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Another Volume of Integration Problem

  1. Jul 30, 2005 #1
    let Q be the solid that is outside both [tex]x^2 +y^2 +z^2=1[/tex] and [tex]z=2x^2 +2y^2 +2[/tex], yet inside [tex]z=4\sqrt{x^2 + y^2}[/tex]

    Could someone give me a little jump start on this problem? I would love a complete solution but anyhelp will do.
  2. jcsd
  3. Jul 30, 2005 #2
    Is that from negative infinity to infinity in both the x and y axes?
  4. Jul 30, 2005 #3
    the bounds are the different surfaces
  5. Jul 30, 2005 #4


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    The plot is a cross-section with y=0. You want the green right?

    What about a triple in spherical coordinates?
    Rock and roll.

    Attached Files:

    Last edited: Jul 30, 2005
  6. Jul 30, 2005 #5
    Yes I need the green part of the graph. SaltyDog, what program did you use to generate that graph?
  7. Jul 30, 2005 #6


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    Mathematica. Here, this is the code:

    Code (Text):

    << Graphics`FilledPlot`
    y1[x_] := (1 - x^2)^(1/2);
    y2[x_] := 2 x^2 + 2;
    y3[x_] := 4 (x^2)^(1/2);
    pf1 = Plot[{y1[x], y2[x], y3[x]}, {x, -1, 1}, PlotRange -> {{-4, 4}, {-4, 4}},
         AspectRatio -> 1]
    FilledPlot[{y2[x], y1[x], y3[x]}, {x, -1, 1}, PlotRange -> {{-4, 4}, {-4, 4}},
       AspectRatio -> 1]
    Of course in Mathematica, you can use real square-roots and exponent symbols. I just used the carrot above so I could cut and paste it here.

    However, I'm having problems converting it to spherical coordinates. Maybe you or someone else can do better or perhaps another method can be used.
    Last edited: Jul 30, 2005
  8. Jul 30, 2005 #7


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    RadiationX, I think I have it but have to evaluate the integral numerically. First of course is to convert the function:


    into spherical coordinates (you get a quadratic which you can then solve for rho using the quadratic formula but you need to know which sign to use). You can do that. Then integrate rho from the sphere which is 1 up to this function you expressed as [itex]\rho(\phi)[/tex]. Theta of course goes from 0 to 2[itex]\pi[/itex]. Now phi only goes from 0 to that cone-shaped function:


    Now, you can figure what the max on [itex]\phi[/itex] has to be from that equation right. Can you then post the expression for [itex]\rho(\phi)[/itex] you determined for (1) above as well as the triple integral?
  9. Jul 31, 2005 #8


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    Hey RadiationX, I'm affraid I suggested a method more difficult that it needs to be. Cylindrical shells is the simplest approach and the integrals can be evaluated analytically. Just need to split it up into two (single) integrals. Talk about over-kill up there. :yuck: Can you do this?

    You should tell me, "you know Salty, just for recommending that ridiculously complex triple integral up there, you need to spend a lot of time trying to evaluate it analytically, yea, and no Mathematica neither!" :surprised

    Edit: Hey, I get the same answer for both though so that's encouraging. :smile:
    Last edited: Jul 31, 2005
  10. Jul 31, 2005 #9
    thanks Saltydog would you evaluate it a little more analyticlly for me?
  11. Jul 31, 2005 #10
    since iam high concerning in math ,iam looking forward to provide me with math books
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  12. Jul 31, 2005 #11


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    Here it is in spherical coordinates:

    The relationship between spherical coordinates and cartesian coordinates is:

    [tex]x=\rho Sin(\phi) Cos(\theta);\quad y=\rho Sin(\phi) Sin(\theta);\quad z=\rho Cos(\phi)[/tex]

    We must first convert the upper function, [itex]z=2x^2+2y^2+2[/itex] to spherical coordinates. Doing this, we obtain:

    [tex]\rho Cos(\phi)=2\left[\rho^2 Sin^2(\phi)\left(Cos^2(\theta)+Sin^2(\theta)\left)+1\right][/tex]

    Solving for [itex]\rho[/itex]:

    [tex]\rho(\phi)=\frac{Cos(\phi)\pm \sqrt{Cos^2(\phi)-16\sin^2(\phi)}}{4Sin^2(\phi)}[/tex]

    I'll leave it to others to show that the positive value does not converge in the interval of [itex]\phi[/itex] we are concerned with. Therefore, this function in spherical coordinates is given as:


    So [itex]\rho[/itex] will be integrated from the sphere which is 1 up to this function. Now,[itex]\phi[/itex] will be integrated from zero to the function [itex]z=4\sqrt{x^2+y^2}[/itex]. It's symmetric about the z-axis. So letting y=0 and just pick a number for z, say z=1, we have 1=4x or x=1/4. Thus:


    We can now set up the integral to determine the volume:

    [tex]V=\int_0^{\phi_{max}} \int_0^{2\pi} \int_1^{\rho(\phi)} \rho^2 Sin(\phi)d\rho d\theta d\phi[/tex]

    Now, how about setting it up in cylindrical shells?
    Last edited: Jul 31, 2005
  13. Jul 31, 2005 #12
    thanks saltydog
  14. Jul 31, 2005 #13


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    Here it is in cylindrical shells:

    The diagram below is a cross-section of the region. Consider thin-walled cylinders circling the z-axis (actually y-axis as used here) and split the region of integration into two sections: Section (1) is from x=0 to x=the point where the circle intersects the line y=4x (first vertical line). Region (2) is from that point to the point where y=4x intersects the top curve [itex]y=2x^2+2[/tex] (second line).

    Considering in general the volume of the i'th cylinder as:

    [tex]V_i=\pi r_i^2(f_a(x_i)-f_b(x_i))-\pi r_{i+1}^2(f_a(x_i)-f_b(x_i))[/tex]



    and forming the Riemann sum and taking the limit as the partition goes to zero we obtain:
    (or just using the formula in the Calculus book)


    in which:

    [tex]V_a=2\pi \int_0^a x\left(2x^2+2-\sqrt{1-x^2}\right)dx[/tex]

    [tex]V_b=2\pi \int_a^b x\left(2x^2+2-4x\right) dx[/tex]




    That is:

    V &= \int_0^{\phi_{max}} \int_0^{2\pi} \int_1^{\rho(\phi)} \rho^2 Sin(\phi)d\rho d\theta d\phi \\
    &= 2\pi \int_0^a x\left(2x^2+2-\sqrt{1-x^2}\right)dx+2\pi \int_a^b x\left(2x^2+2-4x\right) dx\\

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