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Multiple integrals

  1. May 24, 2012 #1
    1. The problem statement, all variables and given/known data

    Screenshot2012-05-24at42118AMcopy.png

    Screenshot2012-05-24at42118AM.png



    3. The attempt at a solution

    I understand the steps, although it took quite a while, but what I don't understand is that a triangle with base 2 and height 2, it's area is 2. With two triangles of that size the area should be 4. The books says the area is 8.
     
  2. jcsd
  3. May 24, 2012 #2
    You see, the book is not calculating the area of the triangle, but the area times two (there's a two in the integrand)
     
  4. May 24, 2012 #3
    one point of the base is -2,-2, the other point is 0,-2 - looks like base 2 to me
     
  5. May 24, 2012 #4

    sharks

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    Consider the 2 right-angled triangles as a whole isosceles triangle. Then, the base of that isosceles triangle is 4 and the height is 2. The formula for calculating the area is still the same, meaning 1/2 x base x height = 1/2 x 4 x 2 = 4. Multiply that 4 by 2 (from the integrand) and you get the answer = 8.

    Now, if you use the double integral given, you should get the same answer.
     
  6. May 24, 2012 #5
    That to me is the answer.

    Why are you doing that?
     
  7. May 24, 2012 #6

    sharks

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    OK, now that you agree that the area of the whole triangle is 4, we go back to look at your problem, which is:
    [tex]\iint 2\,.dpdv[/tex]
    Ignore the limits (just to simplify your understanding). Notice the "2". That's what we call the integrand. Note that the integrand is a constant in this case. The integrand could have been anything. As a general rule, all constants in integrands can be "pulled out" of the integration. So, we get the following which is equivalent:
    [tex]2\iint \,.dpdv[/tex]
    The area of the entire triangle in the figure is given by:
    [tex]\iint \,.dpdv[/tex]
    And you have seen that the area of the entire triangle is 4. What does this mean? It just means that:
    [tex]\iint \,.dpdv=4[/tex]
    If you just put it all back together:
    [tex]\iint 2\,.dpdv=2\iint \,.dpdv=2\times 4 =8[/tex]
     
  8. May 24, 2012 #7
    ok, thanks.
     
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