Results 1 to 12 of 12
Like Tree3Thanks
  • 1 Post By Kmath
  • 1 Post By HallsofIvy
  • 1 Post By Kmath

Math Help - Solve double integral

  1. #1
    Junior Member
    Joined
    Nov 2011
    Posts
    33
    Thanks
    1

    Solve double integral

    Hello!

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

    \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 2\sin{(y^2)}. I tried to guess a couple of antiderivatives, but none of their derivatives were equal to 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

    \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; January 29th 2013 at 09:38 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    16,238
    Thanks
    1795

    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 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 y= \sqrt{\pi} so draw the slant line y= x and the horizontal line 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), (\pi, 0), (\pi, \pi), and (0, \pi). Overall, y goes from to \sqrt{\pi} and, for each y, x goes from x= 0 to x= y.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Nov 2011
    Posts
    33
    Thanks
    1

    Re: Solve double integral

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

    \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 2\sin{(y^2)}.

    I need some more help, please!
    Last edited by Lotte1990; January 29th 2013 at 10:06 AM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Jun 2012
    From
    France
    Posts
    45
    Thanks
    13

    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 (0,0),(\sqrt{\pi},\sqrt{\pi}),(0,\sqrt{\pi}) .
    So the domain can be alternatively determined as follows :
    0\leq y\leq \sqrt{\pi}, 0\leq x\leq y.
    So
    \int_{0}^{\sqrt{\pi}}\int_{0}^{y}2\sin y^2 dxdy. Now it is straightforward.
    Thanks from Lotte1990
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Apr 2005
    Posts
    16,238
    Thanks
    1795

    Re: Solve double integral

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

    2\int_{y= 0}^{\sqrt{\pi}}\int_{x= 0}^y sin(y^2) dxdy
    Last edited by HallsofIvy; January 29th 2013 at 12:10 PM.
    Thanks from Lotte1990
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Jun 2012
    From
    France
    Posts
    45
    Thanks
    13

    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 \sqrt{\pi}, and, for each y, x goes from x= 0 to y."
    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.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Nov 2011
    Posts
    33
    Thanks
    1

    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?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Junior Member
    Joined
    Jun 2012
    From
    France
    Posts
    45
    Thanks
    13

    Re: Solve double integral

    Is it not 2? I got 2. do not forget that you already have 2 in the integral.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Junior Member
    Joined
    Nov 2011
    Posts
    33
    Thanks
    1

    Re: Solve double integral

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

    Update: I forgot the last term, the answer is not -\cos(\pi) = 1 but -\cos(\pi) - (-\cos(0))= 2
    Last edited by Lotte1990; January 29th 2013 at 12:46 PM.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Junior Member
    Joined
    Jun 2012
    From
    France
    Posts
    45
    Thanks
    13

    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 2ysin(y^2) is -cos(y^2) and -cos(\pi) = 1
    note also the lower bound of the integral which is zero the answer will be -(cos(\pi)-\cos0) = 1+1[/QUOTE]
    Thanks from Lotte1990
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Junior Member
    Joined
    Nov 2011
    Posts
    33
    Thanks
    1

    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!
    Follow Math Help Forum on Facebook and Google+

  12. #12
    MHF Contributor

    Joined
    Apr 2005
    Posts
    16,238
    Thanks
    1795

    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.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: April 8th 2012, 09:46 PM
  2. Replies: 2
    Last Post: April 3rd 2011, 12:08 PM
  3. Replies: 1
    Last Post: June 2nd 2010, 03:25 AM
  4. How do I solve this double integral?
    Posted in the Calculus Forum
    Replies: 3
    Last Post: November 8th 2009, 10:37 PM
  5. Replies: 6
    Last Post: May 18th 2008, 07:37 AM

Search Tags


/mathhelpforum @mathhelpforum