Volume of $K = \{(x_1,x_2,x_3) \in \mathbb{R}^3 \mid x_1^2 + x_2^2 \leq x_3 \wedge x_1^2 + x_2^2 + x_3^2 \leq 1 \}$

Question

I wrote an exam and didn't really know how to answer this question which gave $13$ points.

Let

$$K = \{(x_1,x_2,x_3) \in \mathbb{R}^3 \mid x_1^2 + x_2^2 \leq x_3 \wedge x_1^2 + x_2^2 + x_3^2 \leq 1 \}$$

Determine the volume $V$ of the set $K$ (till including the bounds of integration).

I only knew that

$V = \int_K dx$

But I have no idea how one needs to calculate this and I don't know how to write this in Wolfram Alpha either.

It would be awesome if someone could show how it's done because our professor doesn't provide solutions.

Answer 1

The set $K$ is the region in $\Bbb R^3$ below the paraboloid ${x_1}^2+{x_2}^2=x_3$ and inside the sphere ${x_1}^2+{x_2}^2+{x_3}^2=1$.

You can find the volume of $K$ by subtracting the volume of the region above the paraboloid and within the sphere from the volume of the sphere. (Call this region $K'$.) The sphere has radius $1$ so its volume is $\frac{4\pi}3$.

The volume of $K'$ can be computed in cylindrical coordinates:

$$\iiint_{K'}\mathrm dV=\int_0^{2\pi}\int_0^{\sqrt{(-1+\sqrt5)/2}}\int_{r^2}^{\sqrt{1-r^2}}r\,\mathrm dz\,\mathrm dr\,\mathrm d\theta$$

where the upper limit for the integral with respect to $r$ comes from the intersection of the sphere and paraboloid,

$$x^2+y^2=z\implies x^2+y^2+(x^2+y^2)^2=1$$

(Substitute $x^2+y^2=r^2$ and this reduces to the quartic $r^4+r^2=1$; the upper limit is one solution.)

Answer 2

With the cylindrical coordinates, the surfaces are $z=r^2$ and $r^2+z^2=1$. Then, the enclosed volume is

$$V=2\pi \int_0^a \left(\sqrt{1-r^2}-r^2 \right) rdr=\frac{5\pi}6a^4$$

where the radial upper limit is the intersection of the two surfaces, given by $a^2+a^4=1$, or $a^2={\frac{\sqrt5-1}2}$. The integral results in

$$V = \frac{5\pi}{12}(3-\sqrt5)$$