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Homework Help: Volume of a bead

  1. Sep 11, 2006 #1
    We're trying to maximize (then minimize) the volume of a sphere wuth a cylinder drilled through the center.

    So I have a circle with the equation x^2 + y^2 = R^2, and I'm going to rotate it around the y-axis. Actually, I'm only going to rotate the quarter of it that lies in the first quadrant, then multiply its volume by 2.

    Now since the sphere has a cylindrical hole in it, my quarter-circle that I will rotate starts at sqrt(R^2-a^2), where a is half the height of the cylindrical hole.

    (sorry for the mess; I don't know how to draw a graph on the computer)


    My integrand is (2)(pi)(x)(L(x))=(2)(pi)(x)sqrt(R^2-x^2)
    because the equation of a circle is y=sqrt(R^2-x^2)

    My upper and lower limits are R and sqrt(R^2-a^2) respectively.

    I didn't yet learn how to evaluate an integral of this form.
    Is there any way to find the max/min Volumes without evaluating the integal? Or is my integral wrong altogether?
  2. jcsd
  3. Sep 12, 2006 #2


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    No, that's not your integrand. You don't want to find area, you want to find the volume when the circle is rotated about an axis.

    I would be very surprised if you hadn't learned to integrate such square roots long ago- its a simple trig substitution. However, as I pointed out, your integral is altogether wrong. [itex]y= \sqrt{R^2- x^2}[/itex] is a radius of the disk formed when the circle is rotated around the x-axis. The area of such a disk is [itex]\pi y^2[/itex] and the volume is
    [tex]\pi\int y^2dx[/tex]
  4. Sep 12, 2006 #3
    Thank you.

    My Calc II professor loves his subject. Hence our starting with a brief review of the fundemental theorem (the last thing we did in calc I) and moving straight to a day on Areas, then spending time on volumes.
    He says we'll get to methods of integration soon.
  5. Sep 12, 2006 #4
    My professor strongly suggested using the cylindrical shell method, which is how I got my original integrand.

    In my diagram, I set R to be the radius, so wouldn't [tex]y\pm \sqrt{R^2-x^2}[/tex] be the equation of my circle (rather than a radius thereof)?

    Basically, I tried doing it with the disk method and ended up with [tex]\pi\int \sqrt{R^2-x^2}^2dx[/tex] which simplifies to [tex]\pi\int \sqrt{R^2-x^2}dx[/tex], which I still can't evaluate because my class skips around. But is this the integrand I should be looking for?
    Last edited: Sep 12, 2006
  6. Sep 12, 2006 #5


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    No, it doesn't! Squaring a squareroot doesn't leave a squareroot! That simplifies to
    [tex]\pi \int (R^2- x^2) dx[/tex]
  7. Sep 12, 2006 #6
    Would you believe me if I told you that was a typo? Because it was:blushing:

    I'm sorry to be so dense, but I looked it over yet again and still can't follow your reasoning. Doesn't [tex]A(x)=\pi R^2[/tex] rather than [tex]A(x)=\pi y^2[/tex]?
    Why are we putting y (in the form [tex]\sqrt{R^2-x^2}[/tex]) into [tex]\int (A(x)) dx[/tex] instead of putting in R (in the form [tex]\sqrt{y^2+x^2}[/tex])?
    Other than the fact that an integral with x's and y's looks really ugly, that is...
  8. Sep 12, 2006 #7
    yowzers, naps should be mandatoory!
    You're rotating around the xaxis, aren't you? that would make y the radius of that disk... Good Lord. Sheesh.
    Thanks, and good morning!
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