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August 24th, 2009, 06:43 PM   #1
 
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Midpoint Rule?

I got a homework problem that says to use the midpoint rule to approximate a volume of a solid, (finding an area of a given curve then rotated around the x-axis).

I looked up the 'midpoint rule' in my textbook and it only shows how to use it to approximate an area under a curve which I know how to do. But how do I find the volume of a function rotated around the x-axis using the Midpoint Rule.
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August 25th, 2009, 01:46 AM   #2
 
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Re: Midpoint Rule?

When your text uses the midpoint rule to approximate the area under a curve it is (and they may even say so) constructing a rectangle and taking that area. For the case of a solid of revolution, the midpoint rule will give you a value to use as the radius of a cylinder. Just plug that into the formula for the volume of a cylinder. (pi r^2 h)
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August 25th, 2009, 07:25 AM   #3
 
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Re: Midpoint Rule?

So its just like the Simpson's Rule that I learned in high school? Because I knew it was similar but didn't know exactly what it was because i have never heard of it .

Thanks for the reply!
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August 25th, 2009, 07:53 AM   #4
 
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Re: Midpoint Rule?

Okay so I used the midpoint rule, simpsons rule, and the trapezoidal rule to estimate the area under this curve, which was 21, 20, and 20, respectively.

It is rotated around the x-axis, from the intervals: x=2 to x=10

(NOTE: This is not a regular function, just a plain old curve that the author came up with but the range is clearly visible to be able to use the above methods.)

How do I use this to estimate the volume?

I kept trying to use the [pi]r^2 h, but my answer seems to be ALOT larger than the answer provided in the textbook (which says it to be 169).
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August 25th, 2009, 12:54 PM   #5
 
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Re: Midpoint Rule?

Using your numbers:

If the area is 21 (midpoint rule) and the width is 8 (from 10-2), then the 'radius' is 21/8, and the height is 8.

pi * (21/^2 * 8 is approximately 173. I'm assuming you did some rounding which would account for the difference from 169.
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