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: Help with Limits of Integration

  1. Jun 29, 2008 #1
    1. The problem statement, all variables and given/known data
    A Hemispheric bowl has a radius of a and a depth of h. Find the Volume

    2. Relevant equations
    [tex] r= \sqrt{a^2-y^2}

    [tex]\pi \int{(\sqrt{a^2-y^2)^2)}

    3. The attempt at a solution
    I solved the integral using the limits h and 0 and got [tex] \frac{\pi*h(3a^2-h^2)}{3}[/tex].

    But the book used h-a and -a and got [tex]\frac{\pi*h^2(3a-h)}{3}[/tex]

    My main question is if my limits are acceptable and if the book has the correct one why must I use their limits.
  2. jcsd
  3. Jun 29, 2008 #2


    User Avatar
    Homework Helper

    If you think about a sphere of radius a, you found the volume of the spherical cap from the "north pole" at y = a down to y = a-h, while the solver for the textbook chose the "south pole" cap from y = -a up to y = h-a .

    You call your levels y' = 0 and y' = h , which means you are just choosing a different reference level for "zero" and have the direction of y reversed from the textbook's solver. So fundamentally, there is no difference between your approach and theirs. The transformation suggested by the values above to go from their system to yours is y' = y + h . Your choice of coordinates is completely legitimate.

    (Found the problem!)
    EDIT: However, while the expression [tex]r = \sqrt{a^2 - y^2}[/tex] works for the solver's choice of coordinates, which measures from the center of the sphere, for your choice, measurements are made from a pole of the sphere. So you must modify your expression to

    [tex]r = \sqrt{a^2 - (a-y)^2} . [/tex]
    Last edited: Jun 29, 2008
  4. Jun 29, 2008 #3
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook