Results 1 to 7 of 7

Math Help - [SOLVED] Calculate a volume + theoretical question

  1. #1
    MHF Contributor arbolis's Avatar
    Joined
    Apr 2008
    From
    Teyateyaneng
    Posts
    1,000
    Awards
    1

    [SOLVED] Calculate a volume + theoretical question

    1)Identify the region of integration and calculate the following integral using spherical coordinates : \int_{-3}^{3} \int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}} \int _0^{\sqrt{9-x^2-y^2}} z\sqrt{x^2+y^2+z^2} dzdydx.

    My attempt : I notice that the integrand is not the Jacobian of the change of variable spherical to Cartesian but almost. I mean that z\sqrt{x^2+y^2+z^2}=\rho ^2 \cos \phi instead of \rho ^2 \sin \phi. Thus I am at a loss.
    I don't know how to answer the question.

    2) I'd like to know a general method (i.e. applicable in all cases) that transform a complicated solid (like a cone, a paraboloid, hyperboloid or anything different from a cylinder, cube and sphere) into a cube. I'd appreciate any book reference or an example with the above exercise.
    I ask this because even though it might be difficult to find such a transformation, if I success in finding it then its Jacobian shouldn't be a problem to find and so the volume of the complicated solid should be quite easy to solve.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member Random Variable's Avatar
    Joined
    May 2009
    Posts
    959
    Thanks
    3
    You're integrating over a hemisphere of radius 3

     \int^{3}_{0} \int^{\pi /2}_{0} \int^{2 \pi}_{0} \rho^{4} \cos \phi \sin \phi \ d \theta \ d \phi \ d\rho

     = 2 \pi \int^{3}_{0} \int^{\pi /2}_{0} \rho^{4} \cos \phi \sin \phi \ d \phi \ d\rho

    let  u = \sin \phi

     = 2 \pi \int^{3}_{0} \int^{1}_{0} \rho^{4} u \ du\ d\rho

     = \pi \int^{3}_{0} \rho^{4} \ d\rho

     = \frac {243 \pi}{5}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Apr 2005
    Posts
    16,442
    Thanks
    1863
    Quote Originally Posted by arbolis View Post
    1)Identify the region of integration and calculate the following integral using spherical coordinates : \int_{-3}^{3} \int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}} \int _0^{\sqrt{9-x^2-y^2}} z\sqrt{x^2+y^2+z^2} dzdydx.

    My attempt : I notice that the integrand is not the Jacobian of the change of variable spherical to Cartesian but almost. I mean that z\sqrt{x^2+y^2+z^2}=\rho ^2 \cos \phi instead of \rho ^2 \sin \phi. Thus I am at a loss.
    The integrand has nothing to do with the "region of integration". Looking at the "inside integral", z ranges from z= 0 to z= \sqrt{9- x^2-y^2} which is the upper half of the sphere z^2= 9- x^2- y^2 or x^2+ y^2+ z^2= 9. Then, y ranges from y= -\sqrt{9- x^2} to y= \sqrt{9- x^2} so it covers y= 9- x^2, the circle x^2+ y^2= 9, the circle, in the xy-plane with center at (0,0) and radius 3. It should be no surprise that the limits of integration on x are -3 to 3. The region of integration is a hemisphere with center at (0,0,0) and radius 3. Yes, the integrand, in polar coordinates is \rho^2 cos(\phi) so you the integral, in polar coordindates is
    \int_{\phi= 0}^{\pi/2}\int_{\theta= 0}^{2\pi}\int_{\rho= 0}^3 \rho^2 cos(\phi)d\rho d\theta d\phi

    I don't know how to answer the question.

    2) I'd like to know a general method (i.e. applicable in all cases) that transform a complicated solid (like a cone, a paraboloid, hyperboloid or anything different from a cylinder, cube and sphere) into a cube. I'd appreciate any book reference or an example with the above exercise.

    I ask this because even though it might be difficult to find such a transformation, if I success in finding it then its Jacobian shouldn't be a problem to find and so the volume of the complicated solid should be quite easy to solve.
    There is no good way to change a "smooth" surface, like a sphere or paraboloid, into a surface with edges like a cube. And I can't imagine that it is a good idea to do that since, in general, smooth surfaces are better than non-smooth!
    Last edited by Jhevon; July 11th 2009 at 06:24 PM. Reason: fixed LaTeX
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    We have z\sqrt{x^{2}+y^{2}+z^{2}}

    But z={\rho}cos{\phi}, \;\ x={\rho}sin{\phi}cos{\theta}, \;\ y={\rho}sin{\phi}sin{\theta}

    Plugging these all in and simplifying and it whittles down to

    {\rho}^{2}cos{\phi}

    Multiply by the {\rho}^{2}sin{\phi} that comes along with

    spherical integration and get {\rho}^{4}sin{\phi}cos{\phi}

    \int_{0}^{2\pi}\int_{0}^{\frac{\pi}{2}}\int_{0}^{3  }{\rho}^{4}sin{\phi}cos{\phi}d{\rho}d{\phi}d{\thet  a}
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor arbolis's Avatar
    Joined
    Apr 2008
    From
    Teyateyaneng
    Posts
    1,000
    Awards
    1
    Ok, thank you very much to all. I'll take time to think about the problem.
    And about
    Quote Originally Posted by HallsofIvy
    There is no good way to change a "smooth" surface, like a sphere or paraboloid, into a surface with edges like a cube. And I can't imagine that it is a good idea to do that since, in general, smooth surfaces are better than non-smooth!
    Ok. And changing a smooth surface into another one (for example a sphere)?
    For instance, in this thread : http://www.mathhelpforum.com/math-he...lume-cone.html, I don't know how NCA came up with such a transformation to find the volume of a cone. Precisely I don't understand how he found that by putting x=\sqrt2 r \cos \theta, etc. he could find the volume.


    Edit: I understand the first question now. Thank you so much. HallsofIvy forgot to multiply the integrand by \rho ^2 \sin \phi, but otherwise everything's as clear as vacuum. So I'm done with the first question, but the second is still open.
    Last edited by arbolis; July 11th 2009 at 07:42 PM.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member Random Variable's Avatar
    Joined
    May 2009
    Posts
    959
    Thanks
    3
    Quote Originally Posted by arbolis View Post
    Ok, thank you very much to all. I'll take time to think about the problem.
    And about Ok. And changing a smooth surface into another one (for example a sphere)?
    For instance, in this thread : http://www.mathhelpforum.com/math-he...lume-cone.html, I don't know how NCA came up with such a transformation to find the volume of a cone. Precisely I don't understand how he found that by putting x=\sqrt2 r \cos \theta, etc. he could find the volume.
    It is easier to integrate over a circle than over an ellipse.

    If you had  \frac {x^{2}}{a} + \frac {y^{2}}{b} \le c , then you would let  x = \sqrt{a} r \cos \theta and  y = \sqrt{b} r \sin \theta
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor arbolis's Avatar
    Joined
    Apr 2008
    From
    Teyateyaneng
    Posts
    1,000
    Awards
    1
    Quote Originally Posted by Random Variable View Post
    It is easier to integrate over a circle than over an ellipse.

    If you had  \frac {x^{2}}{a} + \frac {y^{2}}{b} \le c , then you would let  x = \sqrt{a} r \cos \theta and  y = \sqrt{b} r \sin \theta
    Ah!!! I understand now. Finally I'm done with this thread. Thanks a million to all!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: May 14th 2010, 05:08 PM
  2. [SOLVED] Theoretical Derivation of Benford's Law
    Posted in the Advanced Statistics Forum
    Replies: 2
    Last Post: May 8th 2010, 11:01 PM
  3. Theoretical Question
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: October 27th 2009, 09:14 PM
  4. Calculate volume
    Posted in the Calculus Forum
    Replies: 9
    Last Post: November 23rd 2008, 11:47 AM
  5. [SOLVED] Volume of Revolution Question
    Posted in the Calculus Forum
    Replies: 2
    Last Post: April 25th 2008, 04:39 AM

Search Tags


/mathhelpforum @mathhelpforum