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

Confused by this integral

  1. Dec 5, 2005 #1

    It has a been a while since college let alone since I've done any serious calculus, so I hope you can help explain this integral in a simple way that I can understand:

    refer to the attached figure. I am trying to find the integral of z dV
    where dV is increment of volume.
    which could be written as integral of z dx dy dz

    now I am trying to understand how to integrate z over dx and dy. the integral will be a circle of a certain radius but how do we exactly integrate z over dx and over dy ?

    please no spherical/cylindrical coordinates. thanks a lot.

    Attached Files:

  2. jcsd
  3. Dec 5, 2005 #2


    User Avatar
    Science Advisor
    Homework Helper

    When you integrate over x and y, the variable z is fixed in the integrand and only comes into play in the limits of integration. You may find it more convenient to use cylindrical coordinats.
  4. Dec 5, 2005 #3
    You must evaluate int(int(int(z)))dV where V is the region enclosed by the positive cone in the xyz plane. In general, if V is defined by: a<x<b, h_1(x)<y<h_2(x), g_1(x,y)<z<g_2(x,y) then int(int(int(f(x,y,z)))dV=int(int(int(f(x,y,z),z,g_1(x,y),g_2(x,y)),y,h_1(x),h_2(x)),x,a,b). In your case, f(x,y,z)=z. g_1(x,y)=0 g_2(x,y)=+sqrt((h/r)(x^2+y^2)) h_1(x)= -sqrt(r^2-x^2) h_2(x)= +sqrt(r^2-x^2) and a=-r b=r. Plug and chug.
  5. Dec 6, 2005 #4
    bear with me here,
    so are you saying that

    [tex] \int z dx = z \int_{z=0}^{z=h} dx [/tex]

    which is equal to [tex] z . \left[ \ x \right]_{z=0}^{z=h} [/tex]


    thanks again
    Last edited: Dec 6, 2005
  6. Dec 6, 2005 #5
    would have to agree with Tide,
    Because you will need to integrate over a region D that is a circular disc centered at the origin, your best bet is to convert to cylindrical coordinates. This will make your life a lot easier.

    To find volume you will need to do a tripple integral.
    Your domain will be theta is between 0 and 2(pi)
    r is between 0 and 1, and z is between 0 and z.
    your tripple integral will simply be,
    the integral (z)dz evaluated from z=0 to z=c.

    The tripple integral comes out to (pi)*(c^2)/2, where c is the height. This is for the volume of a cylinder with radius r = 1 and height z = c.

    Hope this helps a little.

    Best Regards,

  7. Dec 6, 2005 #6
    thanks for your post

    I do understand that using cylindrical coordinates will be much easier but the book I got this example from used rectangular coordinates and I am just trying to understand this integral for the practice and in case I come across something similar in the future. This is why I am sticking to rectangular coordinates.

    thanks again
  8. Dec 7, 2005 #7


    User Avatar
    Science Advisor
    Homework Helper

    If you insist on using cartesian coordinates then start by limiting yourself to one octant (based on symmetry) so your integral should look like this:

    [tex]I = 4\int_{0}^{h} z dz \int_{0}^{\frac {R}{h} z} dy \int_{0}^{\sqrt {(Rz/h)^2-y^2}} dx[/tex]

    As an exercise you may want to try doing it using cylindrical coordinates anyway.
  9. Dec 17, 2005 #8
    Thanks a lot Tide.

    Can you clarify how did you get the limits of integration of the dy and the dx integral? And also where did the 4 come from?
  10. Dec 19, 2005 #9


    User Avatar
    Science Advisor
    Homework Helper

    Gosh, you want me to remember what I did two weeks ago?? ;)

    Seriously, what you want to note is that at a fixed z-level, the surface is bounded by [itex]x^2 + y^2 = r^2[/itex] where r is the radius at that height. The radius, of course, depends on height and is given by [itex]r = R z/h[/itex] being 0 at the vertex and R at the top (z = h).

    To set up the integral, we'll add up the contributions from infinitesimal disks whose volume is dx dy dz (you did insist on using cartesian coordinates!) so the integrand is z dx dy dz. I.e. the volume of each disk is their height (dz) time their cross sectional area (dx dy). We'll focus on the area part first.

    Using symmetry, I chose to integrate over a single quadrant in x-y (hence, the factor of 4). Since [itex]x^2 + y^2 = r^2[/itex] we can integrate first over x from x = 0 to [itex]x = \sqrt {r^2 - y^2}[/itex]. Once you have completed that integration, proceed to the y integration which extends from 0 to r. But [itex]r = R z/h[/itex] which is what I wrote for the upper limit on the y integration.

    Finally, z goes from 0 to h which I use for the limits of the z integration.
    Last edited: Dec 19, 2005
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook