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    (Original post by otrivine)
    ok

    what would you integrate then can you show me
    I gave you the requisite formula in my first reply post - just substitute the expressions for y and x into there.
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    (Original post by Smaug123)
    I gave you the requisite formula in my first reply post - just substitute the expressions for y and x into there.
    oh so we integrate both x and y ! so is d(x) there to mean you integrate x also right
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    (Original post by otrivine)
    ok

    what would you integrate then can you show me
    Smaug123 gave you the formula in post 4 - just plug the theta limits into this.
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    (Original post by otrivine)
    oh so we integrate both x and y ! so is d(x) there to mean you integrate x also right
    Stop guessing and look at the formula that you will have in your book (with examples) and from Smaug earlier on
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    (Original post by otrivine)
    oh so we integrate both x and y ! so is d(x) there to mean you integrate x also right
    Er, sort of. What you have currently is y and x both in terms of some parameter \theta. In order to integrate y with respect to x (which is what you want to find) rather than with respect to theta, you need to use what amounts to the chain rule in order to change variables from theta to x.
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    (Original post by Smaug123)
    Er, sort of. What you have currently is y and x both in terms of some parameter \theta. In order to integrate y with respect to x (which is what you want to find) rather than with respect to theta, you need to use what amounts to the chain rule in order to change variables from theta to x.
    Oh, right cause in the forumula you put on post 4, you put the sub method right, cause I thought that my forumla could work
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    (Original post by otrivine)
    Oh, right cause in the forumula you put on post 4, you put the sub method right, cause I thought that my forumla could work
    It is the same "formula"

    You need to understand what "integration with respect to a variable" means so that you do not think that you can just pretend that x is theta
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    (Original post by otrivine)
    Oh, right cause in the forumula you put on post 4, you put the sub method right, cause I thought that my forumla could work
    Your formula would only work if y were in terms of x, because you're integrating with respect to x. There are two things you can do to fix it: you can actually put y in terms of x (by eliminating theta from the equations) or you can use my post-4 formula, which essentially turns dx into d(theta) and means you integrate with respect to theta.
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    (Original post by Smaug123)
    Your formula would only work if y were in terms of x, because you're integrating with respect to x. There are two things you can do to fix it: you can actually put y in terms of x (by eliminating theta from the equations) or you can use my post-4 formula, which essentially turns dx into d(theta) and means you integrate with respect to theta.
    Bingo! will it be possible to get a cartesian equation to combine x and y and then integrate with the area formula, or you know what Ill stick with the sub method , sounds easier
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    (Original post by otrivine)
    Oh, right cause in the forumula you put on post 4, you put the sub method right, cause I thought that my forumla could work
    Your formula does work - but you have to keep everything in terms of one independent variable. You either work out the Cartesian equation of the curve and write y in terms of x and then integrate y(x) with respect to x, using the x-limits; OR you write y in terms of theta, convert dx to d(theta) using dx/d(theta) and change your limits to the appropriate theta-limits.

    You should have some examples of this technique in your book
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    (Original post by otrivine)
    Bingo! will it be possible to get a cartesian equation to combine x and y and then integrate with the area formula, or you know what Ill stick with the sub method , sounds easier
    Both are possible and, indeed, not hard if you know what sin(arccos(x)) is. The method I called neater is neater, and more generally applicable.
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    (Original post by otrivine)
    Bingo! will it be possible to get a cartesian equation
    If it had wanted you to take this approach it would probably have asked you to find the cartesian equation earlier in the question
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    (Original post by davros)
    Your formula does work - but you have to keep everything in terms of one independent variable. You either work out the Cartesian equation of the curve and write y in terms of x and then integrate y(x) with respect to x, using the x-limits; OR you write y in terms of theta, convert dx to d(theta) using dx/d(theta) and change your limits to the appropriate theta-limits.

    You should have some examples of this technique in your book
    Thank you so much Davros, yes sorry I got it now!
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    (Original post by TenOfThem)
    If it had wanted you to take this approach it would probably have asked you to find the cartesian equation earlier in the question
    I am so awful at maths, I just want to get it done, in Jan 2013 I got 75/100 for C3!
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    (Original post by Smaug123)
    Both are possible and, indeed, not hard if you know what sin(arccos(x)) is. The method I called neater is neater, and more generally applicable.
    Thank you and for the limits! Can you just say how you did cause I did get one of the limit which was 5 root2
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    Okay a lot of you would probably do it differently but I did this to find the intercept of the oval thing...

    y=4sin(angle)
    y=0 a the x intercept
    angle = sin^-1(0)
    angle = 0

    now I can find x... sub the angle into the other formula x=5cos(angle)
    x=5cos(0)
    x=5

    Guys, is there actually an alternative or better way of doing this ?? I know this isn't the most reliable method

    Now you can find area of triangle (well you could do that before without finding the x intercept value):

    area of triangle = 0.5 x 2root(2) x [5root(2) - 5root(2)/2]

    integrate the parametric equations ....

    I would then integrate y dx/d(angle) d(angle)
    I would change the limits by subbing in 5root(2)/2 & 5 into x=5cos(angle)
    This will give me limits

    Finally when I get the area, I will subtract from the area of the triangle found previously

    Okay now I don't know if that's the best way of doing it ! Or if I've just made it sounds complicated

    Sorry I really need to learn how to use latex
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    (Original post by otrivine)
    Thank you and for the limits! Can you just say how you did cause I did get one of the limit which was 5 root2
    Not if you're integrating with respect to theta… if you're integrating with respect to x, then that sounds plausible.
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    (Original post by otrivine)
    Thank you so much Davros, yes sorry I got it now!
    No problem! Once you've done a few of these you'll probably recognize that the curve you're integrating is actually an ellipse...and you've probably seen the cartesian equation for that!
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    (Original post by posthumus)
    Guys, is there actually an alternative or better way of doing this ?? I know this isn't the most reliable method
    Yep, there is; the way we've been prodding OP to do is outlined in broad brushstrokes in post 4 and then elaborated on a bit later (chiefly by davros and me).
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    (Original post by Smaug123)
    Not if you're integrating with respect to theta… if you're integrating with respect to x, then that sounds plausible.
    and like with limits you usually sub values into the parametic equations , so if I integrated the cartesian , I would need to sub detha=pie/4 into the x and y equatons
 
 
 
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