August 24th, 2009, 06:43 PM  #1 
Member Joined: Jul 2009 From: New Jersey Posts: 65 Thanks: 0  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 xaxis). 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 xaxis using the Midpoint Rule. 
August 25th, 2009, 01:46 AM  #2 
Senior Member Joined: Jan 2009 From: Japan Posts: 192 Thanks: 0  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)

August 25th, 2009, 07:25 AM  #3 
Member Joined: Jul 2009 From: New Jersey Posts: 65 Thanks: 0  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! 
August 25th, 2009, 07:53 AM  #4 
Member Joined: Jul 2009 From: New Jersey Posts: 65 Thanks: 0  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 xaxis, 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). 
August 25th, 2009, 12:54 PM  #5 
Senior Member Joined: Jan 2009 From: Japan Posts: 192 Thanks: 0  Re: Midpoint Rule?
Using your numbers: If the area is 21 (midpoint rule) and the width is 8 (from 102), 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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