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Double Integration without Anti-Derivatives

  1. Feb 8, 2014 #1
    1. The problem statement, all variables and given/known data
    Make a good sketch of the plane region D defined by the following simultaneous inequalities:
    D: y >/= -2x, 2y >/= x, 2y </= 4-x.
    Use deep conceptual understanding the insight (and no antiderviative calculations!) to reduce the iterated integral below to a simple algebraic expression depending on the parameters a, b, and c:
    I = ∫∫(a+bx+cy)dA.
    Enter the value of I corresponding to a = -0.8, b = 0.7, and c = -1.2.

    a) -7.04
    b) -6.00
    c) -4.67
    d) 8.07
    e) 2.74


    2. Relevant equations
    None, other than the integral equation provided in the question.


    3. The attempt at a solution
    As asked in the question, I made a quick sketch, shown below:
    30ucktg.png
    Then I used the dot product to figure out that the region in consideration is a right triangle. Hence, it was easy to calculate the area of this region in the 2D plane. I calculated it to be:
    A = 3.097097656
    Now, I don't know how to extend this idea to calculate the volume since f(x,y) is not constant.


    Any help would be greatly appreciated. Thanks!
     
  2. jcsd
  3. Feb 8, 2014 #2

    haruspex

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    Since two of the lines are at right angles, it would be natural to do a change of co-ordinates. This will effectively turn b and c into two different coefficients, i.e. something like ∫∫(a+b'u+c'v)dA. Can you then use a geometric interpretation of ∫∫b'u.dA etc?
     
  4. Feb 8, 2014 #3
    Unfortunately, I don't quite understand the idea that you are suggesting. I am unsure on how switching two co-ordinates will achieve anything in regards to this problem.
     
  5. Feb 8, 2014 #4

    SammyS

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    What does the equation, z = a+bx+cy represent ?
     
  6. Feb 8, 2014 #5
    Hm... it represents a plane. Are you suggesting that I find the volume of the two pieces (triangular base + another triangular piece on top with the slanted plane)?
     
  7. Feb 8, 2014 #6

    haruspex

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    No, SammyS and I are both suggesting you think of it as a volume integral.
     
  8. Feb 8, 2014 #7

    LCKurtz

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    I don't think your area A is correct. I suggest re-calculating it with exact values instead of decimals. Also, it isn't difficult to find the centroid of a right triangle. Your problem can be thought of as the volume of a triangular prism with slanted top.

    Are you sure the values for a,b,c and the answer choices are all typed correctly?

    [Edit] Never mind about the typing, all is correct except your area is only sort of close.
     
    Last edited: Feb 8, 2014
  9. Feb 8, 2014 #8
    Yes, the values are correct (as shown in my question). Althought I must point out that two previous questions have had incorrect sets of answers in this particular problem set. My professor acknowlegded them as well.
     
  10. Feb 8, 2014 #9

    LCKurtz

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    Yes, I had a mistake in calculations I found. Still, you should use exact values, certainly for the area of the triangle. And the answers do have a correct choice.
     
  11. Feb 8, 2014 #10
    Well, I am still a bit confused as to where to go from here. After reading the responses and doing some research online, I stumbled upon something called the "Jacobian." Am I drifting off on a tangent to infinity?
     
  12. Feb 8, 2014 #11

    Ray Vickson

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    For each ##(x,y) \in D##, the integrand ##f = a + bx + cy## is ## f = a + \vec{u} \cdot \vec{r}##, where ##\vec{u} = (c,d)##, ##\vec{r} = (x,y)## and ##\cdot## is the inner product. Since integration is a 'linear' operation, we can pull the ##\vec{u} \cdot (\cdots)## outside the integral, to get
    [tex] \text{integral } = a \int_D\, dA + \vec{u} \cdot \int_D \vec{r} \, dA.[/tex]
    Do you recognize both of these integrals, perhaps from a physics course or somewhere similar?
     
  13. Feb 8, 2014 #12
    No, I don't recognize this from any of my courses. I was able to follow through the logic that you presented. However, shouldn't it be ##\vec{u} = (b,c)##? Also, when you break the integral into two in the end, don't I still have to use anti-derivatives for the second piece in x and y (as contained in the vector r)?
     
  14. Feb 8, 2014 #13

    haruspex

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    Not if you recognize the physical interpretation of the integral and so can apply a well-known formula.
    SammyS and I were pushing you towards thinking of it as a volume integral. LCKurtz also refers to this, specifically as a triangular prism. My rotation of co-ordinates suggestion was to turn it from a triangular prism into something a bit simpler. But the best suggestion is LCKurtz's mention of centroids. (Ray V might also have that in mind.)
    How do you calculate the location of the centroid of a lamina?
     
  15. Feb 8, 2014 #14

    Ray Vickson

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    Yes, it should be ##\vec{u} = (b,c)##. The first term should be easy; the second term will be familiar to you if you have taken a Mechanics course; otherwise, perhaps not.
     
  16. Feb 8, 2014 #15

    Well, for a right triangle, it's simple as the formulas are already dervided online. Or, we could simply integrate using (assuming density is constant):

    x_centroid = 1/A * ∫∫x*dA
    y_centroid = 1/A * ∫∫y*dA

    Suppose I do find the centroid co-ordinates, how does that result into determining a volume? I feel like I am missing something big here but I just can't see it.
     
  17. Feb 8, 2014 #16

    LCKurtz

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    So you could rewrite those as$$
    \bar x A =\iint x~dA,~~\bar y A = \iint y~dA$$Can you use that to rewrite your problem$$
    I = \iint a + bx + cy~dA~\text{?}$$
     
  18. Feb 8, 2014 #17

    SammyS

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    In the original statement of the problem, it's clear that you are not to evaluate the integral by using anti-derivatives.

    I agree with LC Kurtz that you can calculate the volume of a triangular prism -- or maybe two prisms. (I haven't done the calculations.)
     
  19. Feb 8, 2014 #18

    LCKurtz

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    You don't need to do that.
     
  20. Feb 8, 2014 #19

    SammyS

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    Oh, right!

    Using the centroid takes care of all that complication !
     
  21. Feb 8, 2014 #20

    haruspex

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    No, I meant that there are two physical interpretations available for your integral: as the volume of a prism (or tetrahedron if you do the co-ordinate rotation); or as the centroid of a lamina. The second is definitely easier.
     
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