I've got a question where I have to find the area of a semi circle with the centre (2,0), radius 2 by using the Simpson's rule with 4 sub intervals. The problem is i'm having a hard time understanding how to use the simpsons rule. Would someone beable to explain it? thx

This may sound like a stupid question, but if the object is a semi-cirlce and the radius is known to be 2, then (apart from philosophical reasons) why bother with Simpson's rule at all?
What's wrong with: float A = 2.0f*2.0f*3.1415f;

Quote:Original post by Damon Shamkite
This may sound like a stupid question, but if the object is a semi-cirlce and the radius is known to be 2, then (apart from philosophical reasons) why bother with Simpson's rule at all?
What's wrong with: float A = 2.0f*2.0f*3.1415f;

If you want to implement Simpson's rule, how would you test that you are getting reasonable results? I would try to measure the area of some figure for which there is an explicit formula, so I have something to compare my resuts to. Of course, you need to get the formula right. :)

Quote:Original post by Damon Shamkite
What's wrong with: float A = 2.0f*2.0f*3.1415f;

This is based on the maths equation A = pi * r^2. But remember that this would work out the area for the entire circle. For example:

Finding the area of a circle with radius of 5cm would mean that using the formula the following answer would be reached:

A = pi * r^2
= pi * 5 x 5
= pi * 25
= 79 cm^2

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This will work out the area for the entire circle. For the area of half of the circe, just half the area.

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In the most evil scenario of circles, you may have to work out only a sector of a circle. The area for that equation is as follows:

A = (angle of sector / 360) * (pi * r^2)

So that for a 45 degree slice of the circle above would give the answer:

A = (angle of sector / 360) * (pi * r^2)
= (45 / 360) * (pi * 5^2)
= 0.125 * (pi * 25)
= 9.8175 cm^2

This can be proved to be correct as 9.8175 * 8 (for the whole circle) is 79 cm^2

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I hope this answers your question. If I have made an error, or have any questions just ask! [smile]