that same angle t. Now recall the trigonometric identity: tan(t)^2 + 1 = sec^2(t) Putting this all together the limit of the ratio delta-s/delta-x is: ds/dx = sqrt(1+[dy/dx]^2) In general, this is not equal to 1. It is equal to 1 if the tangent line is horizontal throughout the interval, that is, the curve is a horizontal line and the surface is a cylinder. Then the surface area is given by the integral b S = INT [2*Pi*y*ds] (ds, not dx !!) x=a b S = INT [2*Pi*y*(ds/dx)*dx] x=a b S = 2*Pi*INT [y*sqrt(1+[dy/dx]^2)*dx] a If this is not clear, or you want to discuss the matter further, please feel free to write again. - Doctor Rob, The Math Forum http://mathforum.org/dr.math/ |
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