Results 1 to 4 of 4

Math Help - Answer check (spherical coords)

  1. #1
    Senior Member
    Joined
    Jul 2006
    Posts
    364

    Answer check (spherical coords)

    Use spherical coords to evaluate:

    \int_0^1 \int_0^{\sqrt{1-x^2}} \int_0^{\sqrt{1-x^2-y^2}} (2x^2+2y^2+2z^2)^{-1/2} dzdydx

    My sol'n:

    We are dealing with the first octant here, with sphere radius=1. So bounds are:

    0\le r \le 1,  ~0\le\theta\le \frac{\pi}{2}, ~ ~0\le\phi\le \frac{\pi}{2}.

    dV = r^2\sin\phi dr d\theta d\phi.

    So hopefully we have (after some cancelling):

    \frac{\sqrt{2}}{2}\int_0^{\frac{\pi}{2}} \int_0^{\frac{\pi}{2}} \int_0^1 r\sin\phi\ dr d\theta d\phi

    which is

    -\frac{\sqrt{2}\pi}{8}

    I hope that's right
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    16,004
    Thanks
    1660
    Yes, it looks good to me.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Dec 2008
    Posts
    319
    I derived

    \int_0^1\int_0^{\sqrt{1-x^2}}\int_0^{\sqrt{1-x^2-y^2}}(2x^2+2y^2+2z^2)^{-1/2}\,dz\,dy\,dx
    =\int_0^{\frac{\pi}{2}}\int_0^{\frac{\pi}{2}}\int_  0^1 (2\rho^2)^{-\frac{1}{2}}\rho^2\sin \phi\,d\rho\,d\theta\,d\phi
    =\int_0^{\frac{\pi}{2}}\int_0^{\frac{\pi}{2}}\int_  0^1 \frac{1}{\sqrt{2}}\rho^{-1}\rho^2\sin \phi\,d\rho\,d\theta\,d\phi
    =\int_0^{\frac{\pi}{2}}\int_0^{\frac{\pi}{2}}\int_  0^1 \frac{1}{\sqrt{2}}\rho\,\sin \phi\,d\rho\,d\theta\,d\phi
    =\int_0^{\frac{\pi}{2}}\int_0^{\frac{\pi}{2}} \frac{1}{2\sqrt{2}}\sin \phi\,d\theta\,d\phi
    =\int_0^{\frac{\pi}{2}} \frac{\pi}{4\sqrt{2}}\sin \phi\,d\phi
    =\left(-\frac{\pi}{4\sqrt{2}}\cos \phi\right)_0^{\frac{\pi}{2}}
    =-\frac{\pi}{4\sqrt{2}}\cos \left(\frac{\pi}{2}\right)-\left(-\frac{\pi}{4\sqrt{2}}\cos 0\right)
    =0-\left(-\frac{\pi}{4\sqrt{2}}\right)=\frac{\pi}{4\sqrt{2}}  .
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member
    Joined
    Jul 2006
    Posts
    364
    Thank you both very much!

    Scott -- thanks for the detailed working I made a silly sign mistake at the end as you can see :/.

    Your solution is correct (and equivalent to the negative of mine).
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: July 12th 2011, 09:53 PM
  2. Replies: 1
    Last Post: July 4th 2011, 12:16 AM
  3. Cylindrical and Spherical coords volume
    Posted in the Calculus Forum
    Replies: 1
    Last Post: May 7th 2011, 12:02 AM
  4. I want check my answer please >
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: December 8th 2009, 02:02 AM
  5. Triple Integral in Spherical Coords.
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 22nd 2009, 06:11 AM

Search Tags


/mathhelpforum @mathhelpforum