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jedai17 Group Title

what is double integral function? how to solve double integral function?

  • one year ago
  • one year ago

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  1. ViPhy Group Title
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    Double Integral: http://mathworld.wolfram.com/FubiniTheorem.html You solve in the order of the integral, if dy comes first, you solve as simple integral considering x as constant, after that, solve the integral for dx.

    • one year ago
  2. jedai17 Group Title
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    is the value of dy similar to the value of dx?

    • one year ago
  3. yunus Group Title
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    yes they are similar. when you solve the integral according to dy all x values are constant for you. It doesnt matter which one you solve first, dx or dy.

    • one year ago
  4. jedai17 Group Title
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    ahh..ok tnx..can you teach me how to solve this?i have no background in solving this,, \[\int\limits_{0}^{2}\int\limits_{0}^{x ^{2}/2} x/\sqrt{1 + x^2 + y^2} dydx\]

    • one year ago
  5. hartnn Group Title
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    if the limits for both the variables were constant, then we can integrate any variable first., but here since the limits of inner integral is not constant, we need to evaluate the inner integral first, limits are y=0 to y=x^2/2 so, since you are integrating w.r.t 'y', treat x as constant, so you have, \(\int\limits_{0}^{2}x[\int\limits_{0}^{x ^{2}/2} 1/\sqrt{1 + x^2 + y^2} dy]dx\) now the inner integral is od the form, \(\int \dfrac{1}{\sqrt{a^2+y^2}}dy\) do you know how to integrate this ?

    • one year ago
  6. jedai17 Group Title
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    why it become a^2?

    • one year ago
  7. hartnn Group Title
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    i just took 1+x^2 = constant = a^2 because we have a standard integration formula for, \(\int \dfrac{1}{\sqrt{a^2+y^2}}dy\)

    • one year ago
  8. hartnn Group Title
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    then we need to re-substitute back after integration, a^2= 1+x^2 before integrating w.r.t x

    • one year ago
  9. hartnn Group Title
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    *after integration w.r.t y,

    • one year ago
  10. jedai17 Group Title
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    in this we are integrating y and we assume as x as constant?

    • one year ago
  11. jedai17 Group Title
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    what do you mean by w.r.t y, ?

    • one year ago
  12. hartnn Group Title
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    yes, true, so we'll get answer in 'x' after first integrating w.r.t y (with respect to y)

    • one year ago
  13. hartnn Group Title
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    ok, \(\int \dfrac{1}{\sqrt{a^2+y^2}}dy= (1/a)\arctan(y/a)+c\) but then you have to use integration by parts to integrate w.r.t x. so, i suggest we better CHANGE the ORDER of integration, that is, we'll change the limits so that we can first integrate w.r.t 'x' and then 'y'. For that, wee need to plot 2 regions (for 2 limits): (1) y= 0 to y=x^2/2 (2) x=0 to x= 2 can you plot these 4 curves (3 of these are simple straight lines)?

    • one year ago
  14. jedai17 Group Title
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    i'm sorry don't know how to plot this..

    • one year ago
  15. jedai17 Group Title
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    i'm sorry i forgot our previous lessons about that..

    • one year ago
  16. hartnn Group Title
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    so, you say you don't know how to plot lines x=0,y=0,x=2 ? i'll help you with y=x^2/2 but if you don't know how to plot those lines, then it'll be very difficult for you to understand this..

    • one year ago
  17. jedai17 Group Title
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    i know this x=0,y=0,x=2 but i don't know y=x^2/2..i'm sorry

    • one year ago
  18. hartnn Group Title
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    so, plot these x=0,y=0,x=2 using 'Draw' tool here.

    • one year ago
  19. jedai17 Group Title
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    |dw:1362400369683:dw|

    • one year ago
  20. jedai17 Group Title
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    is it correct?

    • one year ago
  21. hartnn Group Title
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    where is the line x=2 ?

    • one year ago
  22. jedai17 Group Title
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    |dw:1362400538076:dw|

    • one year ago
  23. jedai17 Group Title
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    sorry i forgot x = 2..

    • one year ago
  24. hartnn Group Title
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    yes, x=0 is y-axis, y=0 is x axis , and x=2 you have drawn correct, now y= x^2/2 is a parabola which has a plot, |dw:1362400675976:dw| so you are integrating in the region that i am gonna shade

    • one year ago
  25. jedai17 Group Title
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    why it becomes parabola?

    • one year ago
  26. hartnn Group Title
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    |dw:1362400803859:dw| since we were integrating w.r.t 'y' first, i've drawn vertical lines, now we are changing the order of integration, we need to draw horizontal lines,

    • one year ago
  27. hartnn Group Title
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    |dw:1362400889186:dw| to |dw:1362400935932:dw|

    • one year ago
  28. jedai17 Group Title
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    i'm sorry i forgot how to find the equation of the line

    • one year ago
  29. hartnn Group Title
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    oh, its easy, that a horizontal line , so u just need co-ordinates of intersection, which you can find by solving simultaneously, x=2 and x= \(\sqrt{2y}\)

    • one year ago
  30. hartnn Group Title
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    by the way, did you get that final diagram ?

    • one year ago
  31. jedai17 Group Title
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    not yet..

    • one year ago
  32. hartnn Group Title
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    which part you have doubt with ?

    • one year ago
  33. jedai17 Group Title
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    \[x =2 and x = \sqrt{2y}\] i don't know how to fnd the equation of this line..

    • one year ago
  34. hartnn Group Title
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    x= 2, x=\(\sqrt{2y}\) so, 2= \(\sqrt{2y}\) can u find 'y' from here ?

    • one year ago
  35. jedai17 Group Title
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    no..

    • one year ago
  36. jedai17 Group Title
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    \[y=\sqrt{2}/2 ?\]

    • one year ago
  37. hartnn Group Title
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    :O that was simple arithmetic....sorry to say but, if you have problem solving that you should think before dealing double integration, sorry if i am rude, but you are lacking basics ... 2= \(\sqrt{2y}\) 4= 2y (by squaring) y=2 this is the equation of your horizontal line

    • one year ago
  38. jedai17 Group Title
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    yes that's true i almost forgot the topics about integration that's why i am like this..

    • one year ago
  39. hartnn Group Title
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    |dw:1362402005207:dw| so your new limits are, \(\huge \int \limits_0^2\int \limits_{\sqrt{2y}}^2\dfrac{x}{\sqrt{1+x^2+y^2}}dx.dy \) now you have to first integrate w.r.t x to do that put u = 1+x^2+y^2 du=... ?

    • one year ago
  40. jedai17 Group Title
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    du = 2x + 2y

    • one year ago
  41. hartnn Group Title
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    oh, since we are integrating w.r.t 'x' first now, we treat 'y' as constant, so du =2xdx or x dx = du/2 got this ?

    • one year ago
  42. jedai17 Group Title
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    yes..

    • one year ago
  43. hartnn Group Title
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    so, your inner integral will become, \(\int \dfrac{du}{2\sqrt u}\) can you integarte this ?

    • one year ago
  44. jedai17 Group Title
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    1/2..is it correct?

    • one year ago
  45. amistre64 Group Title
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    as long as weve gotten this u substitution correct: \[\int \frac{1}{2\sqrt{u}}du=\sqrt{u}\]

    • one year ago
  46. amistre64 Group Title
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    assuming that u = 1+x^2+y^2, plug back in for u \[\sqrt{ u}=\sqrt{ 1+x^2+y^2}\] from x=sqrt(2y) to 2 \[\sqrt{ 1+2^2+y^2}-\sqrt{ 1+(\sqrt{2y})^2+y^2}\] \[\sqrt{ 5+y^2}-\sqrt{ 1+2y+y^2}\]

    • one year ago
  47. amistre64 Group Title
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    y^2+2y+1 = (y+1)^2 sooo \[\int\sqrt{5+y^2}-(y+1)~dy\]

    • one year ago
  48. jedai17 Group Title
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    why it becomes subtraction?

    • one year ago
  49. jedai17 Group Title
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    why it become y^2+2y+1 = (y+1)^2 ?

    • one year ago
  50. amistre64 Group Title
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    because: \(\Large \int_{a}^{b} f(x)~dx=F(b)-F(a)\)

    • one year ago
  51. jedai17 Group Title
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    ah ok tnx..

    • one year ago
  52. jedai17 Group Title
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    in this time we are integrating y?

    • one year ago
  53. amistre64 Group Title
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    when we replace the \(\sqrt{u}\) with what we defined it to begin with: \(u=1+x^2+y^2\) and input the ab limits of a=\(\sqrt{2y}\) and b=2 we get \[(\sqrt{1+b^2+y^2})-(\sqrt{1+a^2+y^2})\] \[(\sqrt{1+b^2+y^2})-(\sqrt{1+a^2+y^2})\] and yes, then since we are finished with the inside integral, we focus on the next integral

    • one year ago
  54. jedai17 Group Title
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    what's next?

    • one year ago
  55. amistre64 Group Title
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    the next integration is next ... \[\int\sqrt{5+y^2}-(y+1)~dy\]

    • one year ago
  56. jedai17 Group Title
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    \[\sqrt{5 + y} - y\]

    • one year ago
  57. jedai17 Group Title
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    is it correct?i'm sorry have a problem in integration..

    • one year ago
  58. amistre64 Group Title
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    the square root part seems a little off to me

    • one year ago
  59. amistre64 Group Title
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    and it looks like you tried a derivative onthe y+1 parts instead of integeration

    • one year ago
  60. amistre64 Group Title
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    \[-\int y+1~dy=-(\frac{y^2}{2}+y)\]

    • one year ago
  61. jedai17 Group Title
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    yes.. i'm sorry a have a problem in integration..

    • one year ago
  62. jedai17 Group Title
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    why is that u not integrate \[\sqrt{5 + y^2}\]?

    • one year ago
  63. amistre64 Group Title
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    i did not integrate it because I want you to work it out. you are the one that needs the practice at it :)

    • one year ago
  64. amistre64 Group Title
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    if you cant see the integration right off hand, try the "table" method ... which is simply looking up the integral in a table of integrals: look for \(\sqrt{a^2+x^2}\)

    • one year ago
  65. jedai17 Group Title
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    ah.. wat is this \[\sqrt{a^2 + x^2}\]? how will i use this?

    • one year ago
  66. amistre64 Group Title
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    i looked it up in a table, just to be sure. Look at #30

    • one year ago
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  67. amistre64 Group Title
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    a^2 = 5, and x = y

    • one year ago
  68. amistre64 Group Title
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    \[\int \sqrt{5+y^2}dy=\frac12y\sqrt{5+y^2}+\frac12 5^2ln(y+\sqrt{5+y^2})\]from y=0 to y=2

    • one year ago
  69. amistre64 Group Title
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    the 5^2 ln part is a typo; just 5 ln ...

    • one year ago
  70. jedai17 Group Title
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    \[y/2 \sqrt{5 + y^2} + 5/2 \ln (y + \sqrt{5 + y^2}\]

    • one year ago
  71. amistre64 Group Title
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    yes

    • one year ago
  72. jedai17 Group Title
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    what's next?

    • one year ago
  73. amistre64 Group Title
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    plug in the values of your limit to determine the values

    • one year ago
  74. amistre64 Group Title
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    \[(3+\frac52 ln(5))-(\frac54ln(5))-4\] the -4 is from the simpler part we did earlier

    • one year ago
  75. amistre64 Group Title
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    with any luck this simplifies to:\[\frac54ln(5)-1\]

    • one year ago
  76. jedai17 Group Title
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    is this the answer? why -4 ?

    • one year ago
  77. amistre64 Group Title
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    \[\int\sqrt{5+y^2}-(y+1)~dy\] \[\int\sqrt{5+y^2}dy-\int y+1~dy\] \[\int\sqrt{5+y^2}dy-(\frac12y^2+y)\] and since y=0 to 2 \[\int\sqrt{5+y^2}dy-4\]

    • one year ago
  78. amistre64 Group Title
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    i cant verify the steps taken to swap the limits from dy dx to dx dy at the moment, but assuming that was done correctly .... we have come to the end of it

    • one year ago
  79. jedai17 Group Title
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    is that the answer?

    • one year ago
  80. amistre64 Group Title
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    is what the answer?

    • one year ago
  81. jedai17 Group Title
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    i'm sorry..i don't understand it. it's a little bit difficult... :(

    • one year ago
  82. amistre64 Group Title
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    the hardest part was not the y^2/2+y ... that just equals -4 we worked thru the sqrt(5+y^2) from the table, and inputted the limit values to get: \[\frac54ln(5)-4\]as a final result, assuming the swapping of dydx to dxdy was correct to begin with

    • one year ago
  83. jedai17 Group Title
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    i must study first and review about integral in order for me to understand this..

    • one year ago
  84. amistre64 Group Title
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    good luck :)

    • one year ago
  85. jedai17 Group Title
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    please teach me again next time..thanks for your time.. :)

    • one year ago
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