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

Homework Help: More volume

  1. Nov 19, 2007 #1
    1. The problem statement, all variables and given/known data
    Consider two quarters of a cylinder of radius R. What is the volume of a the intersection of a two quarters of two cylinders if the quarters meet at a right angle? Two sides of the bounded region will be rectangular according to this arrangement. (see the diagram when it gets approved)

    2. The attempt at a solution
    I need to do this in Cartesian coordinates so here goes

    let the Z axis be straight up, the X axis out of the page an the Y axis parallel to the page.

    For the cylindrical part parallel to the X axis
    [tex] V = \int_{0}^{r} dx \int_{0}^{r} \int_{0}^{\sqrt{r^2 - y^2}} dy dz [/tex]

    For hte cylindrical parallel to the Y axis
    [tex] V = \int_{0}^{r} dy \int_{0}^{r} \int_{0}^{\sqrt{r^2-x^2}} dx dz [/tex]
    And the find the volume i need to subtract the above two??

    Attached Files:

  2. jcsd
  3. Nov 20, 2007 #2
    Any help would be greatly appreciated!
  4. Nov 21, 2007 #3
    I had simular problem once, but I was given that the volume of the whole region was eight times the part volume.

    [tex] V = 8 \int_{0}^{r} {(r^2-z^2)}dz [/tex]

    [tex] V = 8(r^2z-(z^3/3) |r [/tex]

    [tex] V = (16*r^3)/3 units. [/tex]

    Hope you figure this one out.
  5. Nov 21, 2007 #4
    dont understand how you got your integrand to be r^2 -z^2?

    Adn waht do you mean by 8 times part volume?
  6. Nov 21, 2007 #5
    If you "slice" horizontally at height z above the xy-plane then the slice has a side equal to

    [tex] {\sqrt{(r^2-z^2)}} [/tex] and a area [tex] A(z) = r^2-z^2 [/tex]

    The slice is a rectangle with equal size on all sides. (what is it called in english?)

    where [tex] a = {\sqrt{r^2-z^2} [/tex] is on side of that rectangle.

    But you are wondering why ? I can't really explain it but when you have circular solids you use [tex] {\sqrt(r^2-z^2)} [/tex] where r = radius of the circle.

    From the figure you posted these are two "1/4-th" circles

    I'm sure someone can explain it better then me here.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook