Results 1 to 2 of 2

Thread: Integration Application question

  1. #1
    Newbie br1tt204's Avatar
    Joined
    Nov 2009
    Posts
    3

    Integration Application question

    This may be hard to visualize without a graph, so I apologize in advance, but here is my question:

    There is a circle with radius 'b' that is 'a' units away from the y-axis. if the circle is rotated about the y-axis, how do you find the volume of the doughnut that is formed? (assume that b<a)

    I would even appreciate a jump start! Anything, really! Thank you in advanced!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Jan 2009
    Posts
    715
    Quote Originally Posted by br1tt204 View Post
    This may be hard to visualize without a graph, so I apologize in advance, but here is my question:

    There is a circle with radius 'b' that is 'a' units away from the y-axis. if the circle is rotated about the y-axis, how do you find the volume of the doughnut that is formed? (assume that b<a)

    I would even appreciate a jump start! Anything, really! Thank you in advanced!

    For convinent calculation , I locate the center of the circle at

    $\displaystyle (a,0) $ and thus the equation of the circle is

    $\displaystyle (x-a)^2 + y^2 = b^2 $

    Now , let's go to the integration ,

    We know the volume of the revolution about the y-axis is

    $\displaystyle \pi \int_p^q [ f(y)^2 - g(y)^2 ] ~dy$ , $\displaystyle | f(y) |\geq |g(y) | , y \in [p,q] $

    since $\displaystyle f(y) = $ the distance from the y -axis$\displaystyle (x)$

    we should find x in terms of y , that is

    $\displaystyle (x-a)^2 + y^2 = b^2 \implies x = a + \sqrt{b^2- y^2} $

    and also $\displaystyle x = a - \sqrt{b^2- y^2}$

    But we know $\displaystyle a + \sqrt{b^2- y^2} \geq a - \sqrt{b^2- y^2}$

    For the interval , i choose $\displaystyle (-b,b) $

    Now , we are able to see what the integral looks like !

    It should be

    $\displaystyle \pi \int_{-b}^b \left ( [ a + \sqrt{b^2- y^2} ]^2 - [ a - \sqrt{b^2- y^2}]^2 \right ) ~dy $

    $\displaystyle = \pi \int_{-b}^b 4a\sqrt{b^2-y^2}~dy$

    $\displaystyle = 4a\pi \int_{-b}^b \sqrt{b^2 - y^2} ~dy $

    $\displaystyle = 4a \pi \times ( ~area~of ~a~semicircle ~with~radius~b ~)$

    $\displaystyle = 4a \pi \cdot (\frac{\pi b^2 }{2} ) = 2ab^2\pi^2$
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Application of Integration
    Posted in the Calculus Forum
    Replies: 4
    Last Post: Jul 27th 2011, 06:55 AM
  2. application of integration
    Posted in the Calculus Forum
    Replies: 4
    Last Post: Feb 17th 2010, 10:51 AM
  3. Application of Integration-Mean
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Jul 2nd 2009, 01:46 AM
  4. Help in Application of integration.
    Posted in the Calculus Forum
    Replies: 0
    Last Post: May 17th 2009, 02:55 PM
  5. integration of application
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Mar 25th 2008, 11:50 AM

Search Tags


/mathhelpforum @mathhelpforum