Results 1 to 6 of 6

Math Help - A volume?

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

    A volume?

    Here's the problem :
    Let B_4=\{ (x,y,z,w) \in \mathbb{R}^4 : x^2+y^2+z^2+w^2 \leq 1  \} and B_3=\{  (x,y,z) \in \mathbb{R}^3 : x^2+y^2+z^2 \leq1 \}.
    1)Show by a justification that V(B_4)=2  \iiint _{B_3} \sqrt{1-(x^2+y^2+z^2)}dxdydz.
    2)Deduce that V(B_4)=\frac{\pi ^2}{2}.
    My attempt : almost none. My first problem is : they never defined the function V. If I assume it's a volume function then they seem to ask me to calculate the volume of a 4 dimensional solid (or whatever it is called), which seems senseless. I realize that B_3 is the projection of B_4 in the 3 dimensional space.
    Also, I recognize the integrand to be the positive w that satisfy the first inequation. So V(B_4) really seems a volume since there's the multiplication by 2 in front of the triple integral so that my instinct tells me that this covers the -w. I may not be clear here, but that's how I understand the problem.
    If you understand it better than I, feel free to reformulate it so that I can understand it.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Santa Cruz, CA
    Posts
    2,844
    Thanks
    3
    Quote Originally Posted by arbolis View Post
    Here's the problem :
    Let B_4=\{ (x,y,z,w) \in \mathbb{R}^4 : x^2+y^2+z^2+w^2 \leq 1  \} and B_3=\{  (x,y,z) \in \mathbb{R}^3 : x^2+y^2+z^2 \leq1 \}.
    1)Show by a justification that V(B_4)=2  \iiint _{B_3} \sqrt{1-(x^2+y^2+z^2)}dxdydz.
    2)Deduce that V(B_4)=\frac{\pi ^2}{2}.
    My attempt : almost none. My first problem is : they never defined the function V. If I assume it's a volume function then they seem to ask me to calculate the volume of a 4 dimensional solid (or whatever it is called), which seems senseless. I realize that B_3 is the projection of B_4 in the 3 dimensional space.
    Also, I recognize the integrand to be the positive w that satisfy the first inequation. So V(B_4) really seems a volume since there's the multiplication by 2 in front of the triple integral so that my instinct tells me that this covers the -w. I may not be clear here, but that's how I understand the problem.
    If you understand it better than I, feel free to reformulate it so that I can understand it.
    x^2+y^2+z^2+w^2=1 is a ball in 4-space (or a 4-D sphere).

    If we are to find the volume, we solve for the function w(x,y,z). In this case, w=\sqrt{1-x^2-y^2-z^2}. Recall that when we find volumes in 3-space, we resort to a region in 2-space to determine the limits. In our case, we resort to a region in 3-space in order to find our limits. Then we can find the volume of the ball in 4-space.

    Thus, it would make sense that V\left(B_4\right)=2\iiint\limits_{B_3}\sqrt{1-x^2-y^2-z^2}\,dz\,dy\,dx (we need two triple integrals to take into consideration the two hemispheres of the ball).

    From here, evaluate the integral. A simple change to spherical coordinates will do the job.

    Can you continue? I hope my explanation makes sense...
    Last edited by Chris L T521; July 12th 2009 at 10:11 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor arbolis's Avatar
    Joined
    Apr 2008
    From
    Teyateyaneng
    Posts
    1,000
    Awards
    1
    Ok thanks Chris, I'll try to do it. I was accustomed to speak about volumes only for solids in 3 dimensions. Now I realize that a 3 dimensional sphere has an area ( 4\pi r^2) and with the same idea, a 4 dimensional sphere has a volume!
    I'm curious, does a 4 dimensional sphere also has an area?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,340
    Thanks
    21
    Quote Originally Posted by arbolis View Post
    Ok thanks Chris, I'll try to do it. I was accustomed to speak about volumes only for solids in 3 dimensions. Now I realize that a 3 dimensional sphere has an area ( 4\pi r^2) and with the same idea, a 4 dimensional sphere has a volume!
    I'm curious, does a 4 dimensional sphere also has an area?
    2D surface area (i.e. arc length) \int_R \sqrt{1+y'^2} dx in 3D \iint_R \sqrt{1+z_x^2+z_y^2} \,dA

    What do you think?
    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
    Quote Originally Posted by Danny View Post
    2D surface area (i.e. arc length) \int_R \sqrt{1+y'^2} dx in 3D \iint_R \sqrt{1+z_x^2+z_y^2} \,dA

    What do you think?
    Seems to make sense. Did you deduce it, or saw it in a book/internet?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,340
    Thanks
    21
    Quote Originally Posted by arbolis View Post
    Seems to make sense. Did you deduce it, or saw it in a book/internet?
    Here's an interesting link

    Hypersphere -- from Wolfram MathWorld
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: November 18th 2011, 01:08 PM
  2. volume flux, mass flux and volume flo
    Posted in the Advanced Applied Math Forum
    Replies: 2
    Last Post: June 7th 2011, 06:30 PM
  3. Replies: 1
    Last Post: May 14th 2010, 04:08 PM
  4. divergence = flux / volume is independant of the limiting volume
    Posted in the Differential Geometry Forum
    Replies: 0
    Last Post: April 26th 2010, 07:31 PM
  5. volume
    Posted in the Geometry Forum
    Replies: 3
    Last Post: June 10th 2008, 11:15 AM

Search Tags


/mathhelpforum @mathhelpforum