Results 1 to 3 of 3

Math Help - Find height of liquid inside spherical container as it is being filled

  1. #1
    Newbie
    Joined
    Sep 2010
    Posts
    1

    Find height of liquid inside spherical container as it is being filled

    Here is the problem:

    "A spherical container with a radius of 15cm is being filled at a constant rate of 250╥ ml/sec. What is the height of the liquid from the bottom of the container after 4.5 seconds?"

    I think you need to use disk integration to solve this one, but I am not sure. I could be over thinking this one.


    Also, is it possible to derive an equation representing the height of the liquid as it is being filled vs time?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,423
    Thanks
    1332
    Yes, treating the sphere as a "solid of revolution" and using the disk method to find the volume is probably the best way to do this.

    First start by setting up a coordinate system. You could do this by setting the origin at the center of the sphere but I suspect it will be a little simpler to think about if we set the origin at the bottom of the sphere so that the "height of the liquid" is its z coordinate.. Then the center will be at (0, 0, 15). Looked at from the side, the sphere appears to be a circle, in the xz- plane, with center at (0, 15). Its equation is x^2+ (z- 15)^2= 15^2= 225. Each "disk" at a given z, seen from the side as here, appears to be a line segment from -x to + x where x= \sqrt{225- (z- 15)^2 from the equation of the circle. Rotated around the z-axis, the disk has area \pi r^2= \pi (225- (z- 15)^2). The volume from z= 0 to z= h is given by \pi\int_0^h (225- (z- 15)^2)dz.

    Find the volume of water poured in at 250\pi ml/sec for 4.5 seconds, set that integral equal to it and solve for h.
    Last edited by HallsofIvy; September 15th 2010 at 08:01 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Sep 2010
    Posts
    185
    Thanks
    13
    Quote Originally Posted by surfertmex View Post
    Here is the problem:

    "A spherical container with a radius of 15cm is being filled at a constant rate of 250╥ ml/sec. What is the height of the liquid from the bottom of the container after 4.5 seconds?"

    I think you need to use disk integration to solve this one, but I am not sure. I could be over thinking this one.


    Also, is it possible to derive an equation representing the height of the liquid as it is being filled vs time?
    Liquid inside the sphere would form a spherical cap and the volume of that cap V_{cap} is the volume of the liquid V_{liquid} inside the sphere. Then obviously the height h of the spherical cap is the height of the liquid inside the sphere. Volume of the liquid can be calculated using the information given, and volume of the spherical cap can be calculated using the formula V_{cap}=\frac{h^2 \cdot Pi}{3}(3R-h), where R is the radius and h is the height of the spherical cap. Forming the equation V_{cap}=V_{liquid} makes it possible to find the height h.

    Easy as \pi and no integration whatsoever.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 8
    Last Post: November 18th 2011, 02:03 PM
  2. Replies: 1
    Last Post: March 24th 2011, 11:37 PM
  3. Replies: 13
    Last Post: June 23rd 2008, 09:43 AM
  4. Find Height
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: May 8th 2007, 05:04 AM
  5. find area inside the circle
    Posted in the Calculus Forum
    Replies: 3
    Last Post: February 15th 2007, 12:31 PM

Search Tags


/mathhelpforum @mathhelpforum