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Volumes

 
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Feb3-05, 10:09 AM   #1
 

Volumes


I'm having trouble finding the integral I'm supposed to use for some Volume problems...
Can someone lead me in the direction as to how I should form my integrals to get the solutions?
The below is a scanned page from an AP Calculus textbook, I'm pretty much stumped on how to solve 56-59..
Hope someone can help.
Thanks.

http://www.brokendream.net/xh4/apcalcscan.jpg
 
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Feb3-05, 10:47 AM   #2
 
I'll give you some hints on the first one (56a). They're all pretty much the same. The side of the squares are determined by the difference between the two functions y=x+1 and y=x^2-1. This difference is x-x^2+2, it is zero for x=-1 and x=2. So now you have determined the shape of your base.

With this you can easily find the area of such a square. Integrating over x gives you the total volume.
 
Feb3-05, 01:36 PM   #3
 
OK, so you merely evaluate the integral [tex]A (x) = \int_{-1}^\2 2 x - x^{2} + 2 dx[/tex]?
Does anybody have a clue about the other questions?
Thanks.
 
Feb4-05, 01:31 AM   #4
 

Volumes


Well, if you want to evalute the area enclosed by the two lines (y=..) yes, but...
 
Feb4-05, 03:27 AM   #5
 
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Quote by Pseudo Statistic
OK, so you merely evaluate the integral [tex]A (x) = \int_{-1}^\2 2 x - x^{2} + 2 dx[/tex]?
Does anybody have a clue about the other questions?
Thanks.
[itex]x - x^2+2[/itex] gives the length of one side of the square as a function of x. You need the area of the square.

b) Is somewhat easier, since the height of each rectangle is one, that means the area of a cross sectional rectangle is [itex]x-x^2+2[/itex].
 
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