Disc/Shell Methods

<p>This is not a homework help post. I'm trying to self-study for AP Calc BC.</p>

<p>I am having trouble figuring out when to use the disk method vs. the shell method. I get that, in the disk, the representative rectangle is perpendicular to the axis of revolution, and in the shell it is parallel. However, I don't know how to draw the rectangle - horizontal or vertical? A friend said that it is vertical if f(x) = y (so x is in terms of y). But I found examples contrary to that in my calculus book, so I'm not sure if that is right (or if I'm misunderstanding the examples in the book).</p>

<p>Please help.</p>

<p>How do I know which to use?</p>

<p>You use whichever method will cause you less pain.</p>

<p>It's hard to explain without a picture, but if you're using the disk method you draw the rectangle such that when you rotate it about the axis of revolution you get a disk (think records from the 1960s), and if you're using the shell method you draw the rectangle such that when you rotate it about the axis of revolution you get a shell (really skinny cylinder).</p>

<p>That wasn't very useful, was it :)</p>

<p>I've always found shell easier, but it depends on the problem of course.</p>

<p>I generally always use shells for rotations about the y-axis and discs for the x-axis. I'm not sure what you mean by "draw the rectangle," but shells are nice for the y-axis because that way, you don't have to find f(y) or anything. You know the radius will be x so the volume will be 2<em>pi</em>r<em>h, or 2pi times the integral of x</em>f(x). Then for discs, the volume will be pi*r^2 or pi times the integral of f(x)^2.</p>

<p>Hmm, I hope all of that made sense/is correct. We did volumes of revolution a few chapters ago, so I'm not 100% sure anymore. :P</p>

<p>That makes sense! Thank you (all of you)! :)</p>

<p>I never realized I could really use both. I guess, when presented with a problem, I'll think about which will be easier. And, MathGirl, thanks. If given the option, I will use the disc method for x-axis rotations. Thanks. :]</p>

<p>when you are finding the volume of a solid of revolution, you are simply cutting up the solid into infinitely many differential pieces. To find the volume you simply sum the areas of each differential piece, by taking the integral.</p>

<p>The key to figuring out which method to use is by picturing it in your mind. Suppose you had a sine function from [0, pi] revolved around the x-axis. To find the area below the curve, you would divide the area into differential, vertical rectangles going from 0 to pi. Well now, you're not finding the area; you're finding the volume. So, just as the area is revolved around the x-axis to form a solid with a volume, why not revolve each differential rectangle around to form a circle (a disc), and then sum up all the circles to get the volume of the solid? You see where I'm going here? Each circle would be perpendicular to the x-axis and you'd be summing them up in the x-direction.</p>

<p>Now let's think about it a different way. Instead of dividing the area up into differential, vertical rectangles, why not divide it up into differential horizontal rectangles, and then summing the horizontal rectangles from the absolute max of the function in said interval, down to the x axis? But hey, we're finding volume... so we revolve said rectangles around. Now, can you picture what would happen? You'd get concentric shells, parallel to the x-axis. And to find the volume you'd have to find of the surface area of each shell, and sum them up in the y-direction.</p>

<p>Sorta a long post, but I've always found it easier to visualize this to conceptualize why things are, rather than straight memorization. But yes, shells are used parallel to the axis of revolution, and discs are used perpendicular to the axis of revolution.</p>