## jedai17 Group Title what is double integral function? how to solve double integral function? one year ago one year ago

1. ViPhy Group Title

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 Group Title

is the value of dy similar to the value of dx?

3. yunus Group Title

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 Group Title

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 Group Title

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 Group Title

why it become a^2?

7. hartnn Group Title

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 Group Title

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

9. hartnn Group Title

*after integration w.r.t y,

10. jedai17 Group Title

in this we are integrating y and we assume as x as constant?

11. jedai17 Group Title

what do you mean by w.r.t y, ?

12. hartnn Group Title

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

13. hartnn Group Title

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 Group Title

i'm sorry don't know how to plot this..

15. jedai17 Group Title

i'm sorry i forgot our previous lessons about that..

16. hartnn Group Title

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 Group Title

i know this x=0,y=0,x=2 but i don't know y=x^2/2..i'm sorry

18. hartnn Group Title

so, plot these x=0,y=0,x=2 using 'Draw' tool here.

19. jedai17 Group Title

|dw:1362400369683:dw|

20. jedai17 Group Title

is it correct?

21. hartnn Group Title

where is the line x=2 ?

22. jedai17 Group Title

|dw:1362400538076:dw|

23. jedai17 Group Title

sorry i forgot x = 2..

24. hartnn Group Title

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 Group Title

why it becomes parabola?

26. hartnn Group Title

|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 Group Title

|dw:1362400889186:dw| to |dw:1362400935932:dw|

28. jedai17 Group Title

i'm sorry i forgot how to find the equation of the line

29. hartnn Group Title

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 Group Title

by the way, did you get that final diagram ?

31. jedai17 Group Title

not yet..

32. hartnn Group Title

which part you have doubt with ?

33. jedai17 Group Title

$x =2 and x = \sqrt{2y}$ i don't know how to fnd the equation of this line..

34. hartnn Group Title

x= 2, x=$$\sqrt{2y}$$ so, 2= $$\sqrt{2y}$$ can u find 'y' from here ?

35. jedai17 Group Title

no..

36. jedai17 Group Title

$y=\sqrt{2}/2 ?$

37. hartnn Group Title

: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 Group Title

yes that's true i almost forgot the topics about integration that's why i am like this..

39. hartnn Group Title

|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 Group Title

du = 2x + 2y

41. hartnn Group Title

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 Group Title

yes..

43. hartnn Group Title

so, your inner integral will become, $$\int \dfrac{du}{2\sqrt u}$$ can you integarte this ?

44. jedai17 Group Title

1/2..is it correct?

45. amistre64 Group Title

as long as weve gotten this u substitution correct: $\int \frac{1}{2\sqrt{u}}du=\sqrt{u}$

46. amistre64 Group Title

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 Group Title

y^2+2y+1 = (y+1)^2 sooo $\int\sqrt{5+y^2}-(y+1)~dy$

48. jedai17 Group Title

why it becomes subtraction?

49. jedai17 Group Title

why it become y^2+2y+1 = (y+1)^2 ?

50. amistre64 Group Title

because: $$\Large \int_{a}^{b} f(x)~dx=F(b)-F(a)$$

51. jedai17 Group Title

ah ok tnx..

52. jedai17 Group Title

in this time we are integrating y?

53. amistre64 Group Title

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 Group Title

what's next?

55. amistre64 Group Title

the next integration is next ... $\int\sqrt{5+y^2}-(y+1)~dy$

56. jedai17 Group Title

$\sqrt{5 + y} - y$

57. jedai17 Group Title

is it correct?i'm sorry have a problem in integration..

58. amistre64 Group Title

the square root part seems a little off to me

59. amistre64 Group Title

and it looks like you tried a derivative onthe y+1 parts instead of integeration

60. amistre64 Group Title

$-\int y+1~dy=-(\frac{y^2}{2}+y)$

61. jedai17 Group Title

yes.. i'm sorry a have a problem in integration..

62. jedai17 Group Title

why is that u not integrate $\sqrt{5 + y^2}$?

63. amistre64 Group Title

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 Group Title

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 Group Title

ah.. wat is this $\sqrt{a^2 + x^2}$? how will i use this?

66. amistre64 Group Title

i looked it up in a table, just to be sure. Look at #30

67. amistre64 Group Title

a^2 = 5, and x = y

68. amistre64 Group Title

$\int \sqrt{5+y^2}dy=\frac12y\sqrt{5+y^2}+\frac12 5^2ln(y+\sqrt{5+y^2})$from y=0 to y=2

69. amistre64 Group Title

the 5^2 ln part is a typo; just 5 ln ...

70. jedai17 Group Title

$y/2 \sqrt{5 + y^2} + 5/2 \ln (y + \sqrt{5 + y^2}$

71. amistre64 Group Title

yes

72. jedai17 Group Title

what's next?

73. amistre64 Group Title

plug in the values of your limit to determine the values

74. amistre64 Group Title

$(3+\frac52 ln(5))-(\frac54ln(5))-4$ the -4 is from the simpler part we did earlier

75. amistre64 Group Title

with any luck this simplifies to:$\frac54ln(5)-1$

76. jedai17 Group Title

is this the answer? why -4 ?

77. amistre64 Group Title

$\int\sqrt{5+y^2}-(y+1)~dy$ $\int\sqrt{5+y^2}dy-\int y+1~dy$ $\int\sqrt{5+y^2}dy-(\frac12y^2+y)$ and since y=0 to 2 $\int\sqrt{5+y^2}dy-4$

78. amistre64 Group Title

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 Group Title

80. amistre64 Group Title

81. jedai17 Group Title

i'm sorry..i don't understand it. it's a little bit difficult... :(

82. amistre64 Group Title

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 Group Title

i must study first and review about integral in order for me to understand this..

84. amistre64 Group Title

good luck :)

85. jedai17 Group Title