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    (Original post by TeeEm)
    bottom is

    from 2pi/3 to 5pi/3

    or

    -4pi/3 to -pi/3
    Ah that makes sense. Is there a reason how you knew that the limits in the question would be associated with the upper half of the curve and not the bottom?
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    (Original post by PhyM23)
    Ah that makes sense. Is there a reason how you knew that the limits in the question would be associated with the upper half of the curve and not the bottom?
    absolutely
    the values of theta which produce the top half lie between -pi/3 and 2pi/3
    They are even marked in the picture
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    (Original post by TeeEm)
    absolutely
    the values of theta which produce the top half lie between -pi/3 and 2pi/3
    They are even marked in the picture
    I can see that they are marked in the picture, but what I'm asking is how you knew that, when you were working the limits out, they would be associated with the top half?
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    (Original post by PhyM23)
    I can see that they are marked in the picture, but what I'm asking is how you knew that, when you were working the limits out, they would be associated with the top half?
    because pi/2 which also is marked in the picture lies on the top half and happens to lie between -pi/3 and 2pi/3
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    (Original post by TeeEm)
    because pi/2 which also is marked in the picture lies on the top half and happens to lie between -pi/3 and 2pi/3
    Ah of course. This all seems perfectly clear now. Thank you for bearing with me; this topic has never clicked in my head until now.

    Thank you very much for your help
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    (Original post by PhyM23)
    Ah of course. This all seems perfectly clear now. Thank you for bearing with me; this topic has never clicked in my head until now.

    Thank you very much for your help
    no worries
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    (Original post by TeeEm)
    I have not revised parametrics
    What about your exam?????

    I expect more of you Tee.
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    (Original post by Louisb19)
    What about your exam?????

    I expect more of you Tee.
    I am sorry about that ... Hopefully nobody noticed
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    Why do you integrate the curve between A and B? Wouldn't you cut off part of the curve that is between B and E?
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    (Original post by Ano123)
    Why do you integrate the curve between A and B? Wouldn't you cut off part of the curve that is between B and E?
    you can do it this way too if you are not sure how parametrics work
    look at post 6
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    (Original post by TeeEm)
    you can do it this way too if you are not sure how parametrics work
    look at post 6
    How does parametric integration work exactly?
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    (Original post by Ano123)
    How does parametric integration work exactly?
    read the posts and look at the question and its solution and it should make more sense, assuming you have seen parametric before.
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    (Original post by TeeEm)
    read the posts and look at the question and its solution and it should make more sense, assuming you have seen parametric before.
    So what area would you be calculating if you integrated between A and E parametrically?
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    (Original post by Ano123)
    So what area would you be calculating if you integrated between A and E parametrically?
    the area between the curve and the x axis including the little section on the bottom right which you do not want
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    What I'm saying is why isn't the area calculated this Name:  image.jpg
Views: 16
Size:  56.3 KB
    when you integrate between A and E. If you integrated this using Cartesian coordinates (ignoring the bottom half of the ellipse) between A and E you would calculate the area shown above wouldn't you. But when switching to parametric integration it doesn't seem to be the same. I thought that the only difference between parametric and normal Cartesian integration was that you use the t value for the corresponding c values as the limits. And you integrate with respect to t (and of course multiplying brought by dx/dt). It seems here that there is something fundamentally flawed about how I'm thinking of it.
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    (Original post by Ano123)
    What I'm saying is why isn't the area calculated this Name:  image.jpg
Views: 16
Size:  56.3 KB
    when you integrate between A and E. If you integrated this using Cartesian coordinates (ignoring the bottom half of the ellipse) between A and E you would calculate the area shown above wouldn't you. But when switching to parametric integration it doesn't seem to be the same. I thought that the only difference between parametric and normal Cartesian integration was that you use the t value for the corresponding c values as the limits. And you integrate with respect to t (and of course multiplying brought by dx/dt). It seems here that there is something fundamentally flawed about how I'm thinking of it.
    ...Name:  k2.gif
Views: 11
Size:  12.0 KB
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    (Original post by TeeEm)
    ...Name:  k2.gif
Views: 11
Size:  12.0 KB
    I know.
    It would be fairly long winded to read at the best of times.
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    (Original post by Ano123)
    I know.
    It would be fairly long winded to read at the best of times.
    I am off to bed too now
    I am sure tomorrow many eager people here will help you follow this.
    Goodnight.
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    What area would you be calculating if you we're integrating from θ=-π/3 to say θ=π.
 
 
 
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