Results 1 to 4 of 4

Thread: Volume

  1. #1
    Junior Member
    Joined
    Feb 2008
    Posts
    49

    Volume

    What is the volume of the solid generated by revolving the region enclosed by (x-2)^2 + y^2 = 1 about the y-axis?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    12,028
    Thanks
    848
    Hello, chaddy!

    What is the volume of the solid generated by revolving the region
    enclosed by $\displaystyle (x-2)^2 + y^2 \:= \:1$ about the y-axis?

    Make a sketch. We have a circle with center (2,0) and radius 1
    . . revolved about the y-axis.
    This results in a torus (donut).

    I recommend using Shells: .$\displaystyle V \;=\;2\pi\int^b_a\text{(radius)(height)}\,dx$

    The radius is: $\displaystyle x$

    The height is: .$\displaystyle 2y \;=\;2\sqrt{1-(x-2)^2} $

    We must evaluate: .$\displaystyle V \;=\;4\pi\int^3_1 x\sqrt{1-(x-2)^2}\,dx$

    Good luck!

    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Feb 2008
    Posts
    49
    I worked that integral out and wound up getting zero. That doesn't seem right. Is it?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,656
    Thanks
    14
    Quote Originally Posted by Soroban View Post
    We must evaluate: .$\displaystyle V \;=\;4\pi\int^3_1 x\sqrt{1-(x-2)^2}\,dx$
    Substitute $\displaystyle u=x-2,$

    $\displaystyle \int_1^3 {x\sqrt {1 - (x - 2)^2 } \,dx} = \int_{ - 1}^1 {(u + 2)\sqrt {1 - u^2 } \,du} .$

    We can kill this quickly: $\displaystyle \int_{ - 1}^1 {u\sqrt {1 - u^2 } \,du} = 0$ by symmetry and $\displaystyle \int_{ - 1}^1 {\sqrt {1 - u^2 } \,du}$ which is a half of circle with radius 1, hence area $\displaystyle \frac\pi2.$ Put these things together yields that $\displaystyle \mathcal V=4\pi^2.$
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: Nov 18th 2011, 01:08 PM
  2. volume flux, mass flux and volume flo
    Posted in the Advanced Applied Math Forum
    Replies: 2
    Last Post: Jun 7th 2011, 06:30 PM
  3. Replies: 1
    Last Post: May 14th 2010, 04:08 PM
  4. divergence = flux / volume is independant of the limiting volume
    Posted in the Differential Geometry Forum
    Replies: 0
    Last Post: Apr 26th 2010, 07:31 PM
  5. Volume
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Nov 9th 2008, 04:13 PM

Search Tags


/mathhelpforum @mathhelpforum