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Graphical interpretation of a double integral?

  1. Apr 11, 2015 #1
    Hello,

    I was helping my friend prepare for a calculus exam today - more or less acting as a tutor.

    He had the following question on his exam review:

    ∫∫R y2 dA

    Where R is bounded by the lines x = 2, y = 2x + 4, y = -x - 2


    I explained to him that R is a triangle formed by all three of those lines. The solution to the integral is 192, which is pretty easy to calculate by hand without a calculator.

    He asked me why the solution wasn't just the area of R (two triangles if you split R at the line y = 0) which is 128. I told him its because we are integrating y2 with respect to the given boundaries, and that if we were integrating "1" it would be the area of the region R. However, we then proceeded to integrate "1" with the same boundaries, and the answer was 24, not 128.


    So I ended up coming up with my own question. I can draw a pretty good geometric interpretation of most integration problems including three dimensional ones given in different coordinate systems... but I feel like I'm missing something primitive here. Can anyone help me develop a geometric picture of what this type of double integral is representing?


    Thanks!
     
  2. jcsd
  3. Apr 11, 2015 #2

    robphy

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    How did you get 128?
     
  4. Apr 11, 2015 #3
    That's a fantastic question. I multiplied the "divided region" together, rather than added it. The area of R is 24. :p

    We didn't catch on to that earlier, but it is kind of irrelevant to my primary question.

    Where does y2 fit in - graphically speaking - with the region R?
     
  5. Apr 11, 2015 #4

    robphy

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  6. Apr 11, 2015 #5
    Yes I'm familiar with those concepts. Interestingly some more complex integrals are easier for me to visualize. I'm likely over thinking this and confusing myself.

    y2 represents a parabola (or more technically a parabolic cylinder) that is orthogonally oriented to the triangle formed by the lines given by R. Such a shape would possess no volume in the "real world"... it would be two shapes orthogonal to one another; a triangle and a parabola, whose intersection forms a line.

    That is what is confusing me.
     
  7. Apr 11, 2015 #6
    I drew it out by hand and by golly I think I've got it.

    This is the "projection" of the triangle on to the parabolic cylinder, yes? So the "stretched" area of the triangle as it would appear on the parabola z = y2


    Part of my confusion was probably incorrectly calculating the area of the region with my friend earlier. As integrating the same region with respect to 1 (or any other constant) would project the region on to a flat surface parallel to the region itself, thus yielding a solution to the integral equal to the area of the region itself...

    :oops: Now I feel only slightly ashamed.
     
  8. Apr 12, 2015 #7

    SteamKing

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    This integral can also represent the second moment of area about the x-axis. The second moment of area is sometimes referred to as the moment of inertia.

    In this case, since the region R straddles the x-axis, evaluating the double integral will calculate the combined second moment of area for the two right triangles, referenced to their common base, which are formed when the region R is intersected by the x-axis.

    The moment of inertia for a right triangle about its base Iy = (1/12)bh3

    In this case, the length of the common base for the two triangles is b = 4, and the height of the upper triangle is h = 8 while the lower triangle height is h = 4.

    Thus

    Iy = Iy upper + Iy lower = (1/12)*4*(83 + 43)

    Iy = 192
     
  9. Apr 12, 2015 #8
    What a double integral generally represents is the volume of a solid above a region on a plane. The solid you have here is called a cylindrical surface. The equation of it is [itex]z=y^2[/itex]. This is a parabola. However, since there is no x, we can let x vary from negative infinity to infinity. Doing so moves this parabola forward and backward parallel to the x-axis. The region "carved out" is the surface. And the double integral is the volume under that surface above the triangular region R.
     
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