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    When is it alright to use this formula: \frac{1}{2} \int r^2\ d\theta

    For example, in this question (assuming I want to find the area of the curve r=1.5 +sin(3x) across its points of intersection with r=2) :

    Why do I have to draw lines from the points of intersection to origin and then integrate?
    Even without the lines, wouldn't I be summing up the area of a bunch of sectors?
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    I can't see any requirement to draw the lines in, unless the question asks you to; otherwise it's just an aid to help you see the area (integral) you're dealing with.

    If the shaded area (S) is the one in question, it will be the difference of two integrals.
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    Because it helps.
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    Don't you have to evaluate four integrals? Because you have to do find S from A, B and C (See my bad picture on paint... )
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    (Original post by soutioirsim)
    Don't you have to evaluate four integrals? Because you have to do find S from A, B and C (See my bad picture on paint... )
    A and C don't come into it.

    With appropriate limits:

    Evaluating r=2 function gives you the area B.

    Evaluating the other function gives you the area B+S.

    Don't forget, with polar co-ordindates the area in question is the one between two radii. I think you're getting muddled with cartesian a bit.
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    (Original post by ghostwalker)
    A and C don't come into it.

    With appropriate limits:

    Evaluating r=2 function gives you the area B.

    Evaluating the other function gives you the area B+S.

    Don't forget, with polar co-ordindates the area in question is the one between two radii. I think you're getting muddled with cartesian a bit.
    Aaahh yes I see. Thanks.
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    Thanks!
    and would I be able to use the formula to work this area out (from 0 to pi/2).
    http://www.wolframalpha.com/input/?i...in%283theta%29
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    (Original post by confuzzled92)
    Thanks!
    and would I be able to use the formula to work this area out (from 0 to pi/2).
    http://www.wolframalpha.com/input/?i...in%283theta%29
    You need to be careful here.

    Do you just want the areas under the curve in the upper right quadrant?

    If so, theta is from 0 to pi/3; if you go beyond pi/3, then r will be negative and you are on the bottom loop. Worth trying to sketch it by hand (plotting some points) to get the idea of it.
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    (Original post by ghostwalker)
    You need to be careful here.

    Do you just want the areas under the curve in the upper right quadrant?

    If so, theta is from 0 to pi/3; if you go beyond pi/3, then r will be negative and you are on the bottom loop. Worth trying to sketch it by hand (plotting some points) to get the idea of it.
    Oh right! Its from 0 to pi/3.
    So would it be all right to apply the formula?
    Or to be more specific, in which scenario wouldn't i be able to use it?
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    (Original post by confuzzled92)
    Oh right! Its from 0 to pi/3.
    So would it be all right to apply the formula?
    Yes the formula would be fine

    Or to be more specific, in which scenario wouldn't i be able to use it?
    Obviously you have to be careful with the limits. The only other scenarios I can think of, off hand, are to watch out for r going negative, for loops being traversed more than once; but these are general points that you need to be careful off when dealing with polar curves, rather than specific to working out areas with them.
 
 
 
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