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Double Integrals: Will this solution always give the correct answer?

  1. May 31, 2012 #1
    1. The problem statement, all variables and given/known data
    Here is the problem:
    http://dl.dropbox.com/u/64325990/Photobook/question.PNG [Broken]

    Here is the answer:
    http://dl.dropbox.com/u/64325990/Photobook/solution.PNG [Broken]

    So what the answer says is that you can find the volume under the surface minus the volume of the rectangle with height z=1. However I don't see how this will work for every question. Here I will illustrate why I think their solution will not always be correct:
    http://dl.dropbox.com/u/64325990/Photobook/Photo%202012-05-31%205%2004%2056%20PM.jpg [Broken]

    As you can see, if it so happens that the max height of the surface is less than 1 then subtracting the rectangle will give you an answer too small. Does anyone agree with me?



    For those who are interested, I decided to subtract 1 from the z function and find the surface under z = 1+x^2+(y-2)^2 which also gave the correct answer.
    Thank you!
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. May 31, 2012 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Actually, what I see is that you have the paraboloid upside down! With [itex]z= 2+ x^2+ (y- 1)^2[/itex], and squares never being negative, the smallest value of z will be 2 when x= 0 and y= 1.
     
  4. May 31, 2012 #3
    Ah yes, the illustration I made was just an example that would not work right? But the question given here, because the height is always greater than 2, then subtracting a rectangle would work. Thanks for clearing that up.
     
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