Results 1 to 4 of 4

Math Help - Volume for a cone in cylindrical coordinates.

  1. #1
    Member
    Joined
    May 2010
    Posts
    241

    Volume for a cone in cylindrical coordinates.

    Hi there. I haven't used iterated integrals for a while, and I'm studying some mechanics, the inertia tensor, etc. so I need to use some calculus. And I'm having some trouble with it.

    I was trying to find the volume of a cone, and then I've found lots of trouble with such a simple problem.

    So I thought of using cylindrical coordinates this way:
    \begin{Bmatrix}{ x=r\cos\theta} \\y=r\sin\theta \\z=r\end{matrix}

    And then I've stated the integral this way:

    \displaystyle\int_{0}^{2\pi}\displaystyle\int_{0}^  {r}\displaystyle\int_{r}^{h}rdzdrd\theta=\displays  tyle\int_{0}^{2\pi}\displaystyle\int_{0}^{r}r(h-r)drd\theta=\displaystyle\int_{0}^{2\pi}\displayst  yle\frac{r^2h}{2}-\displaystyle\frac{r^3}{3}=\pi r^2h-\displaystyle\frac{2\pi\r^3}{3}=\pi r^2(h-\displaystyle\frac{2}{3}r)

    But I should get: V_{cone}=\displaystyle\frac{\pi r^2 h}{3}

    I think I'm giving wrong limits for the integration.

    Help pls
    Follow Math Help Forum on Facebook and Google+

  2. #2
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    I think your limits are ok. You have a notational error here:

    \displaystyle\int_{0}^{2\pi}\left(\frac{r^{2}h}{2}-\frac{r^{3}}{3}\right)d\theta=\pi r^{2}h-\frac{2\pi r^{3}}{3}.

    But as that gets fixed later on, that won't explain what's going on. I think part of the issue is that you've determined that z=r. There's no a priori reason to assume that. I would probably have done z=hr/a, where a is the radius of the base, to account for differently "squashed" cones. Be that as it may, I think your formula is actually correct, because in your case, since z=r, you have that r=h, and hence

    \displaystyle \pi r^{2}h-\frac{2\pi r^{3}}{3}=\frac{\pi r^{3}}{3}.

    This formula is correct if r=h.

    Try going through the integration again with z=hr/a, and see what that gets you.

    Something else I would try: don't use r both for the variable of integration and for the radius of the base of the cone. It's confusing.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    May 2010
    Posts
    241
    Thanks for your answer Ackbeet. I thought of z=r because I thought of the cone as z=(x^2+y^2)^{1/2}, thats the cone I have integrated, and as you said, that means that r=h, because the cone in the xz plane goes through the line x=z. The relation I was missing I think that its the one that you gave.

    Thank you again.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    You're welcome. Have a good one!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Covert Cartesian coordinates eq. into Cylindrical coordinates
    Posted in the Differential Geometry Forum
    Replies: 0
    Last Post: August 6th 2010, 06:56 AM
  2. Replies: 0
    Last Post: April 25th 2010, 02:06 PM
  3. Replies: 1
    Last Post: November 1st 2009, 01:11 AM
  4. Replies: 2
    Last Post: April 11th 2009, 06:07 AM
  5. Replies: 3
    Last Post: July 24th 2008, 03:39 PM

Search Tags


/mathhelpforum @mathhelpforum