Calculate the volume of the set M

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In summary, the conversation is centered around calculating the volume of a set $M$ defined by $x^2+y^2-z^2\leq 1$ and $0 \leq z\leq 3$. The speakers discuss using spherical and cylindrical coordinates to determine the boundaries of the integrals needed for the calculation. They also mention using different programs, such as Wolfram and Geogebra, to plot the set in 3D. There is some confusion about the correct input for plotting, but in the end, it is suggested to use $x^2+y^2-z^2=1$ instead of $x^2+y^2-z^2\leq 1$.
  • #1
mathmari
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Hey! :giggle:

We have the set $M=\{(x,y,z)\in \mathbb{R}^3 : x^2+y^2-z^2\leq 1, \ 0 \leq z\leq 3\}$. Draw $M$ and calculate the volume of $M$. I have done the following :

\begin{equation*}\int_M\, dV=\int\int\int\, dx\, dy\, dy\end{equation*} Which are the boundaries of the integrals? Do we have to use spherical coordinates?

Or do we set $x=r\cos\theta$ and $y=r\sin\theta$ and $z$ remains $z$ with $0\leq z\leq 3$ ?
We have that $x^2+y^2-z^2\leq \Rightarrow r^2-z^2\leq 1 \Rightarrow r^2\leq 1+z^2 \Rightarrow -\sqrt{1+z^2}\leq r\leq \sqrt{1+z^2}$ and since $r$ is the radius and so it must be positive we get $0 \leq r\leq \sqrt{1+z^2}$, right? But what about $\theta$ ?

Or is there an other way to calculate that?

:unsure:
 
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  • #2
Hey mathmari!

It's called cilindrical coordinates.
$\theta$ goes all the way around and is between 0 and $2\pi$. 🤔
 
  • #3
Klaas van Aarsen said:
It's called cilindrical coordinates.
$\theta$ goes all the way around and is between 0 and $2\pi$. 🤔

So the boundaries are $0 \leq r\leq \sqrt{1+z^2}$, $0\leq z\leq 3$ and $0\leq \theta\leq 2\pi$, right?

So do we have so far the following?
\begin{align*}\int_M\, dV=\int_0^3\int_0^{\sqrt{1+z^2}}\int_0^{2\pi}\, d\theta \, dr\, dz=\int_0^3\int_0^{\sqrt{1+z^2}}\left [\theta\right ]_0^{2\pi} \, dr\, dz=\int_0^3\int_0^{\sqrt{1+z^2}}2\pi \, dr\, dz=\int_0^32\pi \left [r\right ]_0^{\sqrt{1+z^2}}\, dz=\int_0^32\pi\sqrt{1+z^2} \, dz\end{align*}

:unsure:

How do we know that $\theta$ goes all the way around? Is it because we don't have any restriction? :unsure:
 
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  • #4
mathmari said:
So the boundaries are $0 \leq r\leq \sqrt{1+z^2}$, $0\leq z\leq 3$ and $0\leq \theta\leq 2\pi$, right?

So do we have so far the following?
It should be
\begin{align*}\int_M\, dV=\int_0^3\int_0^{\sqrt{1+z^2}}\int_0^{2\pi}\, r\, d\theta \, dr\, dz\end{align*}
An infinitesimal volume element in cylindrical coordinates is $r\, d\theta \, dr\, dz$.
Or put otherwise, the absolute value of the Jacobian determinant is $r$. 🧐

mathmari said:
How do we know that $\theta$ goes all the way around? Is it because we don't have any restriction?
Indeed.
We should be able to see it in a drawing. 🤔
 
  • #5
Klaas van Aarsen said:
We should be able to see it in a drawing. 🤔

With which programm can I draw that? :unsure:

I tried with Geogebra, but it doesn't work, maybe I write the input in a wrong way.

1630741334781.png


:unsure:
 
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  • #6
mathmari said:
With which programm can I draw that? :unsure:

I tried with Geogebra, but it doesn't work, maybe I write the input in a wrong way.

Wolfram gives me:
1630758717250.png
In Geogebra 3D, I could make:

1630758517380.png


I used Tools / Intersect to intersect the hyperboloid and the planes.
 

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  • #7
Klaas van Aarsen said:
Wolfram gives me:
View attachment 11317In Geogebra 3D, I could make:

View attachment 11318

I used Tools / Intersect to intersect the hyperboloid and the planes.

Ah to plot that do we use $x^2+y^2-z^2=1$ instead of $x^2+y^2-z^2\leq 1$ ? :unsure:
 
  • #8
mathmari said:
Ah to plot that do we use $x^2+y^2-z^2=1$ instead of $x^2+y^2-z^2\leq 1$ ?
I guess so. Geogebra doesn't seem to like the $\le$. :unsure:
 

Related to Calculate the volume of the set M

1. How do you calculate the volume of a set M?

The volume of a set M can be calculated by finding the total amount of space occupied by the set. This can be done by using various mathematical techniques such as integration or by using specific formulas depending on the shape and dimensions of the set.

2. What is the importance of calculating the volume of a set M?

Calculating the volume of a set M is important because it helps in determining the amount of space that is occupied by the set. This information can be useful in various fields such as engineering, architecture, and physics, where precise measurements of volume are necessary for designing and constructing structures or conducting experiments.

3. What are the units of measurement used for calculating the volume of a set M?

The units of measurement used for calculating the volume of a set M depend on the dimensions of the set. For example, if the set is a cube, the volume would be measured in cubic units such as cubic meters or cubic feet. If the set is a cylinder, the volume would be measured in units such as liters or gallons.

4. Can the volume of a set M be negative?

No, the volume of a set M cannot be negative. Volume is a physical quantity that represents the amount of space occupied by an object or a set, and it cannot have a negative value. If the calculation results in a negative value, it means there was an error in the calculation or the set's dimensions were not measured correctly.

5. How can the volume of a set M be used in real-life applications?

The volume of a set M has various real-life applications, such as in construction, where it is used to determine the amount of materials needed for a project. It is also used in manufacturing to calculate the capacity of containers or tanks. In addition, the volume of a set M is used in science and research to analyze and understand the properties of different objects and substances.

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