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Volume: The Disk Method

  1. Aug 7, 2008 #1
    1. The problem statement, all variables and given/known data

    A manufacturer drills a hole through the center of a metal sphere of radius 5 inches. The hole has a radius of 3 inches. What is the volume of the resulting metal ring?


    You can imagine the ring to be generated by a segment of the circle who equation is x^2 + y^2 = 25. Because the radius of the hole is 3 inches, you can let y = 3 and solve the equation x^2 + y^2 = 25 to determine that the limits of integration are x = +- 4 . So, the inner and outer radii are r(x) = 3 and R(x) = sqrt(25-x^2).

    2. Relevant equations

    x^2 + y^2 = r^2 (Standard Form of A Circle)

    3. The attempt at a solution

    This is an example problem in my textbook, Calculus 8th Edition by Larson (pg. 461 Section 7.2)

    My fundamental question: How did they find the limits of integration by substituting 3 with y?

    I just don't see how plugging in the inner radius (y = 3) into the equation of a circle would find the limits of integration? I can see how in a 2D circle, x has solutions when it is -4 and 4.....but why couldn't we plug in radius (y = 5) in instead of 3?

  2. jcsd
  3. Aug 7, 2008 #2
    In this case, take the center corss section of the sphere and look at the top half of it. It will look like a semi-circle with radius 5 correct? (y = sqrt(25-x^2))

    Now the sphere was drilled and the resulting hole has a radius of 3. So now the bottom half of the semi-circle is now removed and the semi-cricle is sitting 3 inches away form the x-axis. Can you visualize this? I wish I could show you a diagram...

    By substituting y = 3 into the circle equation, you will find where the line intersects with the semi-circle. This will give you the end points of the semi-circle and thse end points define the limits of integration.

    I don't know if this helped answer your question.
  4. Aug 7, 2008 #3


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    Homework Helper

    You have draw it out to understand the solution. Draw the equation of the circle and the line y=3. Without the line you can see that the volume of the sphere can be generated by revolving the semicircle about the x-axis. You should see that the volume of solid generated by rotating the graph from the top of the circle to the line y=3 is the remaining solid.

    Plugging in y=5 would mean that the entire sphere was drilled (destroyed).
  5. Aug 7, 2008 #4
    Thank you both!

    Before, what I was doing was:

    1.) Drawing a circle whose center is on (0,5) instead of the origin (0,0). I thought the only way to create a sphere is if the whole circle was above the x-axis (though, I will learn later that a semi-circle is enough to make the sphere with radius 5.

    Basically, I followed your methods: the top semicircle is defined by y = sqrt(25-x^2) and the "hole" will be drilled at y = 3 (and y = -3... radius is still equal to 3).

    Anyway, focusing on one semi circle...to find the limits of integrations, you must equate the two equations (y = sqrt(25-x^2) and y = 3).... which the solution is x = +- 4.

    Thank you. The book's interpretation was far more complicated than you two's.
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