Well, school is out and summer has begun. So I don't let the math part of my brain completely atrophy over the summer, I'm going to go back and fill in the blanks I left during the school year, and then I'll find something else math-related to write about.
I skipped over blogging 6.3 because things got chaotic around that time due to losing a week of school to snow days. So let's take a look at what we missed.
We were talking about finding the volumes of shapes you get by rotating a curve around a given line. We did this by cutting out disc or washer-shaped chunks out of shape by taking a slice that is perpendicular to the axis of rotation. That's all well and good, but we run into a problem if we take a curve like y= 2x^2-x^3. I don't really think there's an easy to draw on Blogger, so use your imagination or plug it into a calculator.
We could technically use the last method to solve this, but figure out the radius of the inner washer, we'd have to solve that equation for x. That's possible, but...ew.
But don't worry, because there's another way to go about this. When we were making the discs, we took a chunk out that was perpendicular to the axis of rotation. This time, let's take out a section that's parallel; from the top of the curve to the base. If we take out that section of a given width from the entire shape, we get a shell that looks like a thick tin can. Since this is a cylinder, we know that if we can find the volume of the entire section, and only take the thickness of the rim (subtract the volume of the big empty part) then we'll have the volume of the shell. Formula-wise, we'd have "(pi)r^2h - (pi)r^2h" (r of the first and r of the second, respectively. I don't know a good way to do subscripts.) With a little algebraic finagling, we can reduce that down to [2(pi)][(r2+r1)/2]h(r2-r1).
Since r2-r1 is the amount of change between the two radii, we can just call it ▲r.
So, we can boil it down to V = 2πrh▲r, or, as the book puts it, the volume is the circumference times the the height times the thickness.
(Remembered the pi code.)
So, let's try this on our scary little curve. We'll take a interval out of it (from a to b) and divide it into some number of equal intervals (n). We'll call the width of these intervals ▲x, the middle of it will be x (Italic x. I don't know how to put the line over it like in the book.) So we'll make a rectangle with the width ▲x and height f(x). Now we'll take that rectangle and spin it around the y-axis. When we do that, we get a neat little cylindrical shell that looks just like the one we were talking about earlier. We can say its average radius is the midpoint of the interval, it's height is f(x), and its width is ▲x.
If we take the sum of the volumes of all of the shells made from our intervals, we get somewhat close to the volume of the shape. If we keep splitting it up into more and more intervals, the estimation gets better and better. So, it seems reasonable to think that the limit of the sum of the volumes as n approaches infinity will be the volume of this shape. This turns out to be true, but we can't prove it yet with the stuff we know.
So, if we don't want to use the disc/washer method, we have a new formula.
V= ⌠(from a to b) (2π)x(f(x)) dx
So there we go. If the disc fails you, try the shell.
It's good to be back.
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