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06B-40
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Re: 06B-40
by thething2 » Tue Mar 11, 2008 8:05 pm
Oh! That one is an elaborate formula. It goes like this:
a^2+b^2=c^2
The long side is the hypotenuse, which is also the diameter of the circle.
The answer is sqrt(3460^2+5710^2).
a^2+b^2=c^2
The long side is the hypotenuse, which is also the diameter of the circle.
The answer is sqrt(3460^2+5710^2).
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Re: 06B-40- 08D59
by cwilli9000 » Wed Mar 12, 2008 10:30 am
sorry
posted wrong question number.
no need to be rude anyway. The question I meant was
08D-59
posted wrong question number.
no need to be rude anyway. The question I meant was
08D-59
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Re: 06B-40
by thething2 » Wed Mar 12, 2008 10:55 am
Well that one is slightly more difficult. The formula when it is rotation parallel to the x-axis is:
([f(x)-b]^2-[g(x)-b]^2)*pi
Integrate that formula and multiply by pi, and that is the answer.
Ok, for the example:
(4/x+5-5)^2-(5-5)^2
Integrate 16/x^2 from 1 to 6, multiply by pi, and that should give you the answer.
If it is rotation parallel to the y-axis, the formula is (x-a)(f(x)-g(x))*2pi
([f(x)-b]^2-[g(x)-b]^2)*pi
Integrate that formula and multiply by pi, and that is the answer.
Ok, for the example:
(4/x+5-5)^2-(5-5)^2
Integrate 16/x^2 from 1 to 6, multiply by pi, and that should give you the answer.
If it is rotation parallel to the y-axis, the formula is (x-a)(f(x)-g(x))*2pi
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