Results 1 to 3 of 3
Like Tree1Thanks
  • 1 Post By HallsofIvy

Math Help - Integral of a hemisphere

  1. #1
    Newbie Naranja's Avatar
    Joined
    Apr 2013
    From
    N/A
    Posts
    13
    Thanks
    2

    Smile Integral of a hemisphere

    Hello everybody,

    I am looking to solve a problem involving a triple (volume) integral. I intend to evaluate: \iiint_{V}z(x^2+y^2+z^2)dxdydz

    where V is the upper hemisphere 0 \leq r \leq a and z \geq 0, with r = \sqrt{x^2+y^2+z^2}.

    I wish to solve using both cartesian coordinates and then spherical polar coordinates, show that the result is equal. What I wish to ask of you is, how can I determine the limits of the integration for the cartesian coordinates. I don't know how I can differentiate between a hemisphere and cylinder in this respect, since in both cases the xy-plane is a circle and then the limits for z is just 0 to a.


    But, to be honest, either way I don't think I can accurately determine the limits. I am having trouble picturing it. I know you have to take one variable at a time, and the limits will be a function of the 'remaining' variables, working down to no variables in the solution.

    I have tried to rearrange 0 \leq r \leq a to get limits with one variable, say x, but then I end up with negative root on the left, which can't be right. So do I leave the lower limit as 0? Either way, I can't justify this. There may be just one simple idea I am missing.


    Thank you
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    14,973
    Thanks
    1121

    Re: Integral of a hemisphere

    If this is the upper hemisphere, you would do better to integrate in the order "dzdydx" rather than "dxdydz". Using Cartesian coordiates, and itegrating in the order "dzdydx", the limits on the outer integral have to cover all possible x-value so run from -a to a. To determine the y limits, observe that the hemisphere projects down to the xy-axis as the region inside x^2+ y^2= a^2 or y= \pm\sqrt{a^2- x^2} so y runs from -\sqrt{a^2- y^2} to \sqrt{a^2- x^2}. Finally, since we have the upper hemisphere of x^2+ y^2+ z^2= a^2, which is equivalent to z= \pm\sqrt{a^2- y^2- z^2}, z will go from 0 to \sqrt{a^2- x^2- y^2}.

    That is, the integral would be \int_{x= -a}^a\int_{y= -\sqrt{a^2- x^2}}^{\sqrt{a^2- x^2}}\int_{z= 0}^{\sqrt{a^2- x^2- y^2}} z(x+ y+ z)dzdydx.

    If you want to use "spherical coorinates" then, at least, you need to understand what the coordinates mean! For any given point, (x, y, z), \rho (not r) is the staright line distance from (0, 0, 0) to (x, y, z), \theta is the angle, in the xy- plane, between the line from (0, 0) to (x, y) and the x-axis, so from 0 to 2\pi to cover the entire plane, and \phi is the angle the line from (0, 0, 0) to (x, y, z) makes with the positive z-axis, so from 0 to \pi to cover all the way from the positive z-axis to the negative z-axis.

    So for the upper hemi-sphere, with center at (0, 0, 0) and radius a, \rho goes from 0 to a, \theta goes from 0 to 2\pi, and \phi goes from 0 to \pi/2.
    \int_{\phi= 0}^{\pi/2}\int_{\theta= 0}^{\pi/2}\int_{\rho= 0}^a \rho^2 cos(\phi)(cos(\theta)sin(\phi)+ sin(\theta)sin(\phi)+ cos(\phi))d\rho d\theta d\phi.
    Last edited by HallsofIvy; April 17th 2013 at 01:25 PM.
    Thanks from Naranja
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie Naranja's Avatar
    Joined
    Apr 2013
    From
    N/A
    Posts
    13
    Thanks
    2

    Re: Integral of a hemisphere

    Hello HallsofIvy,

    Thank you for a great post!

    I am aware the order the variables are integrated is changeable, I posted the problem as it was stated. I wasn't sure which was the best variable to start with. Looking at the limits you gave, I can see where I was going wrong in trying to comprehend it; though I did try to (unsuccessfully) visualise it as such. I can understand where these limits are coming from now, so thanks! So, for any upper hemisphere integral problem, these will remain the same, correct? (Although the 'a' could vary, I would simply substitute any given value in place for this). Regardless of the 'function' being integrated?

    As for the spherical coordinates. I'll be honest and say that I had these limits in my notes! But, you have helped me to further understand the meaning, which will certainly help me to remember.

    I will now solve using both coordinate systems and (hopefully) come up with the same answer in each case!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. centroid of a hemisphere
    Posted in the Calculus Forum
    Replies: 9
    Last Post: April 19th 2012, 01:01 PM
  2. Sphere/hemisphere
    Posted in the Calculus Forum
    Replies: 1
    Last Post: December 12th 2010, 02:43 AM
  3. Temperature in Hemisphere
    Posted in the Differential Equations Forum
    Replies: 0
    Last Post: March 31st 2009, 05:41 AM
  4. How do I integrate this part of a hemisphere
    Posted in the Calculus Forum
    Replies: 9
    Last Post: December 8th 2008, 11:54 PM
  5. Cone and Hemisphere
    Posted in the Trigonometry Forum
    Replies: 4
    Last Post: July 8th 2007, 09:15 AM

Search Tags


/mathhelpforum @mathhelpforum