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    I need to find the area, A, of the shape bounded by the curve, C, where C is given parametrically by

    x(t)=4cos(t)-cos(4t), y(y)=4sin(t)-sin(4t)

    In the working belew, where the integrals have no bounded it means they are over either the region or the curve.

    I used greens theorem with p=x/2, q=-y/2 so

    \displaystyle\int \displaystyle\int \ dx\ dy=\displaystyle\int \displaystyle\int \frac{\partial p}{\partial x}-\frac{\partial q}{\partial y}\ dx\ dy



= \displaystyle\int (p,q) \cdot d\mathbf{r}



=\displaystyle\int^{2\pi}_{0} (2cos(t)-0.5cos(4t),-2sin(t)+0.5sin(4t)) \cdot (-4sin(t)+4sin(4t),4cos(t)-4cos(4t))dt

    and I know I have gone wrong up to here as the above is a straightforward integral which comes to 0 (wolfram confirms this). Where have I gone wrong? I know from looking at the shaoe that it has non zero area.

    EDIT: Anyone know why the latex looks funny? I don't want the last line so spaced out.
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    (Original post by james22)
    ...
    Not an expert on this, but just from googling:

    Comparing your working with Green's theorem, should your integral not be:

    \displaystyle\int (q,p) \cdot d\mathbf{r}

    Also, there's a minor error in that

    \displaystyle - \frac{y}{2}=-2\sin(t)+0.5\sin(4t)
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    (Original post by ghostwalker)
    Not an expert on this, but just from googling:

    Comparing your working with Green's theorem, should your integral not be:

    \displaystyle\int (q,p) \cdot d\mathbf{r}

    Also, there's a minor error in that

    \displaystyle - \frac{y}{2}=-2\sin(t)+0.5\sin(4t)
    Really? That second equals sign is copied directly from my notes (the only difference between mine and the notes one is in the notes it specifies what the integral is over, but the only reason I didn't put that on mine is I don't know the latex for it).

    Fixed that small error.
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    (Original post by james22)
    Really? That second equals sign is copied directly from my notes (the only difference between mine and the notes one is in the notes it specifies what the integral is over, but the only reason I didn't put that on mine is I don't know the latex for it).
    Found a reference Wiki

    PS When you swap it round, it works.
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    Why doesn't my method work though?
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    Where is the q from? Do you have a link to it?
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    (Original post by twig)
    Where is the q from? Do you have a link to it?
    q is just a function, I pick x and y so that I would get dp/dx-dq/dy=1.
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    (Original post by james22)
    Why doesn't my method work though?
    Not sure why you expect it to work. What do your notes say - need a scan/photo, with preamble.
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    http://www.maths.ox.ac.uk/system/fil...lectures10.pdf

    Page 83 theorem 7.5
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    (Original post by james22)
    \displaystyle\int \displaystyle\int \ dx\ dy=\displaystyle\int \displaystyle\int \frac{\partial p}{\partial x}-\frac{\partial q}{\partial y}\ dx\ dy
    You have p and q the other way about to the course material.
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    (Original post by ghostwalker)
    You have p and q the other way about to the course material.
    Thanks for that, I'll give it a go now and see what I get. That may well be what was causing the problems.
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    Yep, that worked. Thanks.
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    (Original post by james22)
    q is just a function, I pick x and y so that I would get dp/dx-dq/dy=1.
    Sorry, by q I meant the question! Haha.
 
 
 
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