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Math Help - double integrals

  1. #1
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    double integrals

    an some1 please help me with the following 2 question, i have attemped both but seem to come out with the correct answer:

    1. ∫dx ∫(x-y)dy the 1st integral sign has the limits a and 0. the 2nd sign has the limits y1 and 0. where y1 = (a^2 -x^2)^0.5
    for this ques the answer should be 0.

    2. by changing the order of integration evaluate:
    ∫dx∫y^-1 sinycos(x/y)dy for the 1st integral sign, limits are 1 and 0. the 2nd sign has limits 1 and x.
    answer: sin1(1-cos1)
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  2. #2
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    Quote Originally Posted by vish1987
    an some1 please help me with the following 2 question, i have attemped both but seem to come out with the correct answer:

    1. ∫dx ∫(x-y)dy the 1st integral sign has the limits a and 0. the 2nd sign has the limits y1 and 0. where y1 = (a^2 -x^2)^0.5
    for this ques the answer should be 0.
    There may be a more elegant way, but I always do these in baby steps.

    \int_0^a dx \int_0^{\sqrt{a^2-x^2}} dy(x-y)
    =\int_0^a dx(xy-(1/2)y^2)|_0^{\sqrt{a^2-x^2}}
    =\int_0^a dx \, x \sqrt{a^2-x^2} \, - \, \int_0^a \, dx(1/2)(a^2-x^2)

    Now, the first integral:
    Use u=a^2-x^2
    du=-2xdx
    \int dx \, x \sqrt{a^2-x^2}=\int du(-1/2) \sqrt{u}=(-1/2)(2/3)u^{3/2}
    =(-1/3)(a^2-x^2)^{3/2}
    Inserting the end limits:
    \int_0^a dx \, x \sqrt{a^2-x^2}=(-1/3)(a^2-x^2)|_0^a=(1/3)a^3

    The second integral is easy:
    \int_0^a \, dx(1/2)(a^2-x^2)=(1/2)[a^2x-(1/3)x^3]|_0^a
    =(1/2)a^3-(1/6)a^3=(1/3)a^3

    So we need to subtract. Finally:
    \int_0^a dx \int_0^{\sqrt{a^2-x^2}} dy(x-y) =(1/3)a^3-(1/3)a^3=0.

    -Dan
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  3. #3
    Forum Admin topsquark's Avatar
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    Someone else is going to have to do 2). I can't figure it.

    -Dan
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    ok thanks for the 1st one neway!
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  5. #5
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    Quote Originally Posted by vish1987
    2. by changing the order of integration evaluate:
    ∫dx∫y^-1 sinycos(x/y)dy for the 1st integral sign, limits are 1 and 0. the 2nd sign has limits 1 and x.
    answer: sin1(1-cos1)
    The integral to be evaluated is:

    <br />
\int_0^1 \int_x^1 \frac{\sin(y) \cos(x/y)}{y} dy\ dx<br />

    Which we may write:

    <br />
\int_A \frac{\sin(y) \cos(x/y)}{y} ds<br />

    where A is the region between the y-axis, and the lines x=1 and x=y,
    and ds the the area element. This now may be written as:

    <br />
\int_0^1 \int_0^y \frac{\sin(y)\cos(x/y)}{y} dx\ dy<br />
,

    which may be integrated to give:

    <br />
\int_0^1 \frac{\sin(y)}{y} \ [ y \sin(x/y)]_{x=0}^{x=y}}\ dy= <br />
\int_0^1 \sin(y) \sin(1) dy=\sin(1)[-\cos(y]_0^1=\sin(1)[1-\cos(1)]<br />

    RonL
    Last edited by ThePerfectHacker; March 1st 2006 at 01:17 PM.
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