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Algebraic equation -- six degree polynomial

  1. May 5, 2016 #1
    1. The problem statement, all variables and given/known data
    Solve the equation for [itex]r,r>0,r<R[/itex].
    [tex]\frac{-2\pi R^3}{3}-\frac{8\pi r^2\sqrt{R^2-r^2}}{3}+2r^2R+\frac{2\pi}{3}R^2\sqrt{R^2-r^2}=\frac{2\pi R^3}{3}[/tex]

    2. The attempt at a solution
    After factoring,
    [tex][/tex]
    [tex]-2R(2\pi R^2-3r^2)=2\pi\sqrt{R^2-r^2}(4r^2-R^2).[/tex]

    After squaring,

    [tex]64\pi^2 r^6+(36R^4-96\pi^2 R^2)r^4+(32\pi^2 R^4-44\pi^2 R^4)r^2+12\pi^2 R^6=0.[/tex]

    [itex]r[/itex] is unknown variable, so the last equation should be the equation of six degree polynomial.
    How to solve it?
     
  2. jcsd
  3. May 5, 2016 #2

    Ray Vickson

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    Try to solve for the new variable ##S = \sqrt{R^2 - r^2}##, by converting your original equation into a cubic in ##S##. This is solvable in principle using Cardan's formula for the solution of a cubic. Furthermore, ##x = S/R## obeys a cubic equation that does not contain the parameter ##R##, so we can take ##R = 1## without loss of generality.

    All solution values for ##r## are complex-valued, so your original equation does not have any real solutions in the region ##-R \leq r \leq R##. Are you sure you have not made any errors in writing down the equation?
     
  4. May 6, 2016 #3

    haruspex

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    The very first equation looks a bit unlikely to me. Are you sure it is correct? If not certain, please post the original problem as given to you.
     
  5. May 7, 2016 #4
    Here is the original problem (the topic should be moved to Calculus Homework):

    Circular tunnel of radius [itex]r[/itex] is punctured through the center of a homogeneous sphere of radius [itex]R, (r<R)[/itex].
    1. What percentage of a sphere is lost?
    2. What should be the value of [itex]r[/itex] such that the sphere maintains exactly the half of it's volume?

    This is a variation of the Napkin ring problem (en.wikipedia.org/wiki/Napkin_ring_problem).
    [itex][/itex]
    1. Height of the spherical cap at the center is [itex]R-\sqrt{R^2-r^2}[/itex].
    Volume of the one spherical cap that is removed is [tex]V_{cap}=\int_{R-\sqrt{R^2-r^2}}^R {(R^2-x^2)}\mathrm dx=\frac{\pi}{3}(r^2 - R^2)(\sqrt{R^2 - r^2} - 3R)[/tex]
    Remaining volume is [tex]V_{remaining}=\frac{4 \pi R^3}{3} - 2V_{cap} - 2 \pi r^2 \sqrt{R^2 - r^2}[/tex]
    Removed volume is [tex]V_{removed}=2(V_{cap}+\pi r^2\sqrt{R^2-r^2})=2\left(\frac{\pi}{3}(r^2 - R^2)(\sqrt{R^2 - r^2} - 3R)+\pi r^2\sqrt{R^2-r^2}\right)[/tex]

    Percentage of the sphere that is lost is [tex]x=\frac{V_{removed}\cdot 100}{V_{total}},V_{total}=\frac{4\pi R^3}{3}\Rightarrow x=\frac{150\left(\frac{\pi}{3}(r^2 - R^2)(\sqrt{R^2 - r^2} - 3R)+\pi r^2\sqrt{R^2-r^2}\right)}{\pi R^3}[/tex]

    2. To find the value of [itex]r[/itex] such that the sphere maintains exactly the half of it's volume we solve the following equation for [itex]r[/itex]:

    [tex]V_{remaining}=\frac{1}{2}V_{total}[/tex]
    [tex]V_{remaining}=V_{total}-2\left(\frac{\pi}{3}(r^2-R^2)(\sqrt{R^2-r^2}-3R)-2\pi r^2\sqrt{R^2-r^2}\right),V_{total}=\frac{4\pi R^3}{3}[/tex]
    [tex]V_{remaining}=\frac{4\pi R^3}{3}-\frac{2r^2\sqrt{R^2-r^2}\pi}{3}+2R\pi r^2+\frac{2R^2\pi\sqrt{R^2-r^2}}{3}-2\pi R^3+4\pi r^2\sqrt{R^2-r^2}[/tex]
    [tex]\frac{4\pi R^3}{3}-\frac{2r^2\sqrt{R^2-r^2}\pi}{3}+2R\pi r^2+\frac{2R^2\pi\sqrt{R^2-r^2}}{3}-2\pi R^3+4\pi r^2\sqrt{R^2-r^2}=\frac{2\pi R^3}{3}\Rightarrow[/tex]
    [tex]-4\pi R^3+10r^2\sqrt{R^2-r^2}\pi+2R^2\sqrt{R^2-r^2}\pi+6R\pi r^2=0[/tex]

    Note: There are some mistakes in the equation of original post.

    After dividing by two:

    [tex]-2\pi R^3+5r^2\sqrt{R^2-r^2}\pi+R^2\sqrt{R^2-r^2}\pi+3R\pi r^2=0[/tex]
    [tex]5r^2\sqrt{R^2-r^2}\pi+R^2\sqrt{R^2-r^2}\pi=2\pi R^3-3R\pi r^2[/tex]

    After squaring:

    [tex]25\pi^2 r^6+2\pi^2 r^4(8R^2-5)+\pi^2 R^2 r^2(10+11R^2)-3\pi^2 R^6=0[/tex]
     
  6. May 7, 2016 #5

    haruspex

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    The final equation looks more likely now in that the π all factor out, but every term should have order 6, e.g. r2R4.
    Those 5 and 10 terms are not possible.
    Why not make use of the napkin result? Think about the length of the cylindrical hole, measured along its side.

    Edit: I suggest you check each equation by considering the extremes of r, namely 0 and R. E.g. take your expression for the volume of a cap and try r=R. It gives 0 instead of 2πR3/3.
     
    Last edited: May 7, 2016
  7. May 7, 2016 #6
    Do you think that the volume of a cap is not correct?
     
  8. May 7, 2016 #7
    Volume of a cap is not correct. It should be [tex][/tex]
    [tex]V_{cap}=\pi\cdot\frac{3R^3-3Rr^2-(R^2-r^2)^{3/2}}{3}[/tex]

    This gives that [tex]V_{removed}=2\left(\pi\left(\frac{3R^3-3Rr^2-(R^2-r^2)^{3/2}}{3}+r^2\sqrt{R^2-r^2}\right)\right)[/tex]

    1. Now [tex]x=\frac{150\left(\pi\left(\frac{3R^3-3Rr^2-(R^2-r^2)^{3/2}}{3}+r^2\sqrt{R^2-r^2}\right)\right)}{\pi R^3}[/tex]

    2. The equation for [itex]r[/itex] is now
    [tex]-3R^3+3Rr^2+(R^2-r^2)^{3/2}-3r^2\sqrt{R^2-r^2}=0[/tex]
     
  9. May 7, 2016 #8

    haruspex

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    Yes, I think it is wrong. Check the integration bounds.
     
  10. May 7, 2016 #9

    haruspex

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    Your Vcap is still wrong. Try the test I suggested, what happens when r=R? What should it give there?
     
  11. May 7, 2016 #10
    If [itex]r=R[/itex], then [itex]V_{cap}=\frac{2\pi R^3}{3}[/itex].
     
  12. May 7, 2016 #11
    After some corrections, I am getting the equation [tex]-4r^6+12R^2r^4-12R^4r^2+3R^6=0.[/tex]
     
  13. May 7, 2016 #12

    Ray Vickson

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    This is a cubic in ##x = r^2##. Furthermore, if you set ##r = Ru## you get an equation for ##u## that does not involve ##R##. In other words, you could have just set ##R = 1## right at the start, since both the questions you want to answer involve percentages of the sphere's volume, rather than actual numerical values.

    The whole analysis would be much simpler if, instead of ##r## (the cylinder's radius) you use as your variable ##h## = height of the spherical cap. Of course, these are related by ##r^2 = R^2 - (R-h)^2##. In terms of ##h## the volume of the removed portion is
    [tex]V_{\text{removed}}= 2 \pi (R-h)[R^2 - (R-h)^2]+ \frac{2 \pi}{3} h^2 (3R-h), [/tex]
    using the formula for the spherical cap's volume from http://mathworld.wolfram.com/SphericalCap.html .
    Setting ##h = Rx## you get the simple cubic equation
    [tex] \frac{4}{3} x^3 - 4x^2 + 4x = \frac{2}{3} [/tex]
    or
    [tex] 4 x^3 - 12 x^2 + 12 x = 2 [/tex]
     
    Last edited: May 7, 2016
  14. May 7, 2016 #13

    haruspex

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    That looks right. Can you see how to 'complete the cube', so as to get an equation in the form (function of r and R)3=(function of R only)?
    I note that Ray's equation is slightly different.

    Edit: maybe not really different. I forgot Ray is not using x to represent r/R.
     
    Last edited: May 7, 2016
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