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    if you take x^2 and rotate it 90 degrees clockwise so you have a sideways quadratic and then shift it to the left two units so it crosses the x axis at -2 and you went onto intergrate between -2 and 0 would you be finding the area above the x axis or below it.

    If you took this graph and moved it up abit so there was more area above the x axis and did the intergral from where it crosses the axis to 0, would you be finding the area bounded by the lines x=(where it crosses) and x=0 and the graph above the axis or the entire area of the graph above the axis.

    Also, if it does give you the area above, how would you find the area below? I know you should integrate with respect to y but can you find it by intergrating with respect to x?
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    I think it'd have to be d/dy since rotating x^2 by 90° gives y^2
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    the function you have described cannot be integrated with respect to x because it is one to many.
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    (Original post by the bear)
    the function you have described cannot be integrated with respect to x because it is one to many.
    Thanks,
    what do you mean by one to many?
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    (Original post by 111davey1)
    Thanks,
    what do you mean by one to many?
    http://images.slideplayer.com/11/303...s/slide_20.jpg
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    Oh, in the question it gave parametric equations. In the solution it started off by saying:

    2 times intergral of y with respect to x. We were not given y in terms of x so had to intergrate para metrically in one direction then times by 2.

    So would the first part be incorrect then (2times intergral of y with respect to x)

    This is for the sideways quadratic crossing at -2.
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    (Original post by 111davey1)
    Oh, in the question it gave parametric equations. In the solution it started off by saying:

    2 times intergral of y with respect to x. We were not given y in terms of x so had to intergrate para metrically in one direction then times by 2.

    So would the first part be incorrect then (2times intergral of y with respect to x)

    This is for the sideways quadratic crossing at -2.
    perhaps you could post the question for us ?

    :holmes:
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    here it is:
    http://pmt.physicsandmathstutor.com/...hapter%202.pdf

    page 64 Excercise E question 8 part c
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    (Original post by 111davey1)
    here it is:
    http://pmt.physicsandmathstutor.com/...hapter%202.pdf

    page 64 Excercise E question 8 part c
    Okay so you need to calculate the finite area on the left of the y-axis. When you are integrating \displaystyle \int_{-2}^0 y.dx you are finding the area above the x-axis because the curve is y^2=x+2 (without checking) so taking square roots of both sides gives y=\sqrt{x+2} which is only the branch above the x-axis (as y\geq0 for this function)

    The reason for 2 lots of it is due to symmetry and how the area above is the same as the area below.

    To find the area below simply say you're integrating -y instead.
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    (Original post by RDKGames)
    Okay so you need to calculate the finite area on the left of the y-axis. When you are integrating \displaystyle \int_{-2}^0 y.dx you are finding the area above the x-axis because the curve is y^2=x+2 (without checking) so taking square roots of both sides gives y=\sqrt{x+2} which is only the branch above the x-axis (as y\geq0 for this function)

    The reason for 2 lots of it is due to symmetry and how the area above is the same as the area below.

    To find the area below simply say you're integrating -y instead.
    but how do you know y=+vert(x+2)
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    (Original post by 111davey1)
    but how do you know y=+vert(x+2)
    vert? I assume you meant sqrt??

    Well y=\pm \sqrt{x+2} so it's EITHER + or - but it cannot be both at the same time (unless you want to jump to piecewise functions with are not needed here), so you need to consider each branch individually.

    For y=-\sqrt{x+2} that's what I was talking about when I told you how to calculate the area below the x-axis with -y if you take y=+\sqrt{x+2} initially
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    (Original post by RDKGames)
    vert? I assume you meant sqrt??

    Well y=\pm \sqrt{x+2} so it's EITHER + or - but it cannot be both at the same time (unless you want to jump to piecewise functions with are not needed here), so you need to consider each branch individually.

    For y=-\sqrt{x+2} that's what I was talking about when I told you how to calculate the area below the x-axis with -y if you take y=+\sqrt{x+2} initially
    Thanks, so how did they know that intergral was going to be positive?
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    (Original post by 111davey1)
    Thanks, so how did they know that intergral was going to be positive?
    They didnt "know" per se, but rather just chose the + because it is more intuitive, makes more sense, and makes up for less confusing answers.

    If they had chosen y=-\sqrt{x+2} then obviously integrating would give negative area so the answer sheet would say something like \displaystyle -2\int_{-2}^0 y.dx instead.
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    (Original post by RDKGames)
    They didnt "know" per se, but rather just chose the + because it is more intuitive, makes more sense, and makes up for less confusing answers.

    If they had chosen y=-\sqrt{x+2} then obviously integrating would give negative area so the answer sheet would say something like \displaystyle -2\int_{-2}^0 y.dx instead.
    Thanks very much
 
 
 
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