Results 1 to 3 of 3

Math Help - Areas of surfaces.

  1. #1
    Super Member Showcase_22's Avatar
    Joined
    Sep 2006
    From
    The raggedy edge.
    Posts
    782

    Arrow Areas of surfaces.

    Here's my question:

    Find the area of the surface created by the complete curve r=a when it is completely revolved about the x axis.Do not assume the formula for the surface area of a sphere.

    Just by looking at it I did:



    Which was incorrect.

    So then I did it the way I think I was supposed to:



    and this answer was also wrong. The book gave the answer as:



    What's going wrong??
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by Showcase_22 View Post
    Here's my question:

    Find the area of the surface created by the complete curve r=a when it is completely revolved about the x axis.Do not assume the formula for the surface area of a sphere.

    Just by looking at it I did:

    Mr F says: *Ahem* You're rotating a circle around the x-axis. This gives a sphere. You do recall that the surface area of a sphere is {\color{red}4 \pi r^2}, don't you?

    Which was incorrect. Mr F says: Obviously.

    So then I did it the way I think I was supposed to:



    and this answer was also wrong. The book gave the answer as:



    What's going wrong??
    The formula is S = 2 \pi \int_{\alpha}^{\beta} r \sin \theta \sqrt{ r^2 + \left( \frac{dr}{d\theta} \right)^2 } \, d \theta.

    So for your problem S = 2 \times 2 \pi \int_{0}^{\pi/2} a \sin \theta \sqrt{ a^2 + 0^2 } \, d \theta = {\color{red}4} \pi \int_{0}^{\pi/2} a^2 \sin \theta \, d \theta.

    That's why your answer is wrong by a factor of 2.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member Showcase_22's Avatar
    Joined
    Sep 2006
    From
    The raggedy edge.
    Posts
    782
    I wasn't visualising it correctly. I was thinking of it in 2D form and not 3D.

    When it's a 2D circle and it's rotated around the x axis, it only needs to be rotated by pi radians and not 2pi.

    Thanks Mr Fantastic!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Surfaces
    Posted in the Calculus Forum
    Replies: 1
    Last Post: May 28th 2010, 11:31 AM
  2. surfaces
    Posted in the Calculus Forum
    Replies: 2
    Last Post: April 6th 2010, 11:24 AM
  3. areas, surfaces and volumes...
    Posted in the Calculus Forum
    Replies: 17
    Last Post: August 8th 2007, 04:28 PM
  4. orthognality of two surfaces.........
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 15th 2007, 11:09 PM
  5. Surfaces
    Posted in the Geometry Forum
    Replies: 2
    Last Post: February 2nd 2007, 05:34 AM

Search Tags


/mathhelpforum @mathhelpforum