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jedai17

  • 2 years ago

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

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  1. ViPhy
    • 2 years ago
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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.

  2. jedai17
    • 2 years ago
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    is the value of dy similar to the value of dx?

  3. yunus
    • 2 years ago
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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.

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

  5. hartnn
    • 2 years ago
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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 ?

  6. jedai17
    • 2 years ago
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    why it become a^2?

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

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

  9. hartnn
    • 2 years ago
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    *after integration w.r.t y,

  10. jedai17
    • 2 years ago
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    in this we are integrating y and we assume as x as constant?

  11. jedai17
    • 2 years ago
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    what do you mean by w.r.t y, ?

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

  13. hartnn
    • 2 years ago
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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)?

  14. jedai17
    • 2 years ago
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    i'm sorry don't know how to plot this..

  15. jedai17
    • 2 years ago
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    i'm sorry i forgot our previous lessons about that..

  16. hartnn
    • 2 years ago
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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..

  17. jedai17
    • 2 years ago
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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

  18. hartnn
    • 2 years ago
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    so, plot these x=0,y=0,x=2 using 'Draw' tool here.

  19. jedai17
    • 2 years ago
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    |dw:1362400369683:dw|

  20. jedai17
    • 2 years ago
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    is it correct?

  21. hartnn
    • 2 years ago
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    where is the line x=2 ?

  22. jedai17
    • 2 years ago
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    |dw:1362400538076:dw|

  23. jedai17
    • 2 years ago
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    sorry i forgot x = 2..

  24. hartnn
    • 2 years ago
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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

  25. jedai17
    • 2 years ago
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    why it becomes parabola?

  26. hartnn
    • 2 years ago
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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,

  27. hartnn
    • 2 years ago
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    |dw:1362400889186:dw| to |dw:1362400935932:dw|

  28. jedai17
    • 2 years ago
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    i'm sorry i forgot how to find the equation of the line

  29. hartnn
    • 2 years ago
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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= \(\sqrt{2y}\)

  30. hartnn
    • 2 years ago
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    by the way, did you get that final diagram ?

  31. jedai17
    • 2 years ago
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    not yet..

  32. hartnn
    • 2 years ago
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    which part you have doubt with ?

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

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

  35. jedai17
    • 2 years ago
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    no..

  36. jedai17
    • 2 years ago
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    \[y=\sqrt{2}/2 ?\]

  37. hartnn
    • 2 years ago
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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

  38. jedai17
    • 2 years ago
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    yes that's true i almost forgot the topics about integration that's why i am like this..

  39. hartnn
    • 2 years ago
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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=... ?

  40. jedai17
    • 2 years ago
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    du = 2x + 2y

  41. hartnn
    • 2 years ago
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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 ?

  42. jedai17
    • 2 years ago
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    yes..

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

  44. jedai17
    • 2 years ago
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    1/2..is it correct?

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

  46. amistre64
    • 2 years ago
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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 \[\sqrt{ 1+2^2+y^2}-\sqrt{ 1+(\sqrt{2y})^2+y^2}\] \[\sqrt{ 5+y^2}-\sqrt{ 1+2y+y^2}\]

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

  48. jedai17
    • 2 years ago
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    why it becomes subtraction?

  49. jedai17
    • 2 years ago
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    why it become y^2+2y+1 = (y+1)^2 ?

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

  51. jedai17
    • 2 years ago
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    ah ok tnx..

  52. jedai17
    • 2 years ago
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    in this time we are integrating y?

  53. amistre64
    • 2 years ago
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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

  54. jedai17
    • 2 years ago
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    what's next?

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

  56. jedai17
    • 2 years ago
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    \[\sqrt{5 + y} - y\]

  57. jedai17
    • 2 years ago
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    is it correct?i'm sorry have a problem in integration..

  58. amistre64
    • 2 years ago
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    the square root part seems a little off to me

  59. amistre64
    • 2 years ago
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    and it looks like you tried a derivative onthe y+1 parts instead of integeration

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

  61. jedai17
    • 2 years ago
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    yes.. i'm sorry a have a problem in integration..

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

  63. amistre64
    • 2 years ago
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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 :)

  64. amistre64
    • 2 years ago
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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 \(\sqrt{a^2+x^2}\)

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

  66. amistre64
    • 2 years ago
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    i looked it up in a table, just to be sure. Look at #30

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  67. amistre64
    • 2 years ago
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    a^2 = 5, and x = y

  68. amistre64
    • 2 years ago
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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

  69. amistre64
    • 2 years ago
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    the 5^2 ln part is a typo; just 5 ln ...

  70. jedai17
    • 2 years ago
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    \[y/2 \sqrt{5 + y^2} + 5/2 \ln (y + \sqrt{5 + y^2}\]

  71. amistre64
    • 2 years ago
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    yes

  72. jedai17
    • 2 years ago
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    what's next?

  73. amistre64
    • 2 years ago
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    plug in the values of your limit to determine the values

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

  75. amistre64
    • 2 years ago
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    with any luck this simplifies to:\[\frac54ln(5)-1\]

  76. jedai17
    • 2 years ago
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    is this the answer? why -4 ?

  77. amistre64
    • 2 years ago
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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 \[\int\sqrt{5+y^2}dy-4\]

  78. amistre64
    • 2 years ago
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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

  79. jedai17
    • 2 years ago
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    is that the answer?

  80. amistre64
    • 2 years ago
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    is what the answer?

  81. jedai17
    • 2 years ago
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    i'm sorry..i don't understand it. it's a little bit difficult... :(

  82. amistre64
    • 2 years ago
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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

  83. jedai17
    • 2 years ago
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    i must study first and review about integral in order for me to understand this..

  84. amistre64
    • 2 years ago
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    good luck :)

  85. jedai17
    • 2 years ago
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    please teach me again next time..thanks for your time.. :)

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