How to calculate the volume given by $(x^2+y^2+z^2)^2 = x^2+y^2$

Question

I am working on the following exercise:

Calculate the volume of the body bounded by the following surface:

$$(x^2+y^2+z^2)^2 = x^2+y^2$$

I would solve this with a triple integral, but I do not see how I can derive the boundaries from $(x^2+y^2+z^2)^2 = x^2+y^2$. Could you help me?

Answer 1

Graphed above is the cross section of the volume in a vertical plane using the cylindrical coordinate $r^2 = x^2+y^2$. Note that

$$(r^2+z^2)^2 = r^2\implies z^2=r-r^2$$

And, $r$ has the lower and upper limits $r=0$ and $r = 1$, respectively, when $z=0$.

Thus, express $z$ in terms of $r$ and integrate the volume as follows,

$$z_1 = -\sqrt{r-r^2},\>\>\>\>\>\>\>z_2 = \sqrt{r-r^2}$$

$$V = \int_0^{2\pi}\int_0^1 (z_2-z_1) rdrd\theta= 4\pi\int_0^1 \sqrt{r-r^2} rdr=\frac{\pi^2}4$$

Answer 2

HINT

Consider the transformation $x = \rho\cos(\theta)$, $y = \rho\sin(\theta)$ and $z = z$. Then you have

\begin{align*} (x^{2} + y^{2} + z^{2})^{2} = x^{2} + y^{2} & \Longleftrightarrow z^{2} = \pm\sqrt{x^{2} + y^{2}} - x^{2} - y^{2}\\ & \Longleftrightarrow z^{2} = \pm\rho - \rho^{2} \Longleftrightarrow z^{2} = \rho - \rho^{2} \end{align*}

since $z^{2} \geq 0$.