Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Find the volume of the resulting solid

  1. Mar 14, 2005 #1
    A ball of radius 11 has a round hole of radius 8 drilled through its center. Find the volume of the resulting solid

    here's what i done:
    y = sqrt(11^2-x^2)
    y = 8 about x-axis

    [tex]\int pi*(8)^2 - pi*(8-sqrt(11^2-x^2))[/tex]

    this is where im stuck at, no clue if my setup is even correct. What would the bounds be and is my setup correct?
  2. jcsd
  3. Mar 14, 2005 #2


    User Avatar
    Science Advisor

    It's not at all clear what you are doing. Where is that "pi*(8)2" from?

    It looks like you are confusing a "ball" (sphere) with a circle.

    Draw a picture: a circle (the sphere seen from the side) with a rectangle inscribed in it (the cylidrical hole). The center of the sphere is at the middle of the rectangle. The top of the rectangle marks where the hole exits the sphere. If you draw a line from the center of the sphere to the center of the top of the rectangle, it has length h/2 (h is the length of the rectangle). Drawing a line from the center of the sphere to the vertex of the rectangle gives a right triangle with legs of length h/2 and 8 (the radius of the hole) and hypotenuse of length 11 (the radius of the sphere). By the Pythagorean theorem h2/4+ 64= 121 so h2= (57)(4) and h= 2√(57).


    The volume of the sphere, before the hole is drilled, is [itex]\frac{4\pi r^3}{3}= \frac{ (4 \pi(11)^3}{3}[/itex].
    The volume of the cylinder removed is [itex]\pi r^2 h= \pi (8)^2(2\sqrt{57})= 128\sqrt{57}\pi[/itex].

    BUT you still need to subtract the volume of those two little "caps" that are removed on each end of the cylinder- that's the calculus part of this problem.
    Since they are clearly the same, you can calculate the volume of one and double it.
    You can do that using the "disk" method. At height "y", The x coordinate is [itex]\sqrt{121- y^2}[/itex] and that becomes the radius of each disk: the area of each such disk is [itex]\pi(121- y^2)[/itex] so the volume is [itex]\pi \int (121- y^2)dy[/itex].
    The limits of integration are y= √(57) (the top of the cylinder) to y= 11 (the top of the sphere.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook