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Homework Help: Setting up Double Intergrals to Calculate Volume

  1. Sep 20, 2012 #1
    1. The problem statement, all variables and given/known data
    Set up the double integral in rectangular coordinates for calculating the volume.
    The solid is created by the equation
    pPQYP.png


    2. Relevant equations
    Fubini's Thereom ∫R∫f(x,y)dA = ∫ab∫f(x,y)dydx


    3. The attempt at a solution
    The shape created by Z = 30 - x2 - y2 is 3d parabola facing downards.
    Since it is bounded by Z = 5, the base shape is a circle at origin with radius 5 since
    5 = 30 - x2 - y2 is equal to x2 + y2 = 25
    As a result x is bounded by -5 and 5 while y is bounded by -√(25-x^2) and √(25-x^2)

    As a result to set up the double integral
    -55√(25-x^2)√(25-x^2)30-x^2-y^2dydx

    Are the integral boundaries in correct?
     
    Last edited: Sep 20, 2012
  2. jcsd
  3. Sep 20, 2012 #2

    uart

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    Yes the boundaries are correct (except for the typo where you left out the "-" in the lower limit).

    However, you really should be making use of the cylindrical symmetry here and changing to cylindrical coordinates. It makes the problem much easier.
     
    Last edited: Sep 20, 2012
  4. Sep 20, 2012 #3
    Thanks for the response! However, that is the answer I entered into Webwork but it states it is incorrect.
     
    Last edited: Sep 20, 2012
  5. Sep 20, 2012 #4
    Okay I managed to figure it out after reading: http://mathinsight.org/double_integral_volume

    My boundaries were correct but the equation I was integrating was not.
    It explained that the integral equation is not f(x,y) but f(x,y) - g(x,y), in my case g(x,y) being the "other surface" or z = 5

    As a result I subtracted 5 from f(x,y) which was 30 - x^2 - y^2 and resulted in a final answer of
    -55-√(25-x^2)√(25-x^2) 25-x^2-y^2dydx
     
  6. Sep 20, 2012 #5

    HallsofIvy

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    Yes, what you had was the volume between the surface [itex]z= 30- x^2- y^2[/itex] and the xy-plane, z= 0. Unfortunately, so many basic problem ask for just the "area below the curve" and "the volume below the surface", we may develop the very bad habit of automatically assuming a lower bound of 0.

    By the way, in cylindrical coordinates, that uart referred to, the upper surface is given by [itex]z= 30- r^2[/tex], the lower surface by z= 5, and the bounding circe by r= 5. The integral would be
    [tex]\int_{\theta= 0}^{2\pi}\int_{r= 0}^5 (30- r^2- 5) rdrd\theta=\int_{\theta= 0}^{2\pi}\int_{r= 0}^5 (25- r^2) rdrd\theta[/tex]
     
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