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Homework Help: The Classical Bead Problem (Volumes of Sphere and Cylinder)

  1. Dec 26, 2008 #1
    Hello. I'm having trouble with the following problem.

    A round hole is drilled through the center of a spherical solid of radius r. The resulting cylindrical hole has height 4 cm. What is the volume of the solid that remains?

    Here's the pic that I drew, I hope it's useful:

    http://img525.imageshack.us/img525/1954/sphereyv4.png [Broken]

    V of sphere = 4/3 πr^3
    V of cylinder = πr^2h
    h = f(r)
    h = 4 cm

    I haven't really gotten anywhere yet, but the following should also be useful.
    I originally tried to use the shells method, the area of the rectangle of which would be A(r) = 2πr * f(r) * dr. I also noticed that as dr/dt increases dh/dt decreases.

    V = (4/3 πr^3) - (πr^2h) = πr^2(4/3r - 4) <--- (I'm not sure if this is useful at all but I did it anyway)

    I don't really know what I should do next. Any suggestions/help is greatly appreciated!
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Dec 26, 2008 #2


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    Welcome to PF!

    Hello Gelo! Welcome to PF! :smile:
    How did you get that? :confused:
  4. Dec 26, 2008 #3


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    Well that's somewhat the right approach. You need not use the shell method of volume rotation. The washer method works fine as well. Draw out a circle with radius r centred on the origin. Then the interval which is rotated about the x-axis is x=-2,+2. You should be able to deduce which is the upper part of the graph and the lower part of the graph and their respective f(x) to substitute into the formula for volume rotation.

    As tiny-tim said, this isn't too helpful since r in the 4/3πr^3 part is different from the 'r' in πr^2h part.
  5. Dec 26, 2008 #4


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    It looks like you just subtracted the volume of the cylinder from the sphere. That's not right. What about the parts of the sphere that were past the ends of the cylinder? Go back to using the shell method and integrating. What is f(r)?
  6. Dec 27, 2008 #5
    I have not been taught with spherical co ordinates. But i just got an ingenious method of doing the sum.

    We can use area under curve. Lets suppose that the centre is at the origin. In Now find the area under curve of [tex] (\pi)r^2 [/tex] from the limits..... The limits are easy enough to find out.
  7. Dec 27, 2008 #6
    Oops sorry. Its a sphere. Not a circle.
  8. May 4, 2009 #7
    Here is what helped me find the answer. First, keep in mind that this problem involves using integration of a line (the line for the positive y-values of a circle) around the x-axis (because the cylinder being cut out has rounded ends!), solving for y at the intersection of the circle and cylinder, and choosing the limits of integration correctly.

    1) rotate the drawing posted in the original question by 90%
    2) draw in an x-axis that goes parallel to and through the center of the cylindrical drilled-out part, so that the drilled-out part is indicated (on our 2-dimensional paper) by 2 lines, equidistant from the x-axis, one with a positive y-value, the other with the corresponding negative y-value
    3) draw in a y-axis that goes perpendicular to the x-axis, and goes through the center of the sphere
    4) label the line designating the above-the-x-axis part of the outline of the sphere (hint: use the formula for a circle (x^2 + y^2 = r^2), solved for y, call it y1, and just use the positive values) (keep in mind, we don't know the value of r!)
    5) label the points at the intersection of the cylinder line and the sphere (the formula you came up with in step 4 above) and determine what Y is at those points (hint: this involves the pythagorean theorem) in terms of x, call it y2 (keep in mind, we don't know the value of r, but r can be used in the answer!)
    6) the limits of integration, as determined by examining the drawing provided above are a=-2 and b=2
    7) This is the step where the "rotation about the x-axis" and solving the "washer" types of problems come in:
    7a) Integrate Y1 that you found in 4, rotated around the x-axis, to get the complete volume of the sphere WITH THE ENDS CUT OFF (Note that you use Y, not r, because you are rotating the Y1 = f(x) line around the x-axis; you are not rotating the radius about the x-axis)
    7b) Subtract out the area of the cylinder (note that you found the radius of the cylinder in step 5) by subtracting the Y2 line that you found in step 5, rotated around the x-axis.
    8) Your integration will be pi, times the integral of the line for the circle, squared (because that y-value becomes the radius for the sphere when rotating around the x-axis), and subtract the line for the cylinder, squared, because it is the radius of the circular part that the cylinder cuts out.

    V = [tex]\pi[/tex]*[tex]\int[/tex][tex]^{2}_{-2}[/tex] [(y on circle)^2 - (y on cylinder)^2] dx

    (I apologise for any deficiency on my part in getting the integral above to display correctly.)

    You'll find that this becomes a classic rotate-the-area-between-2-curves around the x-axis calculus problems when you approach it this way, and that the "tricky stuff" with the rounded ends of the cylinder don't cause any trouble, due to the chosen limits of integration.

    I didn't give the specific formulas for y1 and y2, but it is left to the student (assuming most of the people needing the answer to this question are students) to determine these and then to evaluate the final integral.

    I hope that this answer enables students to understand a thought process that leads to the solution, and that this, in turn, enables them to follow similar thought processing with other types of problems.

    Final hint: you will find that an interesting thing happens to "r" when evaluating the integral!

    Good luck, everyone!
  9. May 4, 2009 #8


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    The radius of the sphere is r but nothing is said about the radius of the cylindrical hole and that's important. If we write r' for the radius of the hole, the drawing a line from the center of the sphere to a point where the cylinder intersects the surface of the sphere gives a right triangle so we can show that [itex]r^2= r'^2+ (h/2)^2[/itex] where h is the height of the cylinder.

    Here is what nkjmom said was "interesting": Since the problem does not give the radius of the cylinder, in order that this be possible to solve, the final answer must not depend on it. Okay, assume a cylinder of infinitesmally small radius. Then the sphere must have diameter equal to the height of the cylinder and no material is removed by the cylinder. Just calculate the volume of such a cylinder.
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