Results 1 to 2 of 2

Thread: Volume of Rotation of a Quarter Circle

  1. #1
    Junior Member
    Joined
    Jan 2010
    Posts
    36

    Volume of Rotation of a Quarter Circle

    I am trying to find the volume of a quarter circle that is rotated about $\displaystyle x=1$. So the equation should look like this:
    $\displaystyle
    \int_0^{\frac {\pi} {2}} \pi(\sqrt {1^2-y^2})^2dy$

    This should simplify to:
    $\displaystyle \int_0^{\frac {\pi} {2}} \pi(1-y^2)dy$

    If I am doing this right thus far, then the integral is:
    $\displaystyle \pi(y-\frac {y^3} {3}) $ from 0 to $\displaystyle \frac {\pi} {2}$

    And when I plug these values in I some how end up with $\displaystyle \frac {\pi^2} {2} - \frac {\pi^4} {24}$

    I feel like I am missing something in between here, or maybe just made an error plugging in the values, although I have tried it a few times and end up with the same answer. Is this right? Thanks for the help.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570
    Thanks
    1
    Hello Spudwad
    Quote Originally Posted by Spudwad View Post
    I am trying to find the volume of a quarter circle that is rotated about $\displaystyle x=1$. So the equation should look like this:
    $\displaystyle
    \int_0^{\frac {\pi} {2}} \pi(\sqrt {1^2-y^2})^2dy$

    This should simplify to:
    $\displaystyle \int_0^{\frac {\pi} {2}} \pi(1-y^2)dy$

    If I am doing this right thus far, then the integral is:
    $\displaystyle \pi(y-\frac {y^3} {3}) $ from 0 to $\displaystyle \frac {\pi} {2}$

    And when I plug these values in I some how end up with $\displaystyle \frac {\pi^2} {2} - \frac {\pi^4} {24}$

    I feel like I am missing something in between here, or maybe just made an error plugging in the values, although I have tried it a few times and end up with the same answer. Is this right? Thanks for the help.
    Get ready to kick yourself:
    The limits of the integral are $\displaystyle 0$ to $\displaystyle 1$, not $\displaystyle 0$ to $\displaystyle \pi/2$.

    Grandad
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 5
    Last Post: Dec 21st 2011, 04:02 AM
  2. Replies: 11
    Last Post: Apr 8th 2011, 01:29 PM
  3. Volume by rotation
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Dec 4th 2009, 07:00 AM
  4. Volume by rotation
    Posted in the Calculus Forum
    Replies: 4
    Last Post: Sep 15th 2008, 09:05 AM
  5. volume of rotation
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Mar 18th 2007, 01:16 PM

Search tags for this page

Click on a term to search for related topics.

Search Tags


/mathhelpforum @mathhelpforum