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Simple double integral

  1. Jun 17, 2016 #1
    1. The problem statement, all variables and given/known data
    ∫∫D x2+y2dA
    where D is the region limited by: y=x2, x=2, y=1 (dA = dxdy)

    2. Relevant equations


    3. The attempt at a solution
    So basically i sketched the area, and i get the area under y=x2 0<x<1 and a square at 1<x<2 , 0<y<1
    So i divded the integrals; for the square ∫0112 of x2+y2dxdy
    for the y=x2 part is the part i need help/confirming x goes from 0 to 1(so integral os x2 with those bounds), and i used substitution for y=x2 and got dy=2xdx so integrated from 0 to 1 x42x dx

    So i was thinking about this in another way, wich gave me the same answer.
    So i integrated x2 and kept the bounds so i got [(x3)/3]10 then integrated y2 and set its bounds to 0 to x2 and substituted so i got [(x3)/3]10 + x^5/3, and moved that inside the bounds.
    Am i correct?
     
    Last edited: Jun 17, 2016
  2. jcsd
  3. Jun 17, 2016 #2

    SteamKing

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    I have a different interpretation of the region D.

    You can always plot these in utilities like Wolfram Alpha.

    I take the upper boundary of D to be y = x2.

    The right hand boundary is going to be the vertical line x = 2.

    The lower boundary will be the horizontal line y = 1. This will also produce the left-hand limit of the region in terms of the x-coordinate, where y = 1 intersects y = x2.

    The region D is roughly in the shape of a triangle, with the hypotenuse being a segment of the parabola y = x2 and the base being the line y = 1.


    https://www4f.wolframalpha.com/Calculate/MSP/MSP641hbce22d448geh87000055267hbac324i9hi?MSPStoreType=image/gif&s=47 [Broken]

    As far as the mechanics of your integration goes, I have no idea what you are trying to do. You don't have the correct region for D and your substitution is frankly a head-scratcher.

    In any event, since you have the double integral of a sum, it would probably be easier to split this into the sum of the integrals of x2 and y2 dA and add the results:

    ##\int \int_D (x^2+y^2)\, dA = \int \int_D x^2\,dA + \int\int_D y^2\, dA##
     
    Last edited by a moderator: May 8, 2017
  4. Jun 17, 2016 #3

    Charles Link

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    The best way to perform this double integral is to do the dy integral first. The dx integral (done second) has two parts that will have different boundaries. So basically divide the double integral into two sections: 1) ## 0<x<1 ## and 2) ## 1<x<2 ##. The dx integral from 0 to 1 is bounded by ## y=x^2 ## for the portion 0<x<1 (as you correctly noticed), and by y=1 for ## 1<x<2 ##. For the first section , you need to carefully look at the dy limits. Do you see that the upper limit here (on the dy integration) needs to be ## x^2 ##? It does take a little effort to get some proficiency with these double integrals.... editing...sorry, my mistake=I see I got the boundary region incorrect. I mistakenly thought you had y=0 as a lower boundary....but I see, yes, I didn't read your problem statement and assumed you had the region correct...
     
    Last edited: Jun 17, 2016
  5. Jun 17, 2016 #4
    ok so my first attempt was incorrect, i see that now...

    The problem with my second attempt was that my teacher defined dA as dxdy so i cant integrate y first, so i have to take the lower limit of x as square root of y.
    If i could integrate dy first i could've done what i did in my second attempt and set the upper limmit of y as x2

    Am i correct?

    so using the now correct boundries ( im leaving the 1<x<2 part out)
    01y½1 x2+y2 dxdy
    so i get
    011/3- (1/3*y3/2)+y2 dy

    is this correct?
     
    Last edited: Jun 17, 2016
  6. Jun 17, 2016 #5

    SteamKing

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    Who says you can't integrate w.r.t. y first? dA = dxdy is the same as dA = dydx. In fact, you should get the same answer to the integral regardless of which variable you choose first.
     
  7. Jun 17, 2016 #6
    in this particular set of problems my teacher said to use dA = dxdy, so i can't use dydx
     
  8. Jun 17, 2016 #7

    SteamKing

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    I don't know why this is, but you should be able to use any order to integrate your function. At least, in the real world you can.
     
  9. Jun 17, 2016 #8
    i'm guessing that its for us to practice integration limits.
    All the other problems i can use whatever i want for dA
     
  10. Jun 17, 2016 #9

    Charles Link

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    Not completely correct. Your upper dx limit needs to be 2. And your y limits are incorrect.
     
  11. Jun 17, 2016 #10
    i was doing that on a separate integral, since i get a square from 1<x<2 and 0<y<1, im suming both up, the one on here, and the double integral with bounds 1-2 on dx and 0 to 1 on dy.
    I can do that right?
     
  12. Jun 17, 2016 #11

    Charles Link

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    Please read SteamKing post #2 and also my edited post #3. Your boundary is a triangular shaped region. (See his graph). Also see my edited post #9.
     
  13. Jun 17, 2016 #12
    ohhh so i thought the lower limit was y=1 as shown in the pic. you are telling me its only the area marked (1) ?
     

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    Last edited: Jun 17, 2016
  14. Jun 17, 2016 #13

    Charles Link

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    Yes...The other calculation would make for an interesting exercise also, but it's not what the problem asked for... The lower limit is y=1, not y=0. With the new area, try to write out the limits. You're on the right track=you simply selected the wrong region...
     
    Last edited: Jun 17, 2016
  15. Jun 17, 2016 #14
    Thanks for your help guys.
    I also realized i forgot to integrate y2 with respect to x getting y2x

    One last question, when no lower bound is mentioned am i to take y=0 as a lower bound?

    and just to check if i have this right, another problem states that the area is limited by y>=x2 and y<2x
    so the lower limit on the dx integral would be y/2 and the upper would be square root of y. correct?
    and the dy is from 0 to 2 ?
     
  16. Jun 17, 2016 #15

    Charles Link

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    When you previously had ## y^2 ## there after the dx integral, I thought you might have already put ## y^2 x ## and evaluated it because your x limits were 0 and 1. For your problem where y<2x, draw the line y=2x and it is the region below it, and above the curve ## y=x^2 ##. It's a crescent shape between x=0 and x=2. The dy is from 0 to 4. (assuming you are doing dx first, yes I think you got te dx limits correct.)
     
  17. Jun 17, 2016 #16
    oh yeah i meant dy 0 to 4 got mixed up there.

    yeah i sketched it, and noticed that as dx "grew" it was limited by y=x^2 , and when dx" went to 0" it was limited by y=2x, so thats where i got lower bound y/2 and upper bound sqrt(y)

    and yes for this set of problems i gotta do dx first, i dont get to chose.
    I think im getting the hang of it. Thanks!
     
  18. Jun 17, 2016 #17

    SteamKing

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    Whew! That's a confusing way to describe the region of interest.

    Let's just say that you draw the line y = 2x and the parabola y = x2.

    You want the area of the region which is between the two curves, so along a line of constant x-value, y ≥ x2 and y < 2x
     
  19. Jun 17, 2016 #18

    Ray Vickson

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    If no lower bound is mentioned there is no justification for taking y = 0 as a lower bound; basically -∞ is your lower bound, and that would make your integration come out infinite because the integration region is unbounded.

    I think that the only reasonable interpretation of this problem is that the integration region is like a triangle with a curved hypotenuse, like the following shaded region.

    upload_2016-6-17_13-45-48.png
     
  20. Jun 17, 2016 #19
    thats what i said... :D
     
  21. Jun 17, 2016 #20

    SammyS

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    What does
    ##y=2x##​
    have to do with anything in this problem?
     
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