Volume of Intersection of Surfaces

Question

I hope you can help me with this:

What's the volume which is enclosed by the equation $(x^2+y^2+z^2)^2=z(x^2+y^2)$?

Whenever I try to calculate the intersection of those surfaces I get lost because I arrive to a 4-degree equation.

Thanks!

Answer 1

First of all, let us define the inside of the volume by inequation:

$$\tag{0}(x^2+y^2+z^2)^2 \ \ < \ \ z(x^2+y^2)$$

Let us work on the slice situated at level $z$:

We are going to establish two results:

• (a) All the volume is in between $z=0$ and $z=\dfrac{1}{4}$ ;

• (b) The horizontal slice at height $z$ is a origin-centered annulus, with smaller circle radius $\sqrt{N_1}$ and larger circle radius $\sqrt{N_2}$ where:

$$\tag{1}\begin{cases}N_1:=\frac{z}{2}(1-2z-\sqrt{1-4z}) \ \ \ \ (a)\\N_2:=\frac{z}{2}(1-2z+\sqrt{1-4z}) \ \ \ \ (b)\end{cases}$$

A plot of $\sqrt{N_1}, \sqrt{N_2}$ as functions of $z$ is given below.

Explanation: Let $N=x^2+y^2$.

Relationship $(0)$ can be written:

$$\tag{2}(N+z^2)^2-z N<0 \ \iff \ N^2-N (2z^2-z)+z^4<0.$$

Considered as a quadratic inequation with respect to variable $N$, we can solve the corresponding equation with roots $N_1, N_2$ given in $(1)$.

Quadratic inequation $(1)$ will hold (be true) iff $N$ is between the roots:

$$0\leq N_1<N<N_2.$$

(explaining point (b)). Due to the domain of validity of $\sqrt{1-4z}$, we must have $z<1/4$, justifying point (a).

Now, how do we compute the area of the slice at height $z$? Without integration! Because it is the difference of two disks' areas, i.e.,

$$\tag{3}A(z)=\pi \sqrt{N_2}^2 - \pi \sqrt{N_1}^2=\pi(N_2-N_1)=\pi z \sqrt{1-4z} $$

Last step: the volume is the integral of $A(z)$:

$$\tag{4}\displaystyle V=\int_{z=0}^{1/4} \pi z \sqrt{1-4z}dz=\dfrac{\pi}{8}\int_{u=0}^1 u^2 (1-u^2) du$$

(due to change of variable $u:=\sqrt{1-4z}$), we obtain finally:

$$\tag{5}V=\frac{\pi}{60}$$

In the following representation of $\sqrt{N_1}$ and $\sqrt{N_2}$, parameter $z$ is the abscissa. The surface is a surface or revolution obtained by rotating the given profile around the horizontal axis:

Answer 2

I think the simplest method is to use spherical coordinates $$ x=r \cos\varphi \sin\theta ,\quad y=r \sin\varphi \sin\theta ,\quad z=r \cos\theta . $$ Then the equation $(x^2+y^2+z^2)^2=z(x^2+y^2)$ becomes $r^4 = r \cos\theta \cdot r^2 \sin^2\theta$, which gives $r=0$ or $$ r=\cos\theta \sin^2 \theta . $$ (This is the polar equation of the curve seen in JeanMarie's answer, with $r$ being the distance from the origin, and $\theta$ the angle from the positive $z$ axis.)

Since it's clear from the original equation that $z$ can't be negative, the angle $\theta$ ranges from $0$ to $\pi/2$.

So if $D$ is the body in question, and $E$ the corresponding region in $(r,\theta,\varphi)$ space, the volume is $$ \iiint_D dxdydz = \iiint_E r^2 \sin\theta \, dr d\theta d\varphi = \int_{\varphi=0}^{2\pi} \left( \int_{\theta=0}^{\pi/2} \left( \int_{r=0}^{\cos\theta \sin^2 \theta} r^2 \sin\theta \, dr \right) d\theta \right) d\varphi , $$ which becomes $\pi/60$ after a few more lines of computation.