Area of a volume of revolutionDate: Tue, 22 Nov 1994 08:05:27 -0800 (PST) From: Uma Chandavarkar Subject: Applications of the Integral Hello! I am a senior at Monta Vista High School in Cupertino, California. I am currently enrolled in Calculus. I am a little confused about finding volumes when graphs are revolved around an axis. I would specifically like to know more about finding the volume of a disk and the volume of a washer. Thank you!!!! - Uma Chandavarkar Date: 23 Nov 1994 16:19:22 GMT From: Ken "Dr." Math Subject: Re: Applications of the Integral Hello there Uma! You've asked about a great problem, that of finding the area of a volume of revolution. Well, there are essentially two different ways to do it, the shell method and the disk method. I'll go over the disk method with you. The first thing you need to do is draw a picture. I'm serious about this. You simply can't learn to do these problems unless you draw pictures of what you're doing. I'm afraid I can't really draw a picture in this message, but I can guide you through making one. Let's take the region below the curve y = x^3 (that means x cubed) and above the curve y = x^2, between the x values of 3 and 5. Draw the picture, and see what the region looks like. We're going to revolve it around the x axis, and find out what the volume of the resulting space figure is. So you should try to draw that too, if you can. Let's take that space figure and chop it up vertically with a ginsu knife, so that we get a bunch of little washers. The thickness of the washers is going to be REALLY SMALL, and we're going to call that thickness dx (as in the dx that sits out to the side of an integral). To find the volume of the washer, we have to remember the formula for the volume of a disk: it's Pi*r^2*h, right? What we'll do is find the volume of the disk made by rotating the region below y=x^3, and then subtract off the region below y=x^2. Does that make sense in your picture? Since the radius of one of these disks is just the height of the function, the volume of one thin disk under y=x^3 is going to be Pi * (x^3)^2 * dx. To get the volume of the whole region, we add up (i.e. integrate) the volumes of all the disks. So we'll have the integral from 3 to 5 of Pi*x^6 dx. I'll let you evaluate it. But we still have to subtract the region below the curve y=x^2. What's the volume of this region? I hope this helps! Ken "Dr." Math |
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