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    Find the area enclosed by the curve r = acos \theta.

     A = \frac{1}{2} \int_0^{2\pi} a^2cos^2\theta d\theta

     A = \frac{a^2}{4} \int_0^{2\pi} cos2\theta + 1 d\theta = \frac{a^2}{4}[\frac{1}{2}sin2\theta + \theta]_0^{2\pi} = \frac{a^2 \pi}{2}

    The answer is half of mine which can be obtained easily using geometry, my question is; where did I miss a half in my integral?
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    (Original post by Ateo)
    Find the area enclosed by the curve r = acos \theta.

     A = \frac{1}{2} \int_0^{2\pi} a^2cos^2\theta d\theta

     A = \frac{a^2}{4} \int_0^{2\pi} cos2\theta + 1 d\theta = \frac{a^2}{4}[\frac{1}{2}sin2\theta + \theta]_0^{2\pi} = \frac{a^2 \pi}{2}

    The answer is half of mine which can be obtained easily using geometry, my question is; where did I miss a half in my integral?
    Just a thought (I haven't tried to draw it!) - do you need to go from 0 to 2pi to get a complete loop or will 0 to pi give you a closed loop?
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    I'd be inclined to go from -pi/2 to pi/2, and avoid the negative r's.

    But as davros said. Going 0 to 2pi is covering the curve twice.

    All hail the Daleks!
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    (Original post by ghostwalker)
    I'd be inclined to go from -pi/2 to pi/2, and avoid the negative r's.

    But as davros said. Going 0 to 2pi is covering the curve twice.

    All hail the Daleks!
    Yep that's probably a better argument - I didn't want to raise the spectre of negative r's as I think we all get confused whenever that comes up
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    (Original post by davros)
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    (Original post by ghostwalker)
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    Ahh, how did I miss this! :facepalm: Thanks.
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    (Original post by davros)
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    (Original post by ghostwalker)
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    One more little thing, how do I find the limits which will provide a single loop?
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    (Original post by Ateo)
    One more little thing, how do I find the limits which will provide a single loop?
    Clearly for theta=0, r > 0, then just work your way round positive/negative theta until you have r=0. Then you have your limits.
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    (Original post by ghostwalker)
    Clearly for theta=0, r > 0, then just work your way round positive/negative theta until you have r=0. Then you have your limits.
    Sorry, I wasn't clear enough, I meant how do I go about it generally. I think I best give the question.

    Find the area enclosed by the curve 2+3cos\theta.

    Some graphs of this:
    (1)(2)(3)

    The question seems to ask for double the area enclosed by graph (3). Here is what I did:

    r = 2 + 3cos\theta = 0
    \displaystyle A = 2 \cdot \frac{1}{2} \int_0^{cos^{-1}(\frac{-2}{3})} (2+3cos\theta)^2 d\theta

    \displaystyle A = \int_0^{cos^{-1}(\frac{-2}{3})} 4 + 12cos \theta + \frac{9}{2}(cos2\theta + 1) d\theta

    \displaystyle A =  [4\theta + 12sin\theta + \frac{9\theta}{2} + \frac{9}{4}sin2\theta]_0^{cos^{-1}(\frac{-2}{3})}

    A = \frac{17}{2}cos^{-1}(\frac{-2}{3}) + 12\sqrt{5/9} + \frac{9}{2} \sqrt{5/9}(\frac{-2}{3})

    A = \frac{17}{2}cos^{-1}(\frac{-2}{3}) + 3\sqrt{5}

    But the book has A = \frac{17}{2}cos^{-1}(\frac{-2}{3}) + 5\sqrt{5} as the answer.


    Also, I was able to find the limit because I knew what the shape of the graph was like.

    In general, when I have some more complex cases, is it safe to say that I should determine any angles at which r=0 in the original function and also along with these would be the points from this:

    y = 0 = rsin\theta \Rightarrow \theta = 0 , \pi, 2\pi, ... ,n\pi

    ...and from here I could decide on the limits?
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    (Original post by Ateo)
    In general, when I have some more complex cases, is it safe to say that I should determine any angles at which r=0 in the original function and also along with these would be the points from this:

    y = 0 = rsin\theta \Rightarrow \theta = 0 , \pi, 2\pi, ... ,n\pi

    ...and from here I could decide on the limits?
    Will look at your example later, as I'm a bit pushed for time at the moment.

    I'm not aware of any particular rules on how to go about this - at least I wasn't taught any.

    I suspect there is little substitute for sketching the curve.

    Noting where r=0 would be very useful. We also need to watch out for r < 0. As in your example.

    Will post on the example later.
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    (Original post by Ateo)
    A = \frac{17}{2}cos^{-1}(\frac{-2}{3}) + 3\sqrt{5}
    Can't see any errors in your working for this answer.

    And checked it on my graph package - your answer is correct.
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    (Original post by ghostwalker)
    Can't see any errors in your working for this answer.

    And checked it on my graph package - your answer is correct.
    Thanks. After a bit of thought I think that the best way to find the limits is to actually sketch the curve and as was pointed out take care when r<0.

    Mind me asking what software you use?
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    (Original post by Ateo)
    Mind me asking what software you use?
    http://www.padowan.dk

    It's free, and you can make a donation.

    Handles standard y=f(x), parametric, and polar, as well as inequalities and relations.

    Can supposedly handle complex numbers as well, but I've never taken the time to suss it out on that score.
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    (Original post by ghostwalker)
    http://www.padowan.dk

    It's free, and you can make a donation.

    Handles standard y=f(x), parametric, and polar, as well as inequalities and relations.

    Can supposedly handle complex numbers as well, but I've never taken the time to suss it out on that score.
    Quite nice and easy to customize. Thank you
 
 
 
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