Results 1 to 8 of 8

Math Help - surface area of a paraboloid

  1. #1
    Newbie
    Joined
    Jul 2010
    Posts
    8

    Question surface area of a paraboloid

    Hi,

    I tried to calculate the area of a paraboloid, say z=x^2+y^2, between z=0 and z=1. I thought that in order to do this calculation, I can think on the paraboloid as consisted of circles with radius sqrt(z), so the surface area would be given by the integral from 0 to 1 on 2*pi*sqrt(z) (the integrand is the perimeter of one circle).

    I know for sure that this result is incorrect, because I have seen one exam in which the calculation is carried on by using surface integral of the first kind, which gives a different solution.
    Last edited by fict; July 1st 2010 at 10:44 AM. Reason: mistake in parameters
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Mar 2010
    From
    Beijing, China
    Posts
    293
    Thanks
    23
    you can do the integration in your way but you need to use "arc length" instead of the z coordinate. That is, the integrand form is not 2*pi*sqrt(z)*dz, but 2*pi*sqrt(z)*ds, where ds is the arc element, you can find an expression of ds by z, but it is not a very simple form.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member AllanCuz's Avatar
    Joined
    Apr 2010
    From
    Canada
    Posts
    384
    Thanks
    4
    Quote Originally Posted by fict View Post
    Hi,

    I tried to calculate the area of a paraboloid, say z=x^2+y^2, between z=0 and z=1. I thought that in order to do this calculation, I can think on the paraboloid as consisted of circles with radius sqrt(z), so the surface area would be given by the integral from 0 to 1 on 2*pi*sqrt(z) (the integrand is the perimeter of one circle).

    I know for sure that this result is incorrect, because I have seen one exam in which the calculation is carried on by using surface integral of the first kind, which gives a different solution.
    Note how,

     1 = x^2 + y^2

     0 = x^2 + y^2

    Which gives rise to  D \to x^2 + y^2 \le 1

    If I recall the equation for surface area is

     \iint_D \sqrt{ ( \frac{ \partial z }{ \partial x } )^2 + (\frac{ \partial z }{ \partial y } )^2 + 1 }

    In which case we get (by switching to polar co-ordinates)

     \int_0^{ 2 \pi } d \theta \int_0^1 r \sqrt{ 4r + 1 } dr

    Simple subsitution will help you complete the second integral
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Jul 2010
    Posts
    8
    Quote Originally Posted by xxp9 View Post
    you can do the integration in your way but you need to use "arc length" instead of the z coordinate. That is, the integrand form is not 2*pi*sqrt(z)*dz, but 2*pi*sqrt(z)*ds, where ds is the arc element, you can find an expression of ds by z, but it is not a very simple form.
    You are correct, but I just wondered why my proposed solution is incorrect; it quite makes sense - building the paraboloid from many circles.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member AllanCuz's Avatar
    Joined
    Apr 2010
    From
    Canada
    Posts
    384
    Thanks
    4
    Quote Originally Posted by fict View Post
    You are correct, but I just wondered why my proposed solution is incorrect; it quite makes sense - building the paraboloid from many circles.
    Your integrand is not correct. You have simplified my double integral, but have not carried through the dS correctly.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Jul 2010
    Posts
    8
    Quote Originally Posted by AllanCuz View Post
    Your integrand is not correct. You have simplified my double integral, but have not carried through the dS correctly.
    Yes, I know that it is incorrect, I'm just wondering from the logical point of view why it is incorrect.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Senior Member AllanCuz's Avatar
    Joined
    Apr 2010
    From
    Canada
    Posts
    384
    Thanks
    4
    Quote Originally Posted by fict View Post
    Yes, I know that it is incorrect, I'm just wondering from the logical point of view why it is incorrect.
    Your idea is correct. It's just your integrand is not correct because it does not represent what you're saying it represents. You still need to project an element of area.

    Your integral is

     \int_0^1 \sqrt{z} dz where  \sqrt{z} = \sqrt{x^2 + y^2 }

    This is a cone....Your integrating a cone from 0 to 1. What does this have to do with surface area? This will give you area under the cone.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Senior Member
    Joined
    Mar 2010
    From
    Beijing, China
    Posts
    293
    Thanks
    23
    AllanCuz is right. Your integrand results in the area of a cone. The problem is, your coordinate curves are not laid down well on the surface to give rise to areas when integrated. For example, the x,y coordinate curves on the xy plane is good, since you can integrate the form dxdy to get areas. In your case, your coordinate curves are circles and radial rays. They're orthogonal but z is not normalized, which means, if you draw the curves z=0, 0.1, 0.2, 0.3... on your surface, they're not evenly distributed.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Surface Area of a Circular Paraboloid
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 11th 2009, 02:14 PM
  2. Area of a portion of a paraboloid
    Posted in the Calculus Forum
    Replies: 1
    Last Post: August 5th 2009, 02:38 AM
  3. Calculate the surface area of the surface
    Posted in the Calculus Forum
    Replies: 2
    Last Post: June 26th 2009, 05:03 AM
  4. Area, C of G of a paraboloid section? ...
    Posted in the Advanced Applied Math Forum
    Replies: 4
    Last Post: November 25th 2008, 02:35 PM
  5. Volume, Surface Area, and Lateral Surface Area
    Posted in the Geometry Forum
    Replies: 1
    Last Post: April 15th 2008, 12:40 AM

Search Tags


/mathhelpforum @mathhelpforum