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John_henderson

  • 2 years ago

V= ∫ ∫ 4dydx + ∫ ∫ 4dydx For the first set of integrals the outer limits are from 1 to 0. The inner limits are from X^2 to 0. For the second set of integrals the outer limits are from 2 to 1. The inner limits are from (X-2)^2 to 0. Sketch the limits of integration, and the shape of the prism, Change the order of limits and solve the volume of the prism. I need help on changing the order of integration and sketching the prism. Please Help.

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  1. John_henderson
    • 2 years ago
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    \[Volume=\int\limits_{0}^{1}\int\limits_{0}^{x^2} 4dydx + \int\limits_{1}^{2}\int\limits_{0}^{(x-2)^2}4dydx\]

  2. John_henderson
    • 2 years ago
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    I need the order to be as dxdy? and sketch the prism ? i guess the vol= pi*y^2

  3. TuringTest
    • 2 years ago
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    the way the problem is given you are not doing a surface revolution, so we can supposed that 4 represents the height, so it's really just about changing the integration order have you got a sketch yet?

  4. John_henderson
    • 2 years ago
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    Oh yes the lengtth is 4cm sorry i missed that out!

  5. John_henderson
    • 2 years ago
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    let me skecth it...

  6. John_henderson
    • 2 years ago
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    |dw:1354460197419:dw|

  7. John_henderson
    • 2 years ago
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    Some thing like that for the first area?

  8. TuringTest
    • 2 years ago
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    you should really draw the two graphs on top of one another to get a feel for the situation|dw:1354460240069:dw|

  9. TuringTest
    • 2 years ago
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    so you can see why they call it a "prism" so, where is the intersection point?

  10. John_henderson
    • 2 years ago
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    Thats what i did , first but the area of region was o the left.. anyway i guess the limits are going to be x=rooty to 0 , 1 to 0 and x=rooty+2 to 2,1 ?

  11. TuringTest
    • 2 years ago
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    almost, but look at a strip for x's bounds...|dw:1354460805961:dw|since x is bounded by the left half of the second graph we need the negative square root now this can be done with just one double integral instead of two; what are the new bounds on y?

  12. John_henderson
    • 2 years ago
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    so do you equate the limits in order to form one double integral? , so i think y= is going to be sqrt(x^2) to 2-sqrt(x-2)^2))

  13. TuringTest
    • 2 years ago
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    why are you taking the square root of x^2 ? just look at the bounds x is bounded below by sqrt(y) and above by 2-sqrt(y), so those are the limits for dx to find the limits on y, all you need is the intersection point of the two graphs. what is the range of y values in the region D ?

  14. TuringTest
    • 2 years ago
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    what is this point?|dw:1354461347743:dw|

  15. John_henderson
    • 2 years ago
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    1 to 0 -root2 to 0?

  16. TuringTest
    • 2 years ago
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    y cannot be negative, and you are giving two sets of bounds. y only has one set of bounds; again find the point of intersection! btw, you should also always say "upper" to "lower" as in "0 to 1", not "1 to 0" unless you specifically intend to integrate negatively

  17. TuringTest
    • 2 years ago
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    I mean lower to upper*

  18. John_henderson
    • 2 years ago
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    Sorry , 0 to 1, . 0 to 3+root2

  19. John_henderson
    • 2 years ago
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    im not sure as there are 2 seperate coordinates ?which do i sub in?

  20. TuringTest
    • 2 years ago
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    okay, I still don't see why you have two sets of bounds, and why they are both numerical. the inner bounds on dx are in terms of y, and we have already found them the outer bounds should be the numerical limits for y in the region

  21. TuringTest
    • 2 years ago
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    |dw:1354461728579:dw|so x is bound as follows\[\sqrt y\le x\le2-\sqrt y\]and y is simply bound between the x-axis and the point of intersection, at which y=1\[0\le y\le1\]so that's your new integration, all one integral with those bounds.

  22. TuringTest
    • 2 years ago
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    one double integral of course*

  23. John_henderson
    • 2 years ago
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    oh right so your dont have to compute the values of x=2,1 why is that?

  24. TuringTest
    • 2 years ago
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    because with the first way we were doing it, we had two different upper bounds for y as we went along x; at first the above for y was bound by x^2, then at x=1 it changes to being bound above by (x-2)^2 this change in our upper bounds forced us to use two integrals when integrating along x last after changing the order, x is bound above the whole time by 2-sqrt(y) and below the whole time by sqrt(y), hence only one integral is necessary.

  25. TuringTest
    • 2 years ago
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    here is the first integral|dw:1354462159166:dw|

  26. TuringTest
    • 2 years ago
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    |dw:1354462244107:dw|so going along x last, this is why we needed two integrals....

  27. TuringTest
    • 2 years ago
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    when integrating along y last we have no such division point, the upper and lower bound on x stay the same

  28. John_henderson
    • 2 years ago
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    right i see so the values of x=1 , 2 are subed into (x-2)^2 to get 0 to1..

  29. TuringTest
    • 2 years ago
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    you could look at it that way, or you could just look at the pic (you may have to run your head sideways to see the shape along y) and see that y is always somewhere between 0 and 1, since the intersection point is (1,1)

  30. TuringTest
    • 2 years ago
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    your way will not always work actually, those values are coincidental to some extent why not sub into the other function, x^2, by your logic?

  31. John_henderson
    • 2 years ago
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    I have finally done got the ans and i got 8/3 i guess i need to multiply this by pi(8/3)^2 =v so get the volume?

  32. TuringTest
    • 2 years ago
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    |dw:1354462650777:dw|to me, the best way is to say "what is this range, here?" and visually you can get it

  33. TuringTest
    • 2 years ago
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    no you don't need to multiply by anything if 4 is the function for the height. This is not a volume of revolution! let me try the integral myself...

  34. John_henderson
    • 2 years ago
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    yes is can clearly see its from o to 1.. its much easir sketching it !

  35. TuringTest
    • 2 years ago
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    yeah I always sketch it anyway, I get the same as you; 8/3 since we are NOT revolving this shape around any axis, this is not going to be one of those cal1-type tricks where we need to use the formula for a circle (in which case we would have needed to have that in the integral, remember?). Actually, the whole point of multivariable calc is that we don't need the revolution trick at all, that is a beginners method only. This shape is not circular; it has an base area D (which we find by integrating dxdy in the graph) and a height of 4 so we just do base times height here, i.e. 4 times the double integral

  36. John_henderson
    • 2 years ago
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    I have got a answer for the voume which is 64/9 pi, does that sound about right. I integrated with respect to dxdy?

  37. TuringTest
    • 2 years ago
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    but the 4 in the problem as you are given it is most likely representing the height, so 8/3 should be the final answer.

  38. TuringTest
    • 2 years ago
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    no, it sounds very wrong, read what I wrote: this is NOT A VOLUME OF REVOLUTION

  39. John_henderson
    • 2 years ago
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    Damn seriously i thought it was the volume of revolution? , right i undertand now basically 8/3 is the base and 4=h , well if you multiply them together dont you get area?

  40. TuringTest
    • 2 years ago
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    |dw:1354463282138:dw|this is what the shape sort of looks like. pi should not be involved as this is not circular

  41. TuringTest
    • 2 years ago
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    whether 8/3 is the final answer or not depends on what the 4 represents in your initial problem

  42. John_henderson
    • 2 years ago
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    I finally get it lol the inregral is the area = 8/3 and the h=4cm so v=a*h v=32/3 cm^3

  43. TuringTest
    • 2 years ago
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    the area of the base D is just\[A=\int_0^1\int_{\sqrt y}^{2-\sqrt y}dxdy\]so I think the 4 in the integral represents height. If not, then why is it there?

  44. TuringTest
    • 2 years ago
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    no, listen to what I am saying, you need to look more closely at what your problem is telling you how exactly is the question worded?

  45. John_henderson
    • 2 years ago
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    The volue of an unusally shaped prism of length 4cm can be evaluated by using the sum of two double integrals as show above. A. Sketch the region bounded by the limts of integration of the ingreals to illustrate the shape of the cross-section of the prism. B. Change the order of integration and adjust the limits appropriately for the integrals in a and express the volume as a single double interal. hence evaluate the integral and obtain the volume of the prism.

  46. TuringTest
    • 2 years ago
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    okay, so the initial two integrals you are given represent the *volume*, not just the base area "The volume of an unusually shaped prism of length 4cm can be evaluated by using the sum of two double integrals as show above." that tells us that the 4 in the integral must be the height (once you get to triple integrals you will see that this is the same as integrating in the z from 0 to 4)

  47. John_henderson
    • 2 years ago
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    Right i see i have never done triple integral before, i dont know anything about that. Anyway i guess the overall answer 32/3 then?

  48. TuringTest
    • 2 years ago
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    no, you are multiplying by 4 again, but the initial set of two integrals you are given, we are told, represent VOLUME not AREA since the area is given by just\[A=\int_0^1\int_{\sqrt y}^{2-\sqrt y}dxdy\] (notice the area does not have 4 in the integrand) the volume must be\[Ah=A(4)=\int_0^1\int_{\sqrt y}^{2-\sqrt y}4dxdy=8/3\]and that's it

  49. John_henderson
    • 2 years ago
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    32/3?

  50. TuringTest
    • 2 years ago
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    no, reread please

  51. John_henderson
    • 2 years ago
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    2/3 divide it by 4

  52. TuringTest
    • 2 years ago
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    why divide?

  53. John_henderson
    • 2 years ago
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    what is is purpose of 4 ? so i guess you leave it as 8/3 =v ,v=AH, well in the question it says volume so i think it remains the same?

  54. TuringTest
    • 2 years ago
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    yes, that is my whole point

  55. John_henderson
    • 2 years ago
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    Oh i see that means the area is 2/3 which is a H=4 , which therefore gives 8/3 , Wow this is real trick question huh? Its so easy to make mistakes!

  56. John_henderson
    • 2 years ago
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    I guess you really need to read the question Lol..!

  57. TuringTest
    • 2 years ago
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    the integral\[A=\iint\limits_D dA=\int_a^b\int_{f(x)}^{g(x)}dydx\](with NO INTEGRAND) is the area, so the 4 MUST be the height, so 8/3 is the final answer. yes, I agree that the wording of the question is meant to throw you off, but I'm quite certain that they mean what you said; the area is 2/3 and height is 4

  58. John_henderson
    • 2 years ago
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    Thank you for your time, i really understand now. I was wondering if your would like to work though another question with me on fourier transform?

  59. TuringTest
    • 2 years ago
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    I don't know Fourier transform. I know Fourier series and Laplace transform, did you mean one of those by chance.

  60. TuringTest
    • 2 years ago
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    and you're welcome :)

  61. John_henderson
    • 2 years ago
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    Yes is mean fourier series sorry.

  62. TuringTest
    • 2 years ago
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    okay, it's been a reeeeeeeally long time for me with Foiurier series, but I'll have a look.

  63. TuringTest
    • 2 years ago
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    you should really post the problem separately, as a new thread

  64. John_henderson
    • 2 years ago
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    A periodic funtion F(t) is defined by f(t)=t^2, over -pi<t<pi A1. state the period of f(t). 2. sketch f(t) over the domain -3pi<t<3pi B. state with reasons, wheter f(t0 is odd , even or neither odd or even. C. dertermine the first six non zero terms of the fourier series represting f(t).

  65. TuringTest
    • 2 years ago
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    I don't think I know how to do this though...

  66. John_henderson
    • 2 years ago
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    For be its meant to be f(t) only not f(t0

  67. John_henderson
    • 2 years ago
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    it is fourier series not transform.

  68. TuringTest
    • 2 years ago
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    It's still been too long for me; my notes are too foreign looking for me to recover how to do this right now. You should post this separately anyway.

  69. John_henderson
    • 2 years ago
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    i got 2 for the period, the sketch is 3times y= pulus or minus y=x^2 the function is even . but i need help onlt of the series please.

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