Results 1 to 9 of 9
Like Tree1Thanks
  • 1 Post By Reckoner

Math Help - Calculate the mass M of the truncated cone

  1. #1
    Newbie
    Joined
    Jul 2012
    From
    germany
    Posts
    5

    Exclamation Calculate the mass M of the truncated cone

    S={(x ,y ,z) ЄR : x + y ≤ (2-z) , 0 ≤ z ≤ 1}

    with variable density(x, y, z)= -ln(z)/√x + y


    I have no clue how to do this...please help...
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    16,206
    Thanks
    1789

    Re: Calculate the mass M of the truncated cone

    Surely you know that the mass of an object is its density integrated over its volume.
    Your volume is bounded by the cone x^2+ y^2= (2- z)^2, below by z= 0 and above by z= 1. The only thing peculiar about this is that the vertex of the cone is at z= 2. Since z<1 2- z is below 2 and we can use z= 2- \sqrt{x^2+ y^2}. Projecting onto the xy-plane, we get circle with center at (0,0) and radii 1 to 2.

    In x, y, z coordinates, we would have to integrate over the 'washer' between the circle with radius 1 and the circle with radius 2. That can be done but would require four separate integrals (from x= -2 to x= -1, from x= -1 to 1 with y above the smaller circle, from x= -1 to 1 with y below the smaller circle, from x= 1 to 2). It is much easier to use cylindrical coordinates. In that case, r goes from 1 to 2, [itex]\theta[/itex] goes from 0 to 2\pi, and, for each r and [itex]\theta[/itex], z goes from the plane, z= 1 up to the cone, [tex]z= 2-\sqrt{x^2+ y^2}= 2- \sqrt{r^2}= 2- r.

    In cylindrical coordinates the density function is \frac{-ln(z)}{\sqrt{x^2+ y^2}}= \frac{-ln(z)}{\sqrt{r^2}}= \frac{-ln(z)}{r}.

    That is your integral is \int_{\theta= 0}^{2\pi}\int_{r= 1}^2\int_{z=1}^{2- r} \frac{-ln(z)}{r}(r drd\theta dz)= -2\pi\int_{r= 1}^{2}\int_{z=1}^{2-r} ln(z)dzdr.
    Last edited by HallsofIvy; July 17th 2012 at 07:45 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor Reckoner's Avatar
    Joined
    May 2008
    From
    Baltimore, MD (USA)
    Posts
    1,024
    Thanks
    75
    Awards
    1

    Re: Calculate the mass M of the truncated cone

    Quote Originally Posted by angel050382 View Post
    S={(x ,y ,z) ЄR : x + y ≤ (2-z) , 0 ≤ z ≤ 1}

    with variable density(x, y, z)= -ln(z)/√x + y
    Cylindrical coordinates would simplify things a good deal. The boundary of the cone, in cylindrical form, is

    r^2 = (2-z)^2

    \Rightarrow r = |2-z| = (2 - z) (since 0\leq z\leq1 and we want r\geq0)

    So we have

    0\leq r\leq 2-z

    0\leq\theta\leq2\pi

    and

    0\leq z\leq 1.

    Now, we integrate the density function over this region to find the mass:

    m = \iiint\limits_S\frac{-\ln z}{\sqrt{x^2+y^2}}\,dV

    =\int_0^1\int_0^{2\pi}\int_0^{2-z}\frac{-r\ln z}r\,dr\,d\theta\,dz

    =-\int_0^1\int_0^{2\pi}\int_0^{2-z}\ln z\,dr\,d\theta\,dz
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Reckoner's Avatar
    Joined
    May 2008
    From
    Baltimore, MD (USA)
    Posts
    1,024
    Thanks
    75
    Awards
    1

    Re: Calculate the mass M of the truncated cone

    Quote Originally Posted by HallsofIvy View Post
    That is your integral is \int_{\theta= 0}^{2\pi}\int_{r= 1}^2\int_{z=1}^{2- r} \frac{-ln(z)}{r}(r drd\theta dz)= -2\pi\int_{r= 1}^{2}\int_{z=1}^{2-r} ln(z)dzdr.
    Not sure that I agree with your lower bound for z. And if I'm not mistaken, I believe your method would neglect the mass in the center of the frustum (within r\leq1).
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Apr 2005
    Posts
    16,206
    Thanks
    1789

    Re: Calculate the mass M of the truncated cone

    Good point. I think I lost track in the middle and started thinking I was only doing the outside! Thanks for the correction.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor Reckoner's Avatar
    Joined
    May 2008
    From
    Baltimore, MD (USA)
    Posts
    1,024
    Thanks
    75
    Awards
    1

    Re: Calculate the mass M of the truncated cone

    Quote Originally Posted by HallsofIvy View Post
    Good point. I think I lost track in the middle and started thinking I was only doing the outside! Thanks for the correction.
    I actually very nearly made the same mistake myself, but I caught it before I posted. It's a tricky one.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Jul 2012
    From
    germany
    Posts
    5

    Re: Calculate the mass M of the truncated cone

    Thank you guys so much!!!! I really appreciate your help. I have another question!

    How should I go on now? Cuz if I start to integrate I have problem at ln(0) what is not defined! How should I handle this now!

    I know I have a lot of stupid questions, but I really try to pass that math class...:-(
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor
    Joined
    Oct 2008
    Posts
    1,035
    Thanks
    49

    Re: Calculate the mass M of the truncated cone

    Just in case a picture helps work the integral from the inside out...



    ... where (key in spoiler) ...

    Spoiler:
    Straight lines differentiate downwards with respect to the indicated variable.



    ... is the product rule.



    ... is lazy integration by parts, doing without u and v.



    __________________________________________________ __________

    Don't integrate - balloontegrate!

    Balloon Calculus; standard integrals, derivatives and methods

    Balloon Calculus Drawing with LaTeX and Asymptote!
    Last edited by tom@ballooncalculus; July 17th 2012 at 11:21 AM.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor Reckoner's Avatar
    Joined
    May 2008
    From
    Baltimore, MD (USA)
    Posts
    1,024
    Thanks
    75
    Awards
    1

    Re: Calculate the mass M of the truncated cone

    Quote Originally Posted by angel050382 View Post
    How should I go on now? Cuz if I start to integrate I have problem at ln(0) what is not defined! How should I handle this now!
    Right, I didn't notice that. This will actually be an improper integral:

    =-\int_0^1\int_0^{2\pi}\int_0^{2-z}\ln z\,dr\,d\theta\,dz

    =-2\pi\int_0^1(2-z)\ln z\,dz

    =-2\pi\lim_{a\to0^+}\int_a^1(2-z)\ln z\,dz

    =-2\pi\lim_{a\to0^+}\left[\left(2z-\frac12z^2\right)^2\ln z+\frac14z^2-2z\right]_a^1 (See tom's post)

    =-2\pi\lim_{a\to0^+}\left[0 + \frac14 - 2 - \left(2a-\frac12a^2\right)\ln a - \frac14a^2 + 2a\right]

    =-2\pi\lim_{a\to0^+}\left[-\frac74+\left(2a-\frac12a^2\right)\ln a - \frac14a^2 + 2a\right]

    =-2\pi\left(-\frac74\right) - 2\pi\lim_{a\to0^+}\left(2a-\frac12a^2\right)\ln a

    =\frac{7\pi}2-\pi\lim_{a\to0^+}\left(4a-a^2\right)\ln a

    =\frac{7\pi}2-\pi\lim_{a\to0^+}\frac{\ln a}{1/\left(4a-a^2\right)}

    This limit produces the indeterminant form \textstyle-\frac\infty\infty, so we can apply L'Hpital's rule,

    =\frac{7\pi}2-\pi\lim_{a\to0^+}\frac{1/a}{-(4-2a)/\left(4a-a^2\right)^2}

    =\frac{7\pi}2+\pi\lim_{a\to0^+}\frac{\left(4a-a^2\right)^2}{a(4-2a)}

    =\frac{7\pi}2+\pi\lim_{a\to0^+}\frac{a\left(4-a\right)^2}{4-2a}

    =\frac{7\pi}2.
    Last edited by Reckoner; July 17th 2012 at 11:53 AM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Mass of Ice Cream Cone using Triple Integrals
    Posted in the Calculus Forum
    Replies: 2
    Last Post: March 23rd 2012, 07:34 AM
  2. Particle of mass, calculate change in K.E.?
    Posted in the Math Topics Forum
    Replies: 1
    Last Post: September 6th 2011, 05:46 PM
  3. calculate mass of the wire
    Posted in the Calculus Forum
    Replies: 3
    Last Post: December 6th 2009, 10:39 PM
  4. Truncated and Non-truncated distribution
    Posted in the Advanced Statistics Forum
    Replies: 7
    Last Post: July 27th 2009, 07:40 AM
  5. center of mass of a cone
    Posted in the Calculus Forum
    Replies: 1
    Last Post: June 18th 2008, 02:04 PM

Search Tags


/mathhelpforum @mathhelpforum