Results 1 to 6 of 6

Math Help - surface area about x-axis

  1. #1
    Member
    Joined
    Jan 2010
    Posts
    79

    surface area about x-axis


    I need to find the surface area of the above curve, about the x-axis. I'm having trouble taking the integral. Once everthing is set up its:

    2pi\int(\sqrt{x})(\sqrt{(1+(1/4x)^2))} from 4 to 9

    Or at least, that's what I think it is..
    Any help would be greatly appreciated.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Mar 2009
    From
    Alberta
    Posts
    173
    Quote Originally Posted by cdlegendary View Post

    I need to find the surface area of the above curve, about the x-axis. I'm having trouble taking the integral. Once everthing is set up its:

    2pi\int(\sqrt{x})(\sqrt{(1+(1/4x)^2))} from 4 to 9

    Or at least, that's what I think it is..
    Any help would be greatly appreciated.
    As you know, the surface area of a revolution is found through the following formula:

     A = 2\pi \int _a ^b y \sqrt{1+(\frac{dy}{dx})^2} dx

    Your integral is correct except for the  (\frac{dy}{dx})^2 term, just need to recheck the square.
    Last edited by Kasper; March 3rd 2010 at 09:23 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Jan 2010
    Posts
    79
    Ah, I made a typo. so the integral would be:

    2pi\int(\sqrt{x})(\sqrt{(1+(1/4x)))}

    since the derivative of \sqrt(x) is 1/(2\sqrt(x)) is it not?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Mar 2009
    From
    Alberta
    Posts
    173
    Quote Originally Posted by cdlegendary View Post
    Ah, I made a typo. so the integral would be:

    2pi\int(\sqrt{x})(\sqrt{(1+(1/4x)))}

    since the derivative of \sqrt(x) is 1/(2\sqrt(x)) is it not?
    You bet.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Jan 2010
    Posts
    79
    thanks, I'm still having trouble taking the integral though. any pointers?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Senior Member
    Joined
    Dec 2007
    From
    Anchorage, AK
    Posts
    276
    How about that \sqrt{x}\sqrt{1+\frac1{4x}}=\sqrt{x\left(1+\frac1{  4x}\right)}=\sqrt{x+\frac14}=\sqrt{\frac{4x+1}4}=\  frac{\sqrt{4x+1}}2?

    --Kevin C.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: March 9th 2011, 01:52 AM
  2. Replies: 3
    Last Post: November 10th 2010, 02:49 PM
  3. Replies: 2
    Last Post: April 19th 2010, 02:54 PM
  4. Replies: 0
    Last Post: December 6th 2009, 12:21 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