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Thread: triple integral on a cone

  1. #1
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    triple integral on a cone

    Hi

    I am trying to solve the following integral, but I am having trouble finding the limits. And first of all sorry for my bad LaTeX.

    $\displaystyle \int{\int{\int{x^2+y^2}}}dV$ over E.
    E is given by $\displaystyle z=x^2+y^2-1$ and $\displaystyle 0<=z<=8$



    My Problem is the lower bound $\displaystyle z=0$. If it was $\displaystyle z=-1$ I would just use polar coordinates $\displaystyle 0<=\theta <=2*Pi$,$\displaystyle 0<=r<=2*\sqrt(2)$ and $\displaystyle -1<=z<=8$.

    Thanks in advance
    Jesper
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  2. #2
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    So, essentially you're integrating your function over the frustrum of an upside-down cone (frustrum in this context = cone with its head chopped off).

    I would still use cylindrical coordinates (I think you meant to say cylindrical, not polar, though I grant you that cylindrical is just polar with an added cartesian coordinate).

    I would split up your integral in two pieces depending on $\displaystyle r$. Integrate on $\displaystyle 0\le r\le 1.$ This would be the cylinder of radius $\displaystyle 1$ from $\displaystyle z=0$ to $\displaystyle z=8$. This corresponds to the "core" of the frustrum. Then, I would integrate on the outer cone part. Here, your $\displaystyle z$ limits would vary from $\displaystyle r^{2}-1$ to $\displaystyle 8$, your $\displaystyle r$ would vary from $\displaystyle 1$ to $\displaystyle 3$, and you should be doing the $\displaystyle z$ integral, in that case, before the $\displaystyle r$ integral.

    Can you put all this together and write down your integrals?
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  3. #3
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    Thank you so much for your fast and very good answer. I think you handle it in a very clever way. Also because both integrals are quite easy to solve. I would call it a paraboloid as we reserve the word cone for expressions of the form: $\displaystyle \frac{z}{c} \neq \sqrt{\frac{x^2}{a}+\frac{y^2}{b}}$.

    However it seems like your answer is correct. As I reed it the integrals you set up are (if my integral is called I):

    $\displaystyle I=\int_0^{2*\Pi}{\int_0^1{\int_0^8{r^2*r}}dz dr d\theta + \int_0^{2*\Pi}{\int_1^3{\int_{r^2-1}^8{r^2*r}}dz dr d\theta$

    Kind regards
    Jesper Bjørnholt
    stud. b.s. Mathematics and Spanish at University of Aarhus, Denmark
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  4. #4
    A Plied Mathematician
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    You're right. It's a paraboloid, not a cone. I was thinking z = r. Your integral looks good to me. As a notational thing, I would use a lower-case pi. The upper-case Pi is reserved for the product notation, usually, though it is pretty clear from the context what you mean.

    Good luck!
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