Results 1 to 9 of 9

Math Help - Surface area

  1. #1
    Member
    Joined
    May 2010
    Posts
    241

    Surface area

    Hi there, I have to compute the surface area for
    V:\{ -2(x+y)\leq{}z\leq{}4-x^2-y^2 \}

    I have a problem on finding the surface area for the paraboloid limited by the plane. I've parametrized the plane in polar coordinates, I thought it would be easier this way, but also tried in cartesian coordinates with the same result. The problem I have is to set the limits of integration on such a way that determines the area of the paraboloid limited by the plane.

    So the paraboloid is parametrized by:

    \begin{Bmatrix}{ x=r\sin \theta}\\y=r \cos \theta \\z=4-r^2\end{matrix}<br />

    Then T_r=( \sin \theata, \cos \theta,-2r);T_{\theta}=(r\cos\theta,-r\sin\theta,0)
    T_r \times T_{\theta}=(-2r^2\sin^2 \theta,-2r^2\cos^2\theta,-r)
    ||T_r \times T_{\theta}||=\sqrt[ ]{4r^4+r^2}

    Now the surface area is determined by: \displaystyle\int_{D} ||T_r \times T_{\theta}||drd\theta
    Now D is the region inside the disc determined by the intersection of the surfaces. So D:
    -2x-2y=4-x^2-y^2\rightarrow (x-1)^2+(y-1)^2=6

    And this is my region of integration for my paraboloid. Now it doesn't seem so easy to express the region for the parametrization I choose.

    So I'm trying to solve this in cartesian coordinates, this is the integral for the surface in cartesian coord:

    \displaystyle\int_{1-\sqrt[ ]{6}}^{1+\sqrt[ ]{6}}\int_{-\sqrt[ ]{6-(x-1)^2}+1}^{\sqrt[ ]{6-(x-1)^2}+1} \sqrt[ ]{4x^2+4y^2+1} dydx

    But this integral isn't so easy to solve. I tried to go from here to cylindrical coordinates, using the substitution x=1+r\cos\theta,y=1+r\sin\theta that helps a bit with the limits of integration.

    Then I get: \displaystyle\int_{1-\sqrt[ ]{6}}^{1+\sqrt[ ]{6}}\int_{-\sqrt[  ]{6-(x-1)^2}+1}^{\sqrt[ ]{6-(x-1)^2}+1} \sqrt[ ]{4x^2+4y^2+1} dydx=\displaystyle\int_{0}^{\sqrt[ ]{6}}\int_{0}^{2\pi}r\sqrt[ ]{16+8r\cos\theta+8r\sin\theta+4r^2}d\theta dr

    Thats the best expression I get, but still too complicated to integrate by hand.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Aug 2007
    From
    USA
    Posts
    3,110
    Thanks
    2

    Re: Surface area

    If it were me, I'd want a rough estimate before doing too much. This might suggest whether one had wandered off.

    The intersection of the two surfaces suggests 12\pi as a lower bound (don't forget the bottom of your container) and not much more than that as an upper bound. Certainly less than 24\pi

    A quick check of your Rectangular Coordinates results in about 25\pi and your last expression comes in at a whopping 30\pi.

    Seems like we don't quite have the right things going on just yet.

    On the other hand, maybe I'm coming off the wrong medication and making far less sense than I think.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    May 2010
    Posts
    241

    Re: Surface area

    You think I've made a mess with the limits of integration or something like that? it's a possibility, I didn't have too much care when doing it, so it's a possibility. But I don't see any mistake. I'll try to look harder. How did you make your estimations?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Aug 2007
    From
    USA
    Posts
    3,110
    Thanks
    2

    Re: Surface area

    Limits: Doesn't strike me, right away either.

    In this case, the estimate seems pretty easy. The intersection is a circle. Beneath the circle is somewhat greater, due to its oblique orientation when compared to the circle. The parabolic piece should be more than that. Since this paraboloid is pretty flat near the top, I would expect it to be only a little larger.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Joined
    Aug 2007
    From
    USA
    Posts
    3,110
    Thanks
    2

    Re: Surface area

    Various other estimates could be made. A sphere with radius \sqrt{6} comes in at 24\pi, for example. There's a better upper bound.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    May 2010
    Posts
    241

    Re: Surface area

    I did it again in a different fashion and I get to:
    \displaystyle\int_{0}^{2\pi}\displaystyle\int_{0}^  {\sqrt[ ]{6}}r\sqrt[ ]{9+4r^2+8r(\sin\theta+\cos\theta)}drd\theta

    What do you think? I don't know how to solve it anyway :P
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    Joined
    Aug 2007
    From
    USA
    Posts
    3,110
    Thanks
    2

    Re: Surface area

    Still 25\pi-ish. I guess the next thing would be to challenge my estimates.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Member
    Joined
    May 2010
    Posts
    241

    Re: Surface area

    Where you think I could be committing the mistake? I've proceeded similarly as I did at first, just changed the parametrization from the beggining, I set x=r\cos\theta +1,y=r\sin\theta+1,z=4-(r\cos\theta +1)^2-(r\sin\theta+1)^2 and proceeded as well.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor
    Joined
    Aug 2007
    From
    USA
    Posts
    3,110
    Thanks
    2

    Re: Surface area

    Okay, one last estimate. The volume of the object is easily calculate to be 56.549 unit^3. A sphere of this same volume has a radius of 2.381 and a surface area of 71.247 or about 22.7\pi. I keep whittling down the upper bound, but still not quite getting anywhere. Any revelations, yet?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Surface Area of Surface
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 1st 2010, 09:53 AM
  2. Calculate the surface area of the surface
    Posted in the Calculus Forum
    Replies: 2
    Last Post: June 26th 2009, 04:03 AM
  3. Lateral Area and Total Surface Area
    Posted in the Geometry Forum
    Replies: 1
    Last Post: May 25th 2009, 04:28 PM
  4. Help finding surface area of a surface
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 3rd 2008, 04:11 PM
  5. Volume, Surface Area, and Lateral Surface Area
    Posted in the Geometry Forum
    Replies: 1
    Last Post: April 14th 2008, 11:40 PM

Search Tags


/mathhelpforum @mathhelpforum