Results 1 to 7 of 7

Math Help - Multiple Integrals (Volume + Centroid)

  1. #1
    Newbie
    Joined
    Nov 2007
    Posts
    10

    Multiple Integrals (Volume + Centroid)

    A) Find the volume bounded by sphere rho=rt(6) and the paraboloid
    z=x^2 + y^2
    B) Locate the centroid of this region

    Any help, tips, work, or similar problems would be greatly appreciated. =)
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    I start you off. The upper surface x^2+y^2+z^2=6^2 and the lower surface z=x^2+y^2. When they intersect the region over we are integrating is the circle x^2+y^2=6^2. Now use polar change of variable.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Nov 2007
    Posts
    10
    Quote Originally Posted by ThePerfectHacker View Post
    I start you off. The upper surface x^2+y^2+z^2=6^2 and the lower surface z=x^2+y^2. When they intersect the region over we are integrating is the circle x^2+y^2=6^2. Now use polar change of variable.

    When you say polar change of variable you mean things like
    x = rcos theta
    y = rsin theta
    z = z
    right?
    and then i would take the triple integral of that?
    I always have trouble setting up the problem. =(
    Like setting the integral ranges.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by numberonenacho View Post
    A) Find the volume bounded by sphere rho=rt(6) and the paraboloid
    z=x^2 + y^2
    B) Locate the centroid of this region
    The upper surface (sphere) is given by z=\sqrt{36 - x^2-y^2}.
    Thus, the volume is,
    \int_A \sqrt{36-x^2-y^2} - (x^2+y^2) \ dy~dx where A is the disk of radius 6.
    Using polar change of variable,
    \int_0^{2\pi} \int_0^6 (\sqrt{36-r^2} - r^2)rdr~d\theta
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Nov 2007
    Posts
    10
    When I tried to evaluate the integral, I can do the R part of it, but theres no place for me to put the theta value. Shouldn't there be two different variables in a double integration problem like this?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    Is that {\rho}=\sqrt{6}?.

    If so, try:

    \int_{0}^{2\pi}\int_{0}^{\sqrt{2}}\int_{r^{2}}^{\s  qrt{6-r^{2}}}rdzdrd{\theta}
    Last edited by galactus; November 24th 2007 at 01:04 PM.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Nov 2007
    Posts
    10
    hmm okay. So when I evaluate it




    how do I do the second part of the integral. That rdr complicates everything.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. multiple integrals
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 14th 2009, 04:53 PM
  2. Multiple integrals
    Posted in the Calculus Forum
    Replies: 5
    Last Post: May 26th 2009, 12:53 AM
  3. multiple integrals
    Posted in the Calculus Forum
    Replies: 3
    Last Post: February 7th 2009, 02:45 PM
  4. Centroid & Volume of region
    Posted in the Calculus Forum
    Replies: 5
    Last Post: September 5th 2006, 07:37 PM
  5. multiple integrals
    Posted in the Calculus Forum
    Replies: 4
    Last Post: April 11th 2006, 08:22 AM

Search Tags


/mathhelpforum @mathhelpforum