<p>On a recent MAO competition, one of the problems asked (paraphrased):</p>
<p>Find the volume of the solid created when the area between circles of radii 2 and 3 both centered at (4,4) is rotated around the y axis.</p>
<p>This is impossible to do in the allotted time (avg of 2 min/problem for this test) using conventional high school disk/washer/shell methods (and you're not allowed a calculator). Obviously there has to be an easier way.</p>
<p>Method 1) Use the formula for the volume of a torus (which is derived from another theorem that 99% of HSers don't know). This is the official explanation, and an unsatisfactory one at that.</p>
<p>Method 2) A friend of mine came up with the right answer by multiplying the inner circumference of the torus by the area of the annulus formed by the concentric circles. Is this coincidence or does it have some mathematical merit?</p>
<p>Thought this would be a good time killer for CC's best and brightest :)</p>
<p>Indeed it does have a mathematical explanation, though it's not really a "trick," just a fairly standard theorem you don't learn til later (multi, maybe? don't really know). Look up the "Theorem of Pappus" about relating the volume of solids formed by moving a cross-sectional area along some path to the distance traveled and the cross-sectional area.</p>
<p>^ Yeah that's the theorem used to derive the formula for the volume of a torus. </p>
<p>Well I'm trying to do this problem with shells, since it's impossible to do with washers. I've split the region horizontally using the line y = 4. If I find the volume of the U-pipe shape top, I can just double it. Basically it's splitting two circles into their respective equations, breaking the annulus into 3 parts, and integrating. Egad.</p>
<p>I think that therom is pretty easy to figure out for this case, since the cross-section is symmetric though, so that's probably the intended solution.</p>
<p>the torus method really isn't that difficult... isn't it just the volume of the bigger one rotated around the axis minus the smaller one... which works out to be V = pi(R^2-r^2) and then intergrate it... but maybe thats just our math's syllabus cause we had a WAAAAY harder question then that in our final... i wanna try and work it out now haha. Actually miss maths... which makes me a complete nerd but oh well.</p>
<p>@beeish: that method would imply washers. Using washers is impossible with this problem. Because these are circles, you can only deal with one half at a time. You can't find the volume of one half rotated (in terms of y) and then doubled because the inner half has a smaller volume than the outer half.</p>
<p>@ajwchin: I'm not sure. You're dealing with three separate regions, all with different limits of integration, then you'd have to double the volume. There's obviously a calculus based method to derive the formula, but it's probably more advanced than the AB/BC curriculum.</p>
<p>Yeah, washer method is normally epic fail on any volume problems on contests...I wonder if there is a volume extension for the shoelace theorem...</p>
<p>It may be easier if you replace X and Y with Sin(theta) and cos(theta) and then integrate with respect to theta, just a thought. No clue if its possible or not though.</p>
<p>like, theta would be the angle as it rotates around the y axis, d(theta)/dr would by a little area slice of the taurus, and then something else...</p>
<p>^ There's a way to get it in terms of just x. Bringing sin and cos into this would make it more complicated. I just want to set up the integral right, then fnInt to see if it's right.</p>
<p>If anyone's curious, this is what I have so far (calculator notation)</p>
<p>fnInt (2 pi r sqrt(9-(x-4)^2), x, 1, 2) + fnInt (2 pi r (sqrt(9-(x-4)^2) - sqrt(4-(x-4)^2)), x, 2, 6) + fnInt (2 pi r sqrt(9-(x-4)^2), x, 6, 7)</p>
<p>I'm not sure exactly how to get r for each subregion. Is it just x - axis of rotation (x - 0 = x) for each one?</p>
<p>It's not that bad, just find the specs of the bigger torus when pulled out into a cylinder, then its volume, then ditto for the little one in it, and subtract.</p>
<p>2 pi [fnInt (x sqrt(9-(x-4)^2), x, 1, 2) + fnInt (x (sqrt(9-(x-4)^2) - sqrt(4-(x-4)^2)), x, 2, 6) + fnInt (x sqrt(9-(x-4)^2), x, 6, 7)]. Not sure how to integrate, I know u-sub won't work and I don't think parts will work either.</p>
<p>Another thing I tried was this (I took one semicircle and subtracted another semicircle):</p>
<p>2 pi [fnInt (x (sqrt(9-(x-4)^2)-4)) - fnInt (x (sqrt(4-(x-4)^2)-4))]</p>
<p>And then multiply by two. See if that makes sense.</p>
<p>@Nick017: Big torus = 2 pi R * r = 2 pi 4 * 3 = 12 pi
Little torus = 2 pi 4 * 2 = 8 pi</p>
<p>So 4 pi.</p>
<p>Wow I feel stupid now. (I think I did that right).</p>
<p>I got 5 pi using the method I talked about before
you just move the circles to centre at (0,0) cause as long as the circles are the same size... their volume would be the same whether they were centered at (0,0) (4,4) or (199999,837483)</p>
<p>What do you get moving the circles to the origin? You still have to rotate them around a vertical line 4 to the left. The formula you posted before is just the area of the bigger one minus the area of the smaller one.</p>