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Double Integrals: Where am I making a mistake?

  1. Apr 13, 2015 #1
    1. The problem statement, all variables and given/known data

    Find the volume of the solid.

    Under the paraboloid z = x^2 + y^2 and above the region bounded by y = x^2 and x = y^2

    Well, those curves only intersects in the xy-plane at (0,0) and (1,1), and in the first Quadrant, and in that first Quadrant y = sqrt(x), and over that interval sqrt(x) >= x^2

    So here is my working.

    3. The attempt at a solution

    [tex] \int \int_D (x^2+y^2) dA = \int_0^1 \int_{x^2}^{\sqrt(x)} x^2+y^2 dy dx [/tex]

    [tex] \int_{x^2}^{\sqrt(x)} x^2+y^2 dy = \left[ yx^2+\frac{y^3}{3} \right]_{x^2}^{\sqrt(x)} [/tex]

    [tex] = x^{1/5} + \frac{x^{1/3}}{3} - x^4 - \frac{x^6}{3} [/tex]

    then

    [tex] \int_0^1 x^{1/5} + \frac{x^{1/3}}{3} - x^4 - \frac{x^6}{3} dx = \left[\frac{5}{6}x^{5/6} + \frac{x^{4/3}}{4} - \frac{x^5}{5} - \frac{x^7}{21}\right]_0^1[/tex]

    [tex] = \frac{5}{6} + \frac{1}{4} - \frac{1}{5} - \frac{1}{21} = \frac{117}{140} [/tex]

    But the textbook gives the answer as [tex] \frac{6}{35} [/tex]

    I can't figure out where I have gone wrong.
     
  2. jcsd
  3. Apr 13, 2015 #2

    HallsofIvy

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    Staff Emeritus
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    I have no idea how you got this last .line. [itex]yx^2+ \frac{y^3}{3}[/itex] evaluated at [itex]y= x^{1/2}[/itex] is
    [itex](x^{1/2}x^2+ \frac{(x^{1/2})^3}{3}= x^{5/2}+ \frac{x^{3/2}}{3}[/itex],, NOT "[itex]x^{1/5}+ \frac{x^{1/3}}{3}[/itex].

    [/QUOTE]
     
  4. Apr 13, 2015 #3
    [/QUOTE]

    Yep, I'm tired. I just realized what I did wrong after running it through MATLAB...I really should just take a break and come back to it, before hastily posting here. Sorry :P
     
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