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Wednesday, March 4, 2009

Cylindrical Shells

Something important came up so this'll be brief and not too fancy, but I'll try my best to make it good (Don't see a reason why to be so detailed when everyone was in class.)

Anyways, we started by reviewing the disk...

and the washer method...
We talked it can be applied to real life situations and the number of questions it can be used. He used Boston as an example as the city grows in semi-circles. He also talked how these are the inverse of related rates problems and how it differs how we're trying to find out "how much" of a thing we have rather than how it's changing.

He returned to the "paper towel" idea so that we'd understand it better. At first we took a solid and rotated it to get a solid, but now we're taking arbitrary solids, and instead we'll get one piece of "paper towel". If we can get the volume of one of them, we can extend the idea of how the it's changing. This is called Cylindrical shells. So we'll just be adding up the shells.

So first we get the volume of one shell. If we open it up, one side of it will be 2*pi*r because it's about half of the circumference. The other will be the function, which is x^2 right now. the depth of it will be dx. To get the volume it will be l*w*h, so...

We have a problem though because one piece is a little like a trapezoid. If he took the end piece and put them together, you'd get a square. So the little piece will be delta little r square. It will get small because the change isn't 1. Even 0.1 is large. That squared is going to be .001, so it's still large, so if you square it again, you'd get .0001. It approaches zero, so the difference between them is negligible. As the change in 'r' gets smaller and smaller, they approach the same length. Because the area is delta r squared, we can regard it as insignificant, because the change is so small. So that difference, "r" can just be called x.

Mr. K tol us that the length will always be circumference (2*pi*r), and x will always be the radius of the shell, and height will be determined by any function that is present.

Anytime you rotate horizontally, circumference will be the length. The radius can be different things, for example, if rotated at -1. radius will be larger. if it's rotated around x = 5, the outside would have curves all around.

Mr. K informed us that the First question on the free response portion of the exam is commonly like this: Find the area between two functions, and that area will be revolved around the x axis, and then find the volume.

All of a sudden we started talking about the 'z' axis, which is an axis coming toward you and out. So the thickness on the z axis is zero... Always. It's the easiest haha.

We find 'S' is the area inbetween the two functions. We'll take S and rotate it. Intersects at zero and four. We then rotated it around the x axis. Then we took a slice and we got a washer and we found the radius was was f(X) and the other part of the washer was g(x) so to find the area we piBIGrsquared minus piSMALLrsquared, and then we found the area. *

We stop here because it's tedious, so then instead we put it into our calculator. Then we got 768pi/5

Next scribe is... Kristina. And remember, you can't talk to me, I'll be silent the whole day. :)

1 comment:

Grey-M said...

Good post, and yes Mr.K is right about that being the sort of question that will be thrown at you first BUT it does not have to be a solid of revolution, it can just be a solid. But try a question like this sometime (I'm sure Mr. K has a better one of these or at least much simpler): You have a solid with a circular base (radius 1) on the x axis, parallel cross sections of the base are equilateral triangles. What is the volume of this object? (This one is on the tougher side, if you can do this one you have no worries) Answer is ((4/3)sqr(3))