Results 1 to 6 of 6

Thread: Surface area of a disc

  1. #1
    Member
    Joined
    May 2011
    Posts
    169

    Surface area of a disc

    I have a surface defined as $\displaystyle z(x^2+y^2)=4$. I have calculated the surface integral but now I just want to calculate the contribution from the disc z=4. How do i do this?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    19,718
    Thanks
    3003

    Re: Surface area of a disc

    Pretty straight forward, isn't it? If z= 4, your equation becomes $\displaystyle 4(x^2+ y^2)= 4$ so $\displaystyle x^2+ y^2= 1$. That's a circle with radius 1.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    May 2011
    Posts
    169

    Re: Surface area of a disc

    I thought so but the answer is 15pi, not pi
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Apr 2005
    Posts
    19,718
    Thanks
    3003

    Re: Surface area of a disc

    The answer to what question?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    May 2011
    Posts
    169

    Re: Surface area of a disc

    The surface integral was 33pi. In part (b), I had to calculate volume integral where V is the region defined by $\displaystyle 0<x^2+y^2<4/z$ and $\displaystyle 1<z<4$. This was 48pi.

    Now I have to verify divegence theorem. Well, I need to show 33pi + contribution from z=4 + contribution from z=1 = 48pi.

    Sorry if I was not clear enough.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Apr 2005
    Posts
    19,718
    Thanks
    3003

    Re: Surface area of a disc

    In cylindrical coordinates, $\displaystyle x^2+ y^2= \frac{4}{z}$ becomes $\displaystyle z= \frac{4}{r^2}$. Of course, in order to have a closed figure, we must have the "top" and "bottom", z= 4, and z= 1.

    The divergence theorem says that $\displaystyle \int\oint\vec{F}d\vec{S}= \int\int\int \nabla\cdot\vec{F} dV$.

    If you want to find the volume of the figure, $\displaystyle \int\int\int dV$, using that, you must have a vector function, $\displaystyle \vec{F}$, such that $\displaystyle \nabla\cdot\vec{F}= 1$. What vector function did you use?

    Of course, the top of the figure, z= 4, is the disk $\displaystyle x^2+y^2\le 1$ while the bottom of the figure, z= 1, is the disk $\displaystyle x^2+ y^2= 4$ which means that, to find the volume directly, you must integrate the constant 1 from z= 0 to z= 4 over the disk $\displaystyle x^2+ y^2= 1$ which is simply 1 times 4 times the area of that disk, $\displaystyle 4\pi$. Then, integrate the constant 1 from z= 0 to $\displaystyle z= \frac{4}{x^2+ y^2}= \frac{4}{r^2}$ (in cylindrical coordinates) for r= 1 to 2 (and $\displaystyle \theta$ from 0 to $\displaystyle 2\pi$, of course).
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 0
    Last Post: Feb 8th 2011, 10:54 AM
  2. Help finding surface area of a surface
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Oct 3rd 2008, 04:11 PM
  3. Non-uniform mass per unit area of a disc?
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Jun 30th 2008, 07:54 AM
  4. Volume, Surface Area, and Lateral Surface Area
    Posted in the Geometry Forum
    Replies: 1
    Last Post: Apr 14th 2008, 11:40 PM
  5. If the area of a disc...
    Posted in the Geometry Forum
    Replies: 3
    Last Post: Oct 27th 2006, 07:52 AM

Search tags for this page

Click on a term to search for related topics.

Search Tags


/mathhelpforum @mathhelpforum