Solids of Revolution

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Published on March 2, 2009

Author: k_ina

Source: slideshare.net

 

Hello everyone. Kristina here! Just thought I’d try a new approach with my usual scribe posts. Since I have more time to do this as, might as well make it fun. *Giggles at 1 st slide picture XD*

Hello everyone. Kristina here! Just thought I’d try a new approach with my usual scribe posts. Since I have more time to do this as, might as well make it fun. *Giggles at 1 st slide picture XD*

A review from last class…

And more review… Here, we reviewed from last class how to find the volume of this cone using calculus. See Paul’s magnificent explanation as he goes into further detail (Since his scribe completes this example XD)

Introduction to Washers! Looking at these two functions from a 3D perspective would show you that there is a hole in the middle. TIP: If you’re having trouble seeing these in the 3 rd dimension, check out Dr. E’s java applet link about two posts below this one :D

Introduction to Washers! What do you do when there’s a hole in the middle? Well, its basically the same idea as the cone example we had. The only difference is the fact that you will now be dealing with two circles. One within the other.

Introduction to Washers! First, cut your washer into pieces, like with the cone. Now you can work with those individual cylinders. We then took a cross section of one of the pieces and examined what we had. As you can see, subtracting the two areas shall give us the area of the washer.

Introduction to Washers! Confused on how to get the radii of the two circles? Well let me show you! Like the cone, the radii for each circle is basically the y-value at whatever point of x you’re looking at. Knowing this, you can now assign R and r their perspective functions. R r

Introduction to Washers! From there it’s a piece of cake. All you have to do is make your definite integral along the interval you are looking for, in this case [2,4]. Now you can solve the area of the washer :D R r

Let’s look at something else… Taking a closer look, you can see that these two functions are sinusoidal, one being a reflection over the x-axis of the other. Since they are reflections of each other, that means that we can use any of the functions and use them as the radius. The only thing that needs to be mentioned is the fact that if you take the negative one, you have to make it positive since radii cannot be negative.

Let’s look at something else… Now you can use the general rule that any function rotated about the x-axis will have circular cross sections. So just use the area of a circle again and then integrate along the interval it is asking for. In this case, A to B. Voila!

Where art thy hole…? Okay, bear with me on this slide… I’m not exactly sure what it was trying to say since I wasn’t sure whether it was using y = 0 as its rotation point, or y = -1. Anywho, if it was indeed y = 0, then the cross section on the side is wrong. There wouldn’t be a hole because if you use your imagination, you’d see that its just a…filled donut..filled..with…stuff..

There it is :D Now, if it had been y = -1, then the diagram on the side is right, as shown in this slide. Holed donut FTW. Anyways, to find the area of this washer, it’s the same routine when working with washers. Take the area of the outer circle and subtract the area of the inner circle.

Intersections :o Then, in this case. Find where y = x² and y = 4 intersect, since the two values you get will be the interval we are going to integrate along. This is like when we were trying to find the area between curves. The only difference is the fact that we are looking at these objects in the 3 rd dimension. Otherwise, same dealio.

Now solve… Now we can do the usual procedure, which is top minus bottom function. Don’t forget that we are working with circular cross sections, so we’ll be using the area of a circle to integrate. Just like in the last washer example we had. From there, its basic calculus integral work.

Boiled egg..yum… Now in this example, has y = x² being rotated over the line y = 4. As for your radius, its basically top minus bottom again. Thus the 4 - x². Make your integral along the interval, which you found by getting the intersections, and then solve. Bon apetite.

Last minute doodles… This example was taken right out of the text. Check out example 5 of chapter 8.2 if you don’t understand my explanation XD. Anyways, using your imagination to picture the revolution over the x-axis, you would see that this is another washer! *Gasp* To begin, you should start by finding the intersections. This will be your interval, like the last examples.

Last minute doodles… Now that you’ve got your interval, cut your washer pieces again and look at the cross sections. Lookie, its outer minus inner again! To find your radius of the outer, its basically x+2. The inner would then be x². Integrate along your interval and solve. Magic, you’re done!

Well, I’m done for today. That was fun, using a slideshow approach and all. Seemed so much neater. Anywho, I’m not sure if you were able to follow me on this scribe post since I was kind of lost on some parts as well, but I tried my best. As for next scribe, I choose you…YINAN!

Well, I’m done for today. That was fun, using a slideshow approach and all. Seemed so much neater. Anywho, I’m not sure if you were able to follow me on this scribe post since I was kind of lost on some parts as well, but I tried my best.

As for next scribe, I choose you…YINAN!

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