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Math Help - Bounded Double Integral Problem.

  1. #1
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    Bounded Double Integral Problem.

    So I'm trying to do a problem for a hw, and I can't seem to get the proper answer

    The problem is giving a triangle, whose side are y from 0 to 2 and x from 0 to 4.

    It wants me to integrate the function f(x,y) = x^2y within this boundary

    So I chose for my constant of this function to be y is less than or equal to 2x and greater than or equal to 0. This is confirmed by an example in the book using an identical triangle

    So i integrated the function with respect to x using these limit and go 8x^4/3
    I then integrated between 0 and 2 with respect to y and go 256/15

    this isnt right, and I cant seem to get a different answer than this

    any help?
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  2. #2
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    Quote Originally Posted by TheUnfocusedOne View Post
    So I'm trying to do a problem for a hw, and I can't seem to get the proper answer

    The problem is giving a triangle, whose side are y from 0 to 2 and x from 0 to 4.

    It wants me to integrate the function f(x,y) = x^2y within this boundary

    So I chose for my constant of this function to be y is less than or equal to 2x and greater than or equal to 0. This is confirmed by an example in the book using an identical triangle

    So i integrated the function with respect to x using these limit and go 8x^4/3
    I then integrated between 0 and 2 with respect to y and go 256/15

    this isnt right, and I cant seem to get a different answer than this

    any help?

    According to what you describe, it seems to me that the triangle's vertices are (0,0), (0,2), (4,0) ==> the straight line joining these last two vertices is y=-\frac{x}{2}+2, so you can put the following limits:

    0\leq x\leq 4\,,\,\,-\frac{x}{2}+2\leq y\leq 0, and thus you want:

    \int\limits_0^4\int\limits_0^{-\frac{x}{2}+2}\!\!x^2y\,dydx=\frac{1}{2}\,\int\lim  its_0^4x^2\left(\left(-\frac{x}{2}+2\right)^2\right)dx=\frac{64}{15} (I think)

    I don't understand why you put 0\leq y\leq 2x when x goes from 0 to 4...

    Tonio
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  3. #3
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    Im not sure why you chose to do -x/2 + 2 when you could of just done x/2, serves the same purpose in my mine since when x = 4, y = 2 and when x = 0, y = 0.

    The boundaries I set were the same as the book had set. Im not trying to be rude but I kinda wanna follow the book seeing as though thats where the problems are from.
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  4. #4
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    Quote Originally Posted by TheUnfocusedOne View Post
    Im not sure why you chose to do -x/2 + 2 when you could of just done x/2, serves the same purpose in my mine since when x = 4, y = 2 and when x = 0, y = 0.

    The boundaries I set were the same as the book had set. Im not trying to be rude but I kinda wanna follow the book seeing as though thats where the problems are from.

    It's not that x/2 "works": it's just incorrect: the boundaries of the problem YOU gave form a triangle with vertices at (0,0), (0,2), (4,0), and these points are joined by the lines x = 0, 0 <= y <=2 (joining (0,0), (0,2)), the line y = 0, 0 <= x <= 4 (joining (0,0), (4,0)), and the line y = -x/2 + 2, joining (0,2) and (4,0).

    The line y = x/2 passes thru the origin and the point (4,2), and this last one is not in your triangle.

    What book do you use, anyway?

    Tonio
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  5. #5
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    Multivariable Calculus by Jon Rogawski

    Its only the last section of a much large book that brings one through calcs 1-3
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  6. #6
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    attaching a photo of the triangle
    Attached Thumbnails Attached Thumbnails Bounded Double Integral Problem.-15-2-figure20b.gif  
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  7. #7
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    Quote Originally Posted by TheUnfocusedOne View Post
    attaching a photo of the triangle

    Ooh, I see! Your first post misled me since it indeed is y between zero and 2 BUT x is between 2y and 2.
    Anyway, I did it in both forms and I get the same result as you did: 256/15, so I can't tell you why the book says otherwise.

    Tonio
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