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Math Help - Using cylindrical coordinates to solve triple integrals

  1. #1
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    Question Using cylindrical coordinates to solve triple integrals

    Hi,

    I'm trying to solve a triple integrals problem. I need to find the volume of the solid delimited by the following functions:

    z=\sqrt{x^2+y^2}\rightarrow z=r
    z=1-2 \sqrt{x^2+y^2}\rightarrow z=1-2r

    Can I do that?
    r=1-2r so that I find out r=z=\frac{1}{3}

    So that I can pose the following triple integral to find the volume of the solid delimited by the 2 functions:
    \int^{2\pi}_{0} \int^{\frac{1}{3}}_{0} \int^{1-2r}_{r}r^2 \mathrm{d}z \mathrm{d}r \mathrm{d}\theta = \frac{\pi }{162}

    Thank you so much for checking out,

    Bazinga
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  2. #2
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    Hi,

    The solid z=r is a cone starting from the origin (0;0;0) and going up with a slope of 1

    The solid z=1-2r is a cone starting from the point (0;0;1) and going down with a slope of 2

    The 2 cones intersect at a circle center (0;0;1/3), radius 1/3 and parallel to the xOy plane

    Can you see all of this ?
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  3. #3
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    Follow-up

    Thanks for taking the time to answer.

    Yes, I see the circle on the Oxy plan and know that z=r is indeed the bottom one as the other is on top. Am I posing the integral the right way? I thought of it as a circle (0 to 2pi) of 0 to 1/3 as radius delimited by both cones from bottom to top in z.

    Regards,

    Bazinga
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  4. #4
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    The "elementary" volume is r \mathrm{d}\theta \mathrm{d}z \mathrm{d}r and not r^2 \mathrm{d}\theta \mathrm{d}z \mathrm{d}r

    The volume of the first cone (at the bottom) is
    \int^{2\pi}_{0} \int^{\frac{1}{3}}_{0} \int^{z}_{0}r \mathrm{d}\theta \mathrm{d}z \mathrm{d}r

    Sorry if my English vocabualry is not the appropriated one ...
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  5. #5
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    Je suis confus. Comment dois-je poser l'intégrale qui résout le problème? Je croyais que le volume du cône était r et l'élément de volume r dr dz d theta. Vous posez une intégrale de 0 à z en d theta dont je ne saisis pas le sens. Merci beaucoup de vos réponses constructives, elles m'aident à comprendre le cheminement de résolution de tels problèmes.

    I am confused. Would you please pose the correct integral so that I understand my mistake. Why is there a 0 to z integral. Thanks for your previous answers, they were much helpful.
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  6. #6
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    Je pense qu'on ne peut pas continuer en Français sur ce forum ou alors éventuellement en MP

    In cylindrical coordinates the elementary volume is r dr d\theta dz
    The volume of the cone is \int \int \int r dr d\theta dz

    Concerning the bounds, \theta varies from 0 to 2 \pi, z varies from 0 to 1/3 and r varies from 0 to a bound which is z-dependent. If you consider a given value for z, you can see that r varies from 0 to z (I am talking about the lower cone)

    Therefore V = \int_{\theta=0}^{\theta=2\pi} \int_{z=0}^{z=\frac{1}{3}} \int_{r=0}^{r=z}r dr d\theta dz
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