Calculate the volume $λ_3(A)$ of $A := \{ (x, y, z) ∈\mathbb{R}^3 : x^2 ≤ y^2 + z^2 ≤ 1\} $

Question

I have a lot of problems with exercises where I must calculate the Volume of a set using integrals.

Here an example:

Le the set $A$ be $A := \{ (x, y, z) ∈\mathbb{R}^3 : x^2 ≤ y^2 + z^2 ≤ 1\} $. Calculate its volume $λ_3(A)$.

So what I thought is to doing something like this:

$λ_3(A)=λ_3(A_1)-λ_3(A_2)$

Where $A_1 := \{ (x, y, z) ∈\mathbb{R}^3 : y^2 + z^2 ≤ 1\}$ and $A_2 := \{ (x, y, z) ∈\mathbb{R}^3 : x^2 ≤ y^2 + z^2 \}$

But here comes my problem for every exercise like this: I dont know how to find the limits of my triple integral.

Can someone help me?

Answer 1

In cylindrical coordinates with $x$ as the height axis the region becomes $$\{(x,r,\theta)\in\mathbb R^3:x^2\le r^2\le1\}$$ This can be simplified a bit since $r$ is positive: $$\{(x,r,\theta)\in\mathbb R^3:|x|\le r\le1\}$$ The region is a cylinder with radius $1$ and height $2$, minus two right circular cones with base radius $1$ and height $1$. The volume is thus $$\pi\cdot2-2\cdot\frac13\pi\cdot1=\frac43\pi$$ We can check this answer by integrating in the aforementioned cylindrical coordinates: $$\int_0^{2\pi}\int_0^1\int_{-r}^r1\cdot r\,dx\,dr\,d\theta$$ $$=2\pi\int_0^12r^2\,dr=4\pi[r^3/3]_0^1=\frac{4\pi}3$$

Answer 2

$\iiint_Adxdydz=\int_{-1}^1C(x)dx$ where $C(x)=\iint_{x^2\le y^2+z^2\le1}dydz=\pi(1-x^2)$ (area of an annulus).

Hence $$\lambda_3(A)=\pi\int_{-1}^1(1-x^2)dx=\pi\left[x-\frac{x^3}3\right]_{-1}^1=\frac{4\pi}3.$$