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!

Finding the volume - Polar coordinates

  1. Jan 2, 2017 #1
    Hello everyone,

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

    Question : Find the volume of the region which remains inside the cyclinder x 2 + y 2 = 2y, and is bounded from above by the paraboloid surface x 2 + y 2 + z = 1 and from below by the plane z = 0

    2. Relevant equations



    3. The attempt at a solution

    This looks like a pretty straightforward question. First thing to do is converting equations to polar coordinates. When they are converted, the cylinder's region can be written as r = 2sinθ. Parabolaid surface's eq. is (1-r2).Through these, a double integral can easily be set up as the following:

    ∫(from 0 to π)∫from 0 to 2sinθ) (1-r2)rdrdθ.

    However, the solution to this problem proposes a different approach which does not return the same answer.

    Here it is:
    Mm1ZSoV.png

    Thanks in advance,
     
    Last edited: Jan 2, 2017
  2. jcsd
  3. Jan 2, 2017 #2

    Mark44

    Staff: Mentor

    No.
    If you convert the cylinder's equation to polar coordinates, you get ##r^2 = 2r\sin(\theta)##, which is different from what you show.
    What you show here is not an equation. What equation do you get?
     
  4. Jan 2, 2017 #3
    Hello Mark,

    I am sorry for the inconvenience I caused.

    I simplified ##r^2 = 2r\sin(\theta)##.

    Then, I converted parabolaid's surface equation.

    ##z = 1 - x^{2}-y^{2}##
    ##z = 1 - r^{2}##​
     
  5. Jan 2, 2017 #4

    Mark44

    Staff: Mentor

    No problem. That's what we're here for.
    You should get a different result from these equations.
     
  6. Jan 2, 2017 #5
    Here is my reasoning :

    By polar coordinates definition ##x = r\cos(\theta)## and ##y = r\sin(\theta)##. Therefore, ##x^{2}+y^{2} = r^{2}## I do not see any problem. Also, the answer confirms my logic. The only problem is that the answer divides the integral into two, one of them is from 0 to ##\pi/6##, and the other is from ##\pi/6## to ##\pi/2##. However, second part of the integral's "r" range is 0 to 1, and I am having difficult times understanding this. My coursebook has a similar question solved in a similar manner as mine.
     
  7. Jan 2, 2017 #6
    You know, I feel like you're missing an extra integral sign. Shouldn't you use cylindrical coordinates, or something?
     
  8. Jan 2, 2017 #7

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Look at the first quadrant in this picture, which shows the circle traces in the ##xy## plane:
    upload_2017-1-2_11-46-6.png
    The brown region is the polar area over which you need to integrate. The ##30^\circ## line divides the area into two parts. In the lower sliver under the line ##r## goes from ##0## to the pink circle. Above the line ##r## goes from ##0## to the brown circle.
     
  9. Jan 2, 2017 #8
    Thank you. I now understand that I did not consider the region which is limited by the surface x 2 + y 2 + z = 1.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted