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

A simple calculation that got me stumped.

  1. Jun 22, 2012 #1


    User Avatar
    Gold Member

    I need to grade some work, and there's an integral to calcaulte:

    compute the area of y^2+z^2=2ax where x is between a and y^2/a.

    I tried using the coordinates:

    [itex]y=\sqrt{2as}\cos(t) \ z=\sqrt{2as}\sin(t) \ x=s[/itex]

    But I don't see how to find the limits of integration with respect to s and t.


    The higher you get in maths, the less you remember what you learned in earlier years. :-(
  2. jcsd
  3. Jun 22, 2012 #2
    I rewrite it as [itex]x^2+y^2=2az[/itex] where z is between a and [itex]x^2/a[/itex]. I just renamed the variables: y as x, z as y and x as z (I am more comfortable that way.)

    Written explicitly as a function of x, y is [itex]y=\sqrt{2az-x^2}[/itex]. We know the limits, and hence this integral becomes

    I think you can go on from there.
  4. Jun 22, 2012 #3


    User Avatar
    Science Advisor
    Homework Helper

    solving your own equations says that a ≤ s = x ≤ y^2/a = 2s cos^2(t). That gives s ≥ a.

    Then canceling s, from s ≤ 2s cos^2(t), we have

    1 ≤ 2cos^2(t), so cos^2(t) ≥ 1/2, so -π/4 ≤ t ≤ π/4.

    if that open ended inequality for s bothers you, note that your surface is unbounded in the x direction.

    I.e. for any x ≥ a >0, it is easy to solve for y such that 2ax ≥ y^2 ≥ ax, and then since 2ax - y^2 ≥ 0,

    one can also solve for z such that z^2 = 2ax - y^2, or z^2 + y^2 = 2ax.
    Last edited: Jun 22, 2012
  5. Jun 22, 2012 #4
    Same with me. Your solution puzzles me. I think there is something I am missing in the question.
  6. Jun 22, 2012 #5


    User Avatar
    Science Advisor
    Homework Helper

    I changed a mistake in mine. (I canceled s/s and got 0.)

    in your integral what does it mean to let x range from a to x^2/a?
  7. Jun 22, 2012 #6
    Hi MP,

    How about this; if [itex]x[/itex] was from 0 to [itex]\frac{y^2}{a}[/itex] then we could use polar coordinates to integrate using the radius value based on θ, so using Pythagoras we have that [itex]cos θ = \frac{y^2}{2ay^{\frac{2}{a}}}[/itex] and therefore [itex]y = (2a\cosθ) ^{\frac{1}{2-\frac{2}{a}}}[/itex] so the squared radius will be

    [tex] r(θ)^2 = 2a((2a\cosθ) ^{\frac{1}{2-\frac{2}{a}}})^{\frac{2}{a}} [/tex]

    Now we simply polar integrate it from 0 to 2π [itex]\frac{1}{2}\int_0^{2π} r(θ)^2 dθ[/itex] but since [itex]x[/itex] is not from 0 to [itex]\frac{y^2}{a}[/itex] but from [itex]a[/itex] to [itex]\frac{y^2}{a}[/itex] we need to take away that part, so finally we have:

    [tex]\frac{1}{2}\int_0^{2π} r(θ)^2 dθ - \frac{1}{2}\int_0^{2π}\min( r(θ)^2 , 2a^2) dθ[/tex]

    I didn't thoroughly check it but I think you get the idea.
    Last edited: Jun 22, 2012
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook