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Negative Area in this problem

  1. Jul 28, 2010 #1
    1. The problem statement, all variables and given/known data
    Don't need help solving just have a question about the question

    Rectangle ABCD has vertices C and d on the x-axis and vertices A and B on part of the parabola y = 9 - x^2 that is above the x-axis.

    a.) Find the coordinates of A, B, C, and D in terms of x.
    b.) Determine the Area of the rectangle as a function of x.
    c.) What is the domain for the area function? Hint: Area cannot be negative

    well I'm completley lost here on how to do this
    also for c. technically speaking can area be negative area is a tensor a magnitude with more then one direciton no? So an area can be negative it would just be in the set of complex numbers no? So yah I'm not buying "area cannot be negative" I'm sure that it can be no? I'm just unsure in this situation becaues this situation refers to a rectangle withing a parabola but still negative area is possible... please I'd like to know the truth, i.e. in real life there is no such thing but in math there is, much like how you are told lies that cos(x)=-2 has no solution untill you find out later it does it's just not in the domain of reals, is this the same case as here, i.e. here the problem is lieing to me as there is such a thing it's just has no application in the real number system or something, lol I'm sure that there is such thing as negative area but it's been so long sense I took calc lol I remeber it coming up and there was such a thing no?

    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jul 28, 2010 #2

    Mark44

    Staff: Mentor

    No, area can't be negative. It's a two-dimensional quantity.
    No.
     
  4. Jul 28, 2010 #3

    Mentallic

    User Avatar
    Homework Helper

    The only time negative area came up in calculus was when integrating in some domain of a function which was under the x-axis.

    For example, [tex]\int^3_{-3}(9-x^2)dx=36 u^2[/tex] and obviously [tex]-\int^3_{-3}(9-x^2)dx=-36u^2[/tex] but this is equivalent to [tex]\int^3_{-3}-(9-x^2)dx=\int^3_{-3}(x^2-9)dx[/tex]
     
  5. Jul 28, 2010 #4
    Thanks!
     
  6. Jul 28, 2010 #5

    Mark44

    Staff: Mentor

    If the integral that represents area is set up correctly, you won't get a negative value.
    If you want to calculate the area between the graph of y = x2 - 9 and the x-axis, the integral should look like this:
    [tex]\int^3_{-3} 0 - (x^2 - 9)dx = \int_{-3}^3 -x^2 + 9~dx = +36[/tex]

    When you set up the integral, the height of the typical area element is ytop - ybottom, or 0 - (x2 - 9), which can be written more simply as -x2 + 9. This gives you a positive value for the height. The other dimension of the typical area element is [itex]\Delta x[/itex], which is already positive.

    BTW, [tex]\int^3_{-3}(9-x^2)dx \neq 36 u^2[/tex]
    Where did the u2 come from?
     
  7. Jul 28, 2010 #6

    Mentallic

    User Avatar
    Homework Helper

    Yes, but I chose not to be picky because of...
    ...and just explain where he could've possibly seen negative area. Many teachers also teach their students to just take the absolute value of the integral as the easy way out.


    The u represents units and is not a variable. Also another common thing to do (back in my class at least) when dealing with integrals that represent areas.
     
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