SamBridge Posted March 30, 2014 Share Posted March 30, 2014 (edited) There's something about this thickness of a material to find the volume that I can't quite figure out. Let's say I have a shape with an Outer surface that's generated by rotating y=x^2 around the x axis, and its 2 inches thick all the way around, meaning that at any given point on the curve y=x^2, you can draw a line perpendicular to that point to reach a secondary curve that is only 2 inches away on the inside that has nearly the exact same shape. How do I describe this volume in terms of using a summation of hollowed cylinders? The greatest mystery of all time, only the best can solve it...cause no one can answer it... Edited March 30, 2014 by SamBridge Link to comment Share on other sites More sharing options...

John Posted March 30, 2014 Share Posted March 30, 2014 Are you sure you worded this problem correctly? I ask because, as written, it seems like a pretty straight-forward problem to solve using the washer method, but you seem to be implying it should be more difficult. Link to comment Share on other sites More sharing options...

SamBridge Posted March 30, 2014 Author Share Posted March 30, 2014 (edited) Are you sure you worded this problem correctly? I ask because, as written, it seems like a pretty straight-forward problem to solve using the washer method, but you seem to be implying it should be more difficult. it is. You can't use the washer method at this poiont because the distance between the large radius and the small radius isn't defined. All you can say is washer radius = x^2-(other radius), we need to use the fact that its always 2 inches thick to develop a formula for the curve of the smaller radius since it isn't defined for us already. Edited March 30, 2014 by SamBridge Link to comment Share on other sites More sharing options...

John Posted March 30, 2014 Share Posted March 30, 2014 (edited) Alright, I see what you mean now. Well, if we parameterize the parabola as x = t and y = t^{2}, then we have this handy formula for finding parallel curves, which gives us (assuming I've made no mistakes, and letting a = 2)[math]x = t - \frac{4t}{\sqrt{1 + 4t^{2}}}[/math] [math]y = t^{2} + \frac{2}{\sqrt{1 + 4t^{2}}}[/math] which is this funny-looking thing. Of course, a = 2 was arbitrary, just letting each unit along the y-axis represent one inch. We could use something like a = 1/18 (thus letting each unit along the y-axis represent a yard), which yields [math]x = t - \frac{\frac{t}{9}}{\sqrt{1 + 4t^{2}}}[/math] [math]y = t^{2} + \frac{\frac{1}{18}}{\sqrt{1 + 4t^{2}}}[/math] and a slightly nicer-looking curve, which when plotted along with the original parabola, seems to be what we want. Converting the parametric equation into a Cartesian equivalent is left as an exercise to the reader. :xEdit: A minor point: Also note that, in the last link, the graph is only plotted for t = -2 to t = 2, rather than t = -4 to t = 4. This is because the uneven scaling of the axes visibly screws with the apparent distance over larger intervals, and AspectRatio doesn't seem to work properly. Edit 2: Also note that our original result, using a = 2, also works, as we can see in this graph, but I think the second result, using the smaller value for a, will work more nicely for the original problem. Edited March 30, 2014 by John 2 Link to comment Share on other sites More sharing options...

SamBridge Posted March 30, 2014 Author Share Posted March 30, 2014 Looks like it's worth a try, thanks. Link to comment Share on other sites More sharing options...

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