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Math Help - Cone volume-triple integral

  1. #1
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    Post Cone volume-triple integral

    My problem appears in solving volume of body, defined by this equation
    (x^2+y^2+z^2) = 4a(x^2+y^2)

    a=constant
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  2. #2
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    Re: Cone volume-triple integral

    I think, that you mean this equation (x^2+y^2+z^2)^{\bold{2}} = 4a(x^2+y^2).

    We consider the case z>0 and multiply by 2 the integral (since this body is symmetrical about the plane z=0).

    Find the body's proection D_{xy} on the plane xOy~(z=0):

    (x^2+y^2)^2 = 4a(x^2+y^2)~~\Leftrightarrow~~ (x^2+y^2)(x^2+y^2-4a)=0~~\Rightarrow~~x^2+y^2=4a

    So, D_{xy}=\{(x,y)\in\mathbb{R}\colon\, x^2+y^2\leqslant(2\sqrt{a})^2\} (As you can see, this projection is a circle whose radius is equal to 2\sqrt{a}).

    We write a set T, which limits the surface (x^2+y^2+z^2)^{\bold{2}} = 4a(x^2+y^2):

    T=\left\{(x,y,z)\in\mathbb{R}^3\colon\, x^2+y^2\leqslant(2\sqrt{a}),~0 \leqslant z\leqslant \sqrt{2\sqrt a \sqrt{x^2+y^2}- (x^2+y^2)}\right\}

    Convert to the cylindrical coordinates \begin{cases}x = r\cos \varphi,\\ y = r\sin \varphi,\\z = z,\end{cases}|J|=r.

    T^{\ast}= \left\{(r,\varphi,z)\in\mathbb{R}^3\colon\, 0 \leqslant r \leqslant 2\sqrt a,~0 \leqslant \varphi  \leqslant 2\pi ,~0 \leqslant z \leqslant \sqrt{2\sqrt a r - r^2}}\right\}

    So, finally you have

    \begin{aligned}V&= 2\iiint\limits_T dxdydz= 2\iiint\limits_{T^{\ast}}|J|\,drd\varphi dz= 2\int\limits_0^{2\pi}d\varphi \int\limits_0^{2\sqrt a}r\,dr \int\limits_0^{\sqrt{2\sqrt a r -r^2}}dz=\\ &=4\pi\int\limits_0^{2\sqrt a}r\sqrt{2\sqrt a r-r^2}\,dr=\ldots  = 2\pi^2a^{3/2}\end{aligned}
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  3. #3
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    Re: Cone volume-triple integral

    thank you very very much, you solved my problem
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