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Trouble finding the integral for volume

by Pseudo Statistic
Tags: volumes
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Pseudo Statistic
#1
Feb3-05, 10:09 AM
P: 390
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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da_willem
#2
Feb3-05, 10:47 AM
P: 599
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.
Pseudo Statistic
#3
Feb3-05, 01:36 PM
P: 390
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.

da_willem
#4
Feb4-05, 01:31 AM
P: 599
Trouble finding the integral for volume

Well, if you want to evalute the area enclosed by the two lines (y=..) yes, but...
Galileo
#5
Feb4-05, 03:27 AM
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Quote 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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