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Thread: Solve double integral

  1. #1
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    Solve double integral

    Hello!

    Can someone please tell me how to solve this double integral?

    $\displaystyle \int_0^{\sqrt{\pi}} \int_x^{\sqrt{\pi}} \! 2\sin{(y^2)} \, \mathrm{d} y\mathrm{d} x$

    The following hint is given: Sketch the domain of integration.

    I don't know the antiderivative of $\displaystyle 2\sin{(y^2)}$. I tried to guess a couple of antiderivatives, but none of their derivatives were equal to $\displaystyle 2\sin{(y^2)}$.

    Is there something else I should do first (and how)?

    Thanks!

    Update: I tried switching dydx to dxdy, which results in

    $\displaystyle \int_x^{\sqrt{\pi}} \int_0^{\sqrt{\pi}} \! 2\sin{(y^2)} \, \mathrm{d} x\mathrm{d} y$

    But this seems even worse, because the variable boundary is now on the outer integral.

    Please help! I still can't solve this one.
    Last edited by Lotte1990; Jan 29th 2013 at 08:38 AM.
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  2. #2
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    Re: Solve double integral

    That has no simple anti-derivative do swap the order of integrals:
    First, you need to be careful about the limits of integration. The best way to do that is to draw a "picture". The outer integral is from x= 0 to $\displaystyle x= \sqrt{\pi}$ so draw vertical lines on an xy- coordinate systems at those values of x. The "inner" integral is from y= x to $\displaystyle y= \sqrt{\pi}$ so draw the slant line y= x and the horizontal line $\displaystyle x= \sqrt{\pi}$. You should see that the region you are to integrate over is the upper half of the square with vertices at (0, 0), $\displaystyle (\pi, 0)$, $\displaystyle (\pi, \pi)$, and $\displaystyle (0, \pi)$. Overall, y goes from to $\displaystyle \sqrt{\pi}$ and, for each y, x goes from x= 0 to $\displaystyle x= y$.
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  3. #3
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    Re: Solve double integral

    I tried to swap the order of the integrals (like I posted in my update), so

    $\displaystyle \int_x^{\sqrt{\pi}} \int_0^{\sqrt{\pi}} \! 2\sin{(y^2)} \, \mathrm{d} x\mathrm{d} y$

    But now the outer integral has a variable boundary (x) I can't get rid off.
    Moreover, I still need to find the antiderivative of $\displaystyle 2\sin{(y^2)}$.

    I need some more help, please!
    Last edited by Lotte1990; Jan 29th 2013 at 09:06 AM.
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  4. #4
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    Re: Solve double integral

    Hi,

    May be you did not figure out HallsofIvy's idea.

    I think his idea is like that, you see that the domain of integration is the triangle whose vertices are the points $\displaystyle (0,0),(\sqrt{\pi},\sqrt{\pi}),(0,\sqrt{\pi}) $.
    So the domain can be alternatively determined as follows :
    $\displaystyle 0\leq y\leq \sqrt{\pi}, 0\leq x\leq y$.
    So
    $\displaystyle \int_{0}^{\sqrt{\pi}}\int_{0}^{y}2\sin y^2 dxdy.$ Now it is straightforward.
    Thanks from Lotte1990
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  5. #5
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    Re: Solve double integral

    Please do not stop after reading just the first sentence!
    My last line was:
    "Overall, y goes from 0 to $\displaystyle \sqrt{\pi}$, and, for each y, x goes from x= 0 to y."

    $\displaystyle 2\int_{y= 0}^{\sqrt{\pi}}\int_{x= 0}^y sin(y^2) dxdy$
    Last edited by HallsofIvy; Jan 29th 2013 at 11:10 AM.
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  6. #6
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    Re: Solve double integral

    Quote Originally Posted by HallsofIvy View Post
    Did you not see the last line of my previous post?
    "Overall, y goes from 0 to $\displaystyle \sqrt{\pi}$, and, for each y, x goes from x= 0 to y."
    $\displaystyle 2\int_{y= 0}^\sqrt{\pi}\int_{x= 0}^y sin(y^2) dxdy$
    Yes of course, but I think he did not figure out your post because he asked for further explanations.
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  7. #7
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    Re: Solve double integral

    Thank you both for your help! I (think I) get it now.

    My solution for the double integral is 1. Is this correct?
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  8. #8
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    Re: Solve double integral

    Is it not 2? I got 2. do not forget that you already have 2 in the integral.
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  9. #9
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    Re: Solve double integral

    I figured the 2 disappears while calculating the last antiderivative because the antiderivative of $\displaystyle 2y\sin(y^2)$ is $\displaystyle -\cos(y^2)$ and $\displaystyle -\cos(\pi) = 1$

    Update: I forgot the last term, the answer is not $\displaystyle -\cos(\pi) = 1$ but $\displaystyle -\cos(\pi) - (-\cos(0))= 2$
    Last edited by Lotte1990; Jan 29th 2013 at 11:46 AM.
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  10. #10
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    Re: Solve double integral

    Quote Originally Posted by Lotte1990 View Post
    I figured the 2 disappears while calculating the last antiderivative because the antiderivative of $\displaystyle 2ysin(y^2)$ is $\displaystyle -cos(y^2)$ and $\displaystyle -cos(\pi) = 1$
    note also the lower bound of the integral which is zero the answer will be $\displaystyle -(cos(\pi)-\cos0) = 1+1$[/QUOTE]
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  11. #11
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    Re: Solve double integral

    You are right! I forgot to substract the lower boundary! I get it now! Thank you so much for your help!
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  12. #12
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    Re: Solve double integral

    Quote Originally Posted by Kmath View Post
    Yes of course, but I think he did not figure out your post because he asked for further explanations.
    It was to Lotte1990 I was responding, saying essentially the same thing you did.
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