1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Spherical Coordinates - Help me find my bounds

  1. Apr 2, 2015 #1

    RJLiberator

    User Avatar
    Gold Member

    1. The problem statement, all variables and given/known data

    A vase is filled to the top with water of uniform density f = 1. The side profile of the barrel is given by the surface of revolution obtained by revolving the graph of g(z) = 2 + cos(z) over the z-axis, and bounded by 0 ≤ z ≤ π. Find the mass of the vase.

    2. Relevant equations


    3. The attempt at a solution

    So I know that Mass = the triple integral of density dV.
    I need to find my bounds.

    I would like to use spherical coordinates as it seems to make sense with this problem.

    Theta goes from 0 to 2pi.
    phi goes from 0 to pi ? since it is bounded by 0 and pi? Or am I wrong here?
    and rho goes from 0 to something involving the g(z) equation, but I am not sure how to manipulate it :/.

    Any tips to get me started or confirmation on the phi bounds would be great.
     
  2. jcsd
  3. Apr 2, 2015 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Cylindrical coordinates make much more sense since you have the radial limit as an explicit function of z.
     
  4. Apr 2, 2015 #3

    Svein

    User Avatar
    Science Advisor

    Therefore, in cylindrical coordinates:
    Yes, and r=g(z).
     
  5. Apr 2, 2015 #4

    RJLiberator

    User Avatar
    Gold Member

    Ah, so I assume the wrong coordinates. I guess this makes sense since a barrel is a cylinder... Sigh. 0_o. Let me try some work now.
     
  6. Apr 2, 2015 #5

    RJLiberator

    User Avatar
    Gold Member

    So I am getting:

    Theta: from 0 to 2pi.
    r from 0 to 2+cos(z)
    z from 0 to pi.

    This would mean we integrate in the order dr dz dtheta
     
  7. Apr 2, 2015 #6

    RJLiberator

    User Avatar
    Gold Member

    So I performed the triple integral with the bounds:
    Theta: from 0 to 2pi.
    r from 0 to 2+cos(z)
    z from 0 to pi.

    using cylindrical coordinates and received the answer of (9pi^2)/2

    It seems to be right, my one concern is with the bounds on r. It's clear to me that it starts at 0, but why am I allowed to put 2+cos(z) as the upper limit. So z=r in cylindrical coordinates, and the function was g(z)=2+cos(z) so it was that simple?
     
  8. Apr 2, 2015 #7

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Sound right. I'm not sure what you are saying with 'z=r in cylindrical coordinates'. The function g(z) defines the largest value of r for a given value of z, isn't that what the problem description implies?
     
  9. Apr 2, 2015 #8

    RJLiberator

    User Avatar
    Gold Member

    Hm. I see.

    The function g(z) definitely defines the largest value for r, that makes more sense to me. r is the distance from the z-axis in cylindrical coordinates and that is what is implied by the question (bounds).

    That makes sense to me. The other bounds are rather obvious, so that seems to work.

    Thank you for your guidance.
     
  10. Apr 2, 2015 #9

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    You could also have done it by cross sections as a single integral since it is a solid of revolution.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted