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AP Calculus AB

An interactive log for students and parents in my AP Calculus class. This ongoing dialogue is as rich as YOU make it. Visit often and post your comments freely.

Friday, March 03, 2006

Solid by Revolution

In today's class we started with a quiz of course, just like every day. We then went on to practice integrals by solving problems using disks and cylindrical shells. One such problem is what would be the volume of a solid obtained by revolving the region bounded by the lines y=9-x^2 and y=-7 around the line y=-7. The points of intersection of the two functions are at x=-4 and x=4.

Using the disk method is the easier method for this particular problem. Because a disk can be thought of as a cylinder and the equation for a cylinder is V=(pi)hr^2, if we were to think of the radius of the disk as the value of the function +7 and if the height of each disk was an infinitely small amount, we could piece together a function for each individual disk as V=(16-x^2)^2 dx.
Now if we were to take the integral of that function with the limits of integration as the points of intersection of the functions (-4 to 4), it would give us the volume of the solid.

The cylindrical shell method is a little more difficult than that however. To get a cylindrical shell, we would have to evaluate the integral as a function of y not a function of x. Therefore we should rearrange our parabola for y, which is y=sqrt(9-y). So now that we have that we need to think of the equation of the volume of a cylindrical shell, the volume being the circumference multiplied by the height and width. Because the distance between the large and small radii will be so insignificant, we can think of the width as dy. The equation for circumference is 2(pi)r, and because r is the distance between y=-7 and the function value, r can be thought of as (y+7). The height of each individual shell will be the value of x at each y value multiplied by two because it is a parabola and it's simetrical, so h=2sqrt(9-y).
So now if we put the two together, we get the equation V=2(pi)(y+7)2sqrt(9-y) dy
Now we integrate the function with the limits of integration being all the y-values between the line y=-7 and the maximum function value (-7 to 9) and that will give us the volume of the solid.

Closer to the end of class Mr. K gave us a stencil with similar problems on it.
(Sorry about my lack of the proper characters, the system I'm on at the moment has very limited resources.)

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  • At 5:28 PM CST, Blogger Kramer said…

    You know, I actually remember doing this stuff in my Calculus class when I was in high school. Of course, that was the end of my senior year, so I don't remember many of the specifics, but I'm impressed by the way you seem to grasp the content. I'm curious to know how you all feel about the quizzes you do at the beginning of each class. Do you find these helpful in that you need to remember what you have been working on in class, or do you simply find them a hassle that hurts your grade?


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