Tuesday, February 06, 2007

Lesson 6.2: Volumes

I would like to apologize in advance for my computer being a pain in the neck. I spent like an hour typing up all these amazing equations in Word and the darn thing won't let me upload. Grr. I'm sorry if it gets confusing without graphs and stuff...I really am...

Ok, so here goes. So far with integrals, we have learned how to find the area under a curve (between the curve and the x-axis) and the area between curves. And now, for something completely different...we are going to fine the volume of solids using integrals.

Don't panic! It's not so bad...

Ok so firstly, let's think of the awesome $500 shape thingies that Mr. French showed us. Remember the one that had a circular base and had sides consisting of squares? Visualize it...because I don't have one for you...

Let's think way back to geometry for a moment. Remember how we learned how to find the volume of stuff like cylinders and prisms? When the volumes had fixed side lengths, it was simple. However, since this is calculus, it needs to be a little bit more complex.

With geometry, we could find the area of, say, a cylinder by multiplying the area of the circular base by the height of the cylinder. We can apply this same principle to calculus using integrals.

If our base is a circle, we can use the formula for area (A = pi x radius^2) and multiply it by the height...but what about when our height isn't a constant? This is where integrals come into play. For simple area of a region, the base was a constant, but the height was constantly changing, so we used the small rectangles to compute the area. With solids, it's similar, but we're using prisms.
The book definition for the volume of a solid (S) that lies on the xy - graph between x = a and x = b is as follows:

V = the limit as n approaches infinity is the sum of n where i =1 of A(x sub i) times delta x = integral from a to b of A(x)dx

if the cross-sectional area of S that lies in the plane Px, through x and perpendicular to the x-axis is A(x), where A is a continuous function.

Whew! That's a lot. But what does it mean? Well think back to the solid that was a circle with square cross-sections. The area of the base never changed, but the height was not a constant. The height was a function of the length of the cross section, which was a function of x. So we can find the area of the solid by finding the integral of the height and multiplying it by the base. Then we integrate to find the volume.
V = (integral from a to b) A(x)dx

So let's think back to the shapey thingy. The area of the base is [x^2 + y^2 = r^2] and the area of a square is s^2. s = 2y where y = plus-or-minus (root of r^2 - x^2). So s = 2 times (root of r^2 - x^2), and s^2 = 4(r^2 - x^2)
We then integrate:
integral from -r to r of s^2 dx
integral from -r to r of 4(r^2 - x^2)dx
2 [integral from 0 to r of 4 r^2 - 4 x^2 dx]
2(4 r^2 x - 4/3 x^3) from 0 to r
8 r^3 - 8/3 r^3

And voila!

We also learned about solids of revolution. Let's start with a graph of y = x^2
[sorry about not having the actual graph!]
And let's take this graph from zero to one.
We also need to set a y-boundary, so let's set it at y = 1

To create the solid of rotation, we set an axis of rotation (such as x = 0) and rotate the area of the graph around it. We will get a funky shape that resembles a ring. But it's not exactly cylindrical, so how can we calculate the volume? Same way as we did with the other shapes...we can break it into infinitely small prisms.

So with solids of rotation, we first need to ask one question: does the original shape share a side with the axis of rotation? Usually if the axis of rotation also acts as a boundary of the given area, the answer is yes, then we are dealing in terms of disks.
The general formula for solids of rotations with disks is:
integral from a to b of pi r^2 dx

It's the same principle as that behind the previous solid, just with a different basic formula (pi - r - squared times heights instead of length-times-width-times-height). But what if the graph doesn't use the axis of rotation as a boundary? Like...with the graph of x^3, when it is bound by x = 1 and y = 1

As you can imagine, when you rotate the area it creates a big empty space in the middle, and if you divided it into cross-sections you would end up like something resembling a very well-sliced bagel. That's ok, though, because we can still find the volumes because of the differences in radii length. So, when we're using washers, it looks like this.
integral from a to b of pi R^2 - r^2 dx
when R is the outer radius and r is the inner radius.

See, when finding the volume of the little cylinders, we can subtract the volume of the inner cylinder (aka the hole) from that of the big cylinder. ***Big important note: Subtracting the volumes is NOT AT ALL the same as using the difference between radii to find a volume. You WILL get the wrong answer if you do this*** (This is just in your best interest, because I know I've tried it...)

The last thing I think I need to remind you of is the use of axes and dx vs. dy. You can do this in terms of either, since your formulas are going to be set equal to each other (y = x^3, x^2 + y^2 = 4), but your "slices" (which are expressed by the derivative) should always run perpendicular to the axis of rotation. That will help you determine which one to use.

In closing, since I cannot upload a comic like I usually do, I leave you with this:

http://www.math.hmc.edu/calculus/tutorials/volume/

This is probably a better tutorial than I can accomplish in my current state, so please use this well. It also includes shell stuff...

Oh, and some of you may have overheard me whining about my existential crisis brought on by physics, so as if I can't get any more confused about what else I thought were constants in life, check this out:

http://www.math.hmc.edu/funfacts/ffiles/30001.1-3-8.shtml

Crazy stuff...Lauren, you're up next...oh wait, you already know, never mind...

This is priceless...I couldn't resist...this one goes out to Gianna especially...

Two mathematicians were having dinner in a restaurant, arguing about the average mathematical knowledge of the American public. One mathematician claimed that this average was woefully inadequate, the other maintained that it was surprisingly high.
"I'll tell you what," said the cynic. "Ask that waitress a simple math question. If she gets it right, I'll pick up dinner. If not, you do." He then excused himself to visit the men's room, and the other called the waitress over. "When my friend returns," he told her, "I'm going to ask you a question, and I want you to respond 'one third x cubed.' There's twenty bucks in it for you." She agreed.
The cynic returned from the bathroom and called the waitress over. "The food was wonderful, thank you," the mathematician started. "Incidentally, do you know what the integral of x squared is?"
The waitress looked pensive, almost pained. She looked around the room, at her feet, made gurgling noises, and finally said, "Um, one third x cubed?"
So the cynic paid the check. The waitress wheeled around, walked a few paces away, looked back at the two men, and muttered under her breath, "...plus a constant."

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