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Math Help - trig substitution and integration by parts

  1. #1
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    trig substitution and integration by parts

    how do i find the area of the ellipse 4x^2 + y^2 = 9
    using some form of integration by parts?
    trig substitution?
    Last edited by razorfever; January 14th 2009 at 10:27 PM.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by razorfever View Post
    how do i find the area under the circle 4x^2 + y^2 = 9
    using some form of integration by parts?
    trig substitution?
    firstly, this is not a circle. secondly, what do you mean by "under"? below the curve but above the x-axis?
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    yeah ... its an ellipse ... what was i thinking ... and i meant to say area of the circle ... i'm just used to finding the area under the circle ... wait i think u just answered my question ... would i have to integrate from 0 to the radius of the circle and multiply by four ??? but i still don't understand how to use trig substitution ... cause thats what i've been taught and i'm sure it fits in this question somewhere ...
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    If you put this equation in terms of y you get y = sqrt(9-4x^2)
    To get the interval of this equation find out where y = 0, the answer is +- 2/3
    So the integral is [sqrt(9-4x^2) dx from -2/3 to 2/3
    Use 9/4sin(theta) as a substitution for x and then integrate.
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  5. #5
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by razorfever View Post
    how do i find the area under the circle 4x^2 + y^2 = 9
    using some form of integration by parts?
    trig substitution?
    You can, to do this solve for y and notice that you will get a \pm...you can disregard this by doubling the area you get with + (why?). You can then make a trigonometric sub of the form 2x=3\sin(\vartheta)

    Alternatively we can parametrize this by x=\frac{3}{2}\cos(\vartheta) and y=3\sin(\vartheta). So now we can use the parametrization to find that the area is equal to \frac{9}{2}\int_0^{2\pi}\cos^2(\vartheta)~d\varthe  ta=\frac{9\pi}{2}

    Why does this work? Well if you know parametrization of curves find out that the area given by a counterclockwise oriented curve is \int_{\alpha}^{\beta}x(t)\cdot y'(t)~dt where [\alpha,\beta] is the interval on which the curve takes to make a full tracing.
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by pberardi View Post
    If you put this equation in terms of y you get y = sqrt(9-4x^2)
    To get the interval of this equation find out where y = 0, the answer is +- 2/3
    So the integral is [sqrt(9-4x^2) dx from -2/3 to 2/3
    Use 9/4sin(theta) as a substitution for x and then integrate.
    Right idea, but be careful with your solving of this...it should be x=\pm\frac{3}{2}
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  7. #7
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    This isn't a circle. But yes you can use trig substitution. You have y= \pm \sqrt{3^2-(2x)^2} Let the hypotenuse of a right triangle be 3, the vertical leg be 2x, and the other leg be 9-4x^2.

    edit: well i am very far behind
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  8. #8
    Member pberardi's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    Right idea, but be careful with your solving of this...it should be x=\pm\frac{3}{2}
    ack! Just for the record I have +-3/2 written in my notebook . I welcome the corrections.
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