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Lecture

7.2 volumes washers.pdf

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Department
Mathematics
Course
MAT237Y1
Professor
Dan Dolderman
Semester
Fall

Description
Calc 1 Wed 12/3/08 7.2 - Volumes of Revolution And now, just like how when after you ▯rst learned derivatives, you went on to applications of derivatives, we go on to applications of integrals. 7.2 is all about volumes of revolution using disks and washers. Before we talk about revolution, let’s ▯rst talk about volumes. Firstly, think about, say, an apple. It’s spherical, essentially. Now take a slice out of the apple. I don’t mean cut it into quarters or something as you normally would in order to eat it. I mean cut vertically (or horizontally) and then make another cut right next to the ▯rst so that you get out a thin piece of the apple. ...Or, if you’ve ever made an apple pie before, think about if you used an apple corer. What happened? As the apple goes through, it slices it up. In what shape are these slices? It’s a circle (essentially...it just has a little thickness to it that’s all)! If i were to keep slicing up that apple the same way (or just let it go all the way through the apple corer), what i’d end up with is a whole bunch of circles. That is to say, now if i work backwards, a sphere is made up of a whole bunch of circles stuck together. In a more general case, volume of any object can be thought of a bunch of slices (ie area) put together to form the entire object (its height). So, the \volume" of each of these slices is dV = [area of cross section] ▯ [height] Keeping this in mind, let’s think about a sphere of radius r. If you were to take a slice out of the sphere, it would look like a circle. Well, this slice is what happens when the sphere meets the xy-plane. Pretend I were to take a sphere and start to move it through the chalkboard. Half way through, I stop. Where the chalkboard intersects the sphere would be my slice, and it would be a perfect circle! What radius would this circle be? r (because it’s exactly half way through)! So, on the xy-plane, if i go out x and y, we have 2 2 2 x + y = r Well, now let’s take a look at the disk that would form the entire sphere. It goes veritcally and has a radius of y, and would be out x along the x-axis inside the circle. Why radius y? Because r is the radius of the sphere, and as we look at the di▯erent disks, they’re all changing in size, so the radius of each disk will be y (a variable) rather than r (a constant). So, imagine this disk with radius y. From our equation above, we have p y = r ▯ x 2 We only want the positive half because then we’d end up doubling everything from that disk. So, anyway, what would the disk thickness be (ie, height)? It would be dx. It’s in▯nitesimally small, but it still has some thickness to it (think aboutpwhen we did area and ▯x. Same deal, basically). The radius of this disk i already mentioned is y = r ▯ x . So, the volume of this disk is: 2 2 2 dV = ▯y dx = ▯(r ▯ x ) dx Why ▯y ? Because that’s the area of a circle! Our radius is y, remember? Why is it dV ? Because that’s only the volume of the disk, not the sphere (V would be the volume of the sphere). It’s considered to be changing as you move along in the x-direction. Ah, well, we have a di▯erential equation now! Let’s integrate! Z r 2 2 V = ▯(r ▯ x ) dx ▯r 1 We go from ▯r to r because that’s the radius of the sphere. Z r 2 2 = ▯ r ▯ x dx ▯r ▯ x3▯r = ▯ r x ▯ 3 ▯r ▯ 3 ▯ 3▯▯ 3 r 3 ▯r = ▯ r ▯ ▯ ▯r ▯ ▯ 3 ▯ 3 2r3 = ▯ 2r ▯ 3 4 = ▯r3 3 which is good, because that’s what we know as the volume of a sphere! So, let’s just recap a little here of what just happened. To ▯nd the volume of a sphere, we ▯rst looked at the area of a cross section (the disk) and then integrated it. In fact, that’s essentially what volume is: it’s the antiderivative of the area, or Z b V = A(x) dx a for A(x) the area function. The key here, however, is that your cross-sectional area has to be perpendicular to the x-axis. Or, in other words, the width has to be dx. If your width were dy, then you would have to think of everything in terms of y. But this is also thinking in terms of rectangles. It’s better to think in terms of circles. Why? Because when you revolve something about the x-axis, you’re dealing with circles all the time. So, let’s change that equation above just a little bit and instead think of this stu▯ in terms of f(x), and we get Z b 2 V = ▯[f(x)] dx a This is called ▯nding the volume using the disk method. Why? Because when you draw it, it looks like a disk! Notice how we have f(x) is the radius of our disk (if we were to call it r, we’d have 2 ▯r !). How can you picture volumes of revolution? Well, think about those halloween decorations where you have some cardboard shape and then you open it up and there’s that tissue paper stu▯ in between and you fasten the cardboard parts together? What did you start out with? area (the cardboard shape). what did you end up with? volume (all that tissue paper in between). How did you get to that end? By revolving the carboard shape! Example: Find the volume that results when f(x) = 1 is revolved about the x-axis between x x = 1 and x = 2. 1 Think about how this looks for a second. We start o▯ with our functx. We revolve this sucker around the x-axis, so our slice is going to have width dx and height f. Or, if you just want x to think of everything as plugging into formulas, then we get ▯ ▯ Z 2 1 2 V = ▯ dx 1 x ▯▯ ▯ ▯▯
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