Cone volume-triple integral

**Vektor** · June 12th 2012, 02:35 PM

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

**DeMath** · June 12th 2012, 06:43 PM

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

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

**Vektor** · June 13th 2012, 06:04 AM

thank you very very much, you solved my problem

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