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Math Help - double integral

  1. #1
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    Thumbs up double integral

    My problem is like this:
    Evaluate the double integral of x(1+y^2)^(-1/2)dA where R is the region in the first quadrant enclosed by y=x^2, y=4 and x =0.

    I try to solve as type I, that is I set the limit of integration with respect to y from x^2 to 4 and then limit of ontegration with respect to x from 0 to 2.

    I use trigonometric subtitution which yields

    y=tan(delta) and dy=sec^2(delta)d(delta)

    then I dont know how to change the limit integration which respect to y into the limit integration which respect to delta.

    Please help,
    Thanks

    P/s: this is my attached fomula:double integral.doc
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  2. #2
    Senior Member yeKciM's Avatar
    Joined
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    did u draw how it looks ?
    and when u do that ... u'll see how u should do integration.... limit on x (dx) and y (dy) depending how u gonna do it... first by x or y ?




    P.S. U don't need that subtitution to solve this one ...

    if u change order of integration it's much much easier to solve .... look at it like this



    I= \int_{0} ^{4} dy \int_{0}^{\sqrt{y}} \frac{x}{\sqrt{1+y^2}}dx

    I= \int_{0} ^{4} \frac{dy}{\sqrt{1+y^2}} \int_{0}^{\sqrt{y}}xdx

    I=\frac {1}{2} \int_{0} ^{4} \frac {ydy}{\sqrt{1+y^2}}

    1+y^2=t^2 \to 2ydy=2tdt \to ydy =tdt

    I=\frac {1}{2} \int_{0} ^{4} \frac {dt}{t} ....
    Last edited by yeKciM; July 23rd 2010 at 03:51 PM. Reason: suggestion
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  3. #3
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    Quote Originally Posted by yeKciM View Post
    did u draw how it looks ?
    and when u do that ... u'll see how u should do integration.... limit on x (dx) and y (dy) depending how u gonna do it... first by x or y ?




    P.S. U don't need that subtitution to solve this one ...

    if u change order of integration it's much much easier to solve .... look at it like this



    I= \int_{0} ^{4} dy \int_{0}^{\sqrt{y}} \frac{x}{\sqrt{1+y^2}}dx

    I= \int_{0} ^{4} \frac{dy}{\sqrt{1+y^2}} \int_{0}^{\sqrt{y}}xdx

    I=\frac {1}{2} \int_{0} ^{4} \frac {ydy}{\sqrt{1+y^2}}

    1+y^2=t^2 \to 2ydy=2tdt \to ydy =tdt

    I=\frac {1}{2} \int_{0} ^{4} \frac {dt}{t} ....
    I got it, basically the only thing I have to do is to switch to type II which integrate with respect to x first...lol...wow, it is much easier, I din't think about type II !
    Thanks you so much !
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