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Double integral evaluation.

  1. Nov 26, 2011 #1
    1. The problem statement, all variables and given/known data

    Evaluate the integral:

    integral D of y sin (x+y^2) dA where

    D = [0x2] x [0x2] U [1,3] x [1,3]

    2. Relevant equations



    3. The attempt at a solution

    So D is basically a square which simplifies to D = [1,2] x [1,2] since that is the portion of both rectangles that overlap.

    So then my integral becomes

    integral from 1 to 2 (integral from 1 to 2 of y sin (x+y^2) dy) dx

    So the indefinite inner integral is:

    -1/2(cos(x+y^2))

    So here I am supposed to evaluate y at 2 and 1 right? If that's correct, I am getting a wrong answer for some odd reason I think. I'm using wolfram alpha to check and their answer for the inner integral is:

    http://www4d.wolframalpha.com/Calculate/MSP/MSP46819i63f8i8617ih80000063597a8d4817fhid?MSPStoreType=image/gif&s=41&w=259&h=36 [Broken]

    Am i doing something wrong?
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Nov 26, 2011 #2
    Okay this is crazy, but, I think you're okay. Did you get [itex] -\frac{1}{2}cos(x+4)+\frac{1}{2}cos(x+1) [/itex] when you solved the definite by hand? That's what I got. Apparently (according to my calculator's compare function) this is equivalent to wolframs answer.
     
  4. Nov 26, 2011 #3
    Yes that's what i got. Ok I guess I'll just use my calculator to compare the answers from now on to wolfram. Thanks!
     
  5. Nov 27, 2011 #4

    Mark44

    Staff: Mentor

    No, that is incorrect, assuming that the region is [0, 2] X [0, 2] U [1, 3] X [1, 3]. I don't know what [0x2] means.

    If I understand what you meant to write, region D consists of the square [0, 2] X [0, 2] together with the square [1, 3] X [1, 3].
     
  6. Nov 27, 2011 #5
    Yeah that was a typo. Thanks.
     
  7. Nov 27, 2011 #6

    Mark44

    Staff: Mentor

    Just to be clear, D does NOT simplify to the square [1, 2] X [1, 2], which you said in your first post.
     
  8. Nov 27, 2011 #7
    Hmm. Why is that? I drew out D and since its the intersection of the two rectangles given, the only overlapping portion is the square [1,2] x [1,2]. So isn't D the same thing as that?
     
  9. Nov 27, 2011 #8
    You wrote the "union" of the two areas in your original post. If you meant intersection, its [itex]\bigcap[/itex]
     
  10. Nov 27, 2011 #9
    Nevermind, you're right the question says union. I mistook it for intersection. My answer is wrong. Thanks for pointing that out.

    So then it is two rectangles with one overlapping portion. I guess the way to solve this is evaluate the integral of both the rectangles and subtract the overlapping portion?

    Also I'm guessing both the rectangles should have the same area since they are the same size. So would i just be able to do 2x the integral of one of them to find both?
     
  11. Nov 27, 2011 #10
    You can do 2x one rectangle and subtract the overlapping portion.

    Or do each rectangle and subtract the overlapping portion. Since you already probably found the overlapping area, you're partly done! :)
     
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