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Find double integral over region R

  1. Dec 18, 2012 #1

    sharks

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    1. The problem statement, all variables and given/known data
    17053353dc2e7873bcdde1279d8f98229ca7635d.png
    170533065419ba669070324486b749108e957894.png

    2. Relevant equations
    For example, for f(x,y)=x+y-2
    1705331137368ef375f83d511e0553a6bbe9ab03.png


    3. The attempt at a solution
    I've figured out part (a) which is quite simple. I simply used the relevant equations above for ##f(x,y)= 3(x^2+y^2)##

    I know i should use the given hint to figure out the value of I, which is ##L_f(P)\leq I\leq U_f(P)##

    But i'm not sure how to solve for the value of I, which i've calculated to be:
    $$\frac{3}{4}\sum_{i=1}^m \sum_{j=1}^n \left[ (x^2_i+x^2_{i-1}) + (y^2_j+y^2_{j-1})\right]\Delta x_i \Delta y_j (wrong)$$
    $$(Fixed) \frac{3}{4}\sum_{i=1}^m \sum_{j=1}^n \left[ (x_i+x_{i-1})^2 + (y_j+y_{j-1})^2\right]\Delta x_i \Delta y_j\\ $$
    I've tried the old-fashioned way of expanding the terms to find the sum, but it gets messy after expanding all the terms in i and j. But in this case, since a hint is given, i guess i should be using it to simplify the inequality. But how?

    Here's my next attempt:

    Using the given hint: ##0\leq s \leq t## and applying it to the given limits for x and y:

    ##0\leq x \leq b## gives the corresponding inequality: ##3x^2\leq b^2+bx+x^2 \leq 3b^2##

    ##0\leq y \leq d## gives the corresponding inequality: ##3y^2\leq d^2+dy+y^2 \leq 3d^2##

    I tried adding both of these inequalities but the answer for the middle term, I, doesn't make sense, as it retains x and y, along with b and d.

    ##3x^2 + 3y^2\leq b^2+d^2+bx+dy+x^2+y^2 \leq 3b^2+3d^2##

    I'm stuck. The final answer is not given in my notes, but based on previous calculations of the same type, the answer should be only in terms of b and d, meaning no x and y terms are involved.
     
    Last edited: Dec 18, 2012
  2. jcsd
  3. Dec 18, 2012 #2

    haruspex

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    I would guess that the hint should be applied in individual ranges of the partitions, not to the whole intervals. So in the range [itex]\check{x}\leq x\leq \hat{x}[/itex]
    [itex]\check{x}^2+x\check{x}+x^2\leq3x^2\leq \hat{x}^2+x\hat{x}+x^2 [/itex]
    [itex]\check{x}^2+x\check{x}\leq2x^2\leq \hat{x}^2+x\hat{x}[/itex]
     
  4. Dec 18, 2012 #3

    pasmith

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    I think the point of the hint is that [itex]t^3 - s^3 = (t - s)(t^2 + st + s^2)[/itex].

    Now
    [tex]L_f(P) = \sum_i \sum_j 3(x_{i-1}^2 + y_{j-1}^2)(x_i - x_{i-1})(y_j - y_{j-1}) \\
    \leq \sum_i \sum_j (x_{i-1}^2 + x_{i-1}x_i + x_i^2+ 3y_{j-1}^2)(x_i - x_{i-1})(y_j - y_{j-1}) \\
    = \sum_i \sum_j (x_i^3 - x_{i-1}^3)(y_j - y_{j-1}) + \sum_i \sum_j 3y_{j-1}^2 (x_i - x_{i-1})(y_j - y_{j-1})
    [/tex]
    etc.

    Similarly for [itex]U_f(P)[/itex], but the inequalities will be reversed.
     
    Last edited: Dec 18, 2012
  5. Dec 18, 2012 #4

    sharks

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    I don't get what you did.
    OK, i agree with the first part about $$L_f(P) = \sum_{i=1}^m \sum_{j=1}^n 3(x_{i-1}^2 + y_{j-1}^2)(x_i - x_{i-1})(y_j - y_{j-1})$$Then,
    $$U_f(P) = \sum_{i=1}^m \sum_{j=1}^n 3(x_{i}^2 + y_{j}^2)(x_i - x_{i-1})(y_j - y_{j-1})$$Since,
    $$L_f(P)\leq I\leq U_f(P)$$##I## becomes (i made a mistake in my first post. i fixed it here):
    $$\frac{3}{4}\sum_{i=1}^m \sum_{j=1}^n \left[ (x_i+x_{i-1})^2 + (y_j+y_{j-1})^2\right]\Delta x_i \Delta y_j\\
    =\frac{3}{4}\sum_{i=1}^m \sum_{j=1}^n (x_i+x_{i-1})^2\Delta x_i \Delta y_j + \frac{3}{4}\sum_{i=1}^m \sum_{j=1}^n (y_j+y_{j-1})^2 \Delta x_i \Delta y_j\\$$Evaluating the first sum:
    $$\frac{3}{4}\sum_{i=1}^m \sum_{j=1}^n (x_i+x_{i-1})^2\Delta x_i \Delta y_j\\
    =\frac{3}{4}\sum_{i=1}^m \sum_{j=1}^n (x_i+x_{i-1})^2 (x_i - x_{i-1}) \Delta y_j\\
    =\frac{3}{4}\sum_{i=1}^m \sum_{j=1}^n (x^2_i+2x_ix_{i-1}+x^2_{i-1}) (x_i+x_{i-1}) \Delta y_j\\$$I expanded all the terms in ##i## but how to proceed?
     
    Last edited: Dec 18, 2012
  6. Dec 18, 2012 #5

    pasmith

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    where does the factor of 1/4 come from?

    EDIT: Nevermind - I see now you're taking the average value on [itex][x_{i-1},x_{i}] \times [y_{j-1},y_j][/itex]. Unfortunately that doesn't assist you - see further below.

    Go back to my calculation:

    The etc. here is that:
    [tex]L_f(P) \leq \sum_i \sum_j (x_i^3 - x_{i-1}^3)(y_j - y_{j-1}) + \sum_i \sum_j 3y_{j-1}^2 (x_i - x_{i-1})(y_j - y_{j-1}) \\
    \leq \sum_i \sum_j (x_i^3 - x_{i-1}^3)(y_j - y_{j-1}) + \sum_i \sum_j (x_i - x_{i-1})(y_j^3 - y_{j-1}^3) \\
    = \sum_i (x_i^3 - x_{i-1}^3) \sum_j (y_j - y_{j-1}) + \sum_i (x_i - x_{i-1})\sum_j (y_j^3 - y_{j-1}^3) \\
    = b^3d + d^3b[/tex]
    by telescoping.

    Hence [itex]L_f(P) \leq b^3d + d^3b[/itex]. But the right hand side is independent of [itex]P[/itex], so in fact [itex]\sup_P L_f(P) \leq b^3d + d^3b[/itex].

    If you carry out the same calculation for [itex]U_f(P)[/itex], you should find that [itex]\inf_P U_f(P) \geq b^3d + d^3b[/itex].

    So [itex]\sup_P L_f(P) \leq b^3d + d^3b \leq \inf_P U_f(P)[/itex] and if you invoke the proposition that a continuous function is integrable you are done:
    [tex]\sup_P L_f(P) = b^3d + d^3b = \inf_P U_f(P)[/tex]

    EDIT: The alternative way to use the hint is to invoke the intermediate value theorem to show that there exists [itex](\zeta_i,\eta_j) \in [x_{i-1},x_i] \times [y_{j-1},y_j][/itex] such that
    [tex]3(\zeta_i^2 + \eta_j^2) = (x_{i-1}^2 + x_{i-1}x_i + x_i^2 + y_{j-1}^2 + y_{j-1}y_j + y_j^2)[/tex]
    so that the resulting Riemann sum is
    [tex]
    \sum_i \sum_j 3(\zeta_i^2 + \eta_j^2)(x_i - x_{i-1})(y_j - y_{j-1}) \\
    = \sum_i \sum_j ((x_i^3 - x_{i-1}^3)(y_j - y_{j-1}) + (x_i - x_{i-1})(y_j^3 - y_{j-1}^3)
    = b^3d + d^3b[/tex]
    by telescoping as before. Since the result is independent of [itex]P[/itex], it must be the value of the integral.
     
    Last edited: Dec 18, 2012
  7. Dec 18, 2012 #6

    sharks

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    Here's an example from my notes:
    http://img2.uploadhouse.com/fileuploads/17053/170536720903325240d5cea4d8f9181b2dd80f5a.png
    From the last line, the middle terms in ##i## and ##j## appear to be the mid-points of ##x_i## and ##x_{i-1}##, and ##y_i## and ##y_{i-1}##.
    This is the same method that i've applied to finding ##I##, related to the current problem ##f(x,y)= 3(x^2+y^2)##.
     
  8. Dec 18, 2012 #7

    pasmith

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    Yes: that works for [itex]f(x,y) = x + y - 2[/itex] because [itex](x_{i} + x_{i-1})(x_{i} - x_{i-1}) = x_{i}^2 - x_{i-1}^2[/itex], which gives a telescoping sum whose value is independent of the particular partition. In fact it's that line of thought (finding a telescoping sum) which leads one to consider [itex]x_i^3 - x_{i-1}^3 = (x_i^2 + x_ix_{i-1} + x_{i-1}^2)(x_i - x_{i-1})[/itex].
     
  9. Dec 18, 2012 #8

    sharks

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    OK, i get the idea of telescoping sum and how that would ignore the partition size, since only the upper and lower bounds are taken into account when evaluating the sum, but on line 2, how did you get ##x_{i-1}^2 + x_{i-1}x_i + x_i^2## from ##3x_{i-1}^2##?

    From what i understand, ##t = x_i## and ##s=x_{i-1}##, so that should give:
    $$(x_i - x_{i-1})(x_{i}^2 + x_{i-1}x_i + x_{i-1}^2)$$
    How is that relevant to the calculations? I guess my confusion arises from the beginning of the problem. The one and only example in my notes used the method for finding the mid-points of x and y (according to me, the reason for using that method is that finding the mid-points would automatically allow the middle term of the inequality to lie in-between the P lower and upper sums of f), but if i use the same principle in the current problem, it doesn't work. Why? What method are you using at that particular step in solving the problem?

    From this point forwards, i get the solution as you did:
    So, i think i should work backwards, in order to understand how you arrived at that point. I still don't get it.

    EDIT: I believe that i finally figured out what the hint really meant!

    Given: ##0\leq s \leq t, 3s^2 \leq t^2 +ts +s^2 \leq 3t^2##

    In this case, let ##s = x_{i-1}## and let ##t = x_i##, which makes sense in the inequality.

    So, i get: ##0\leq x_{i-1} \leq x_i## which corresponds to:
    $$3x^2_{i-1} \leq x^2_i +x_ix_{i-1} + x^2_{i-1} \leq 3x^2_i$$
    Similarly, for ##0\leq y_{j-1} \leq y_j##:
    $$3y^2_{i-1} \leq y^2_j +y_jy_{j-1} + y^2_{j-1} \leq 3y^2_j$$
    Adding up those 2 inequalities:
    $$3x^2_{i-1} + 3y^2_{j-1} \leq x^2_i +x_ix_{i-1} + x^2_{i-1} + y^2_j +y_jy_{j-1} + y^2_{j-1}\leq 3x^2_i + 3y^2_j$$

    Now, if we go back to the original definition:
    $$L_f(P)\leq I\leq U_f(P)
    \\L_f(P) = \sum_{i=1}^m \sum_{j=1}^n 3(x_{i-1}^2 + y_{j-1}^2)(x_i - x_{i-1})(y_j - y_{j-1})
    \\U_f(P) = \sum_{i=1}^m \sum_{j=1}^n 3(x_{i}^2 + y_{j}^2)(x_i - x_{i-1})(y_j - y_{j-1})$$
    Therefore, the middle term of the inequality becomes:$$I = \sum_{i=1}^m \sum_{j=1}^n (x^2_i +x_ix_{i-1} + x^2_{i-1} + y^2_j +y_jy_{j-1} + y^2_{j-1})(x_i - x_{i-1})(y_j - y_{j-1})$$
    Then, simplifying gives:
    $$\sum_i \sum_j (x_i^3 - x_{i-1}^3)(y_j - y_{j-1}) + \sum_i \sum_j (x_i - x_{i-1})(y_j^3 - y_{j-1}^3) \\
    = \sum_i (x_i^3 - x_{i-1}^3) \sum_j (y_j - y_{j-1}) + \sum_i (x_i - x_{i-1})\sum_j (y_j^3 - y_{j-1}^3) \\
    = b^3d + d^3b$$

    This is a much simpler method!
     
    Last edited: Dec 18, 2012
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