Volume of $M := \left \{ (x,y,z)\in \mathbb{R^3}: 0 \leq z \leq 2 \text{ and } x^2 + y^2 \leq z \right \}$

Question

I want to know how one can calculate the volume of

$$M := \left \{ \begin{pmatrix} x \\ y \\ z \end{pmatrix} \in \mathbb{R^3}: 0 \leq z \leq 2 \text{ and } x^2 + y^2 \leq z \right \} \text { in } \mathbb{R^3}$$

I have seen a similar question here, but I don't quite understand how you get the function here.

The boundaries would be

$$ V = \int _0 ^2$$

I have also looked at this wikipedia article https://en.wikipedia.org/wiki/Disc_integration, and it says $R(y)$ is the distance between the function and the axis of rotation, but what does that mean?

Answer 1

In cylindrical coordinates, the domain is $0 \leq z \leq 2 \text{ and } z\geq r^2$, whose volume can be integrated as $$V =\int_0^2 \pi r^2 (z)\>dz=\pi\int_0^2 z\>dz=2\pi $$ where $\pi r^2(z)$ is the area at $z$.

Answer 2

Direct approach

$$V=\iiint\limits_V dxdydz=\int_0^2\int_{y_{min}(z)}^{y_{max}(z)}\int_{x_{min}(y,z)}^{x_{max}(y,z)} dx\, dy\, dz$$

So we need to express minimum and maximum possible values of $y$ and $x$.

At certain $z$: $x^2+y^2 \leq z \Rightarrow y^2 \leq z-x^2 \leq z$. Then $y_{min}=-\sqrt{z}$, $y_{max}=\sqrt{z}$.

At certain $z$ and $y$: $x^2+y^2\leq z\Rightarrow x^2\leq z-y^2$. Then $x_{min}=-\sqrt{z-y^2}$, $x_{max}=\sqrt{z-y^2}$.

Then

$$V=\int_0^2\int_{-\sqrt{z}}^{\sqrt{z}}\int_{-\sqrt{z-y^2}}^{\sqrt{z-y^2}} dx\, dy\, dz$$

Other approach is

$$V=\iiint\limits_V dxdydz=\int_0^2\iint\limits_{S(z)}dxdy\, dz$$

At certain $z$: $S(z)$ is region $x^2+y^2\leq z$, which is interior of circle with radius $r=\sqrt{z}$, then $$\iint\limits_{S(z)}dxdy=\pi r^2=\pi z$$

$$V=\int_0^2 \pi z\, dz=\left.\frac{\pi z^2}{2}\right|_0^2=2\pi$$