Region of Cone in Cylindrical Coordinates: Wondering

[mathmari]

Hey!

Which region does the set $D=\{(x, y, z) \mid \sqrt{x^2+y^2}\leq z\leq 1\}$ represent?

The surface $z= \sqrt{x^2+y^2}$ is a cone, or not? But why when we want to calculate an integral over $D$ we use cylindrical coordinates? (Wondering)

 

[I like Serena]

Hey!

Which region does the set $D=\{(x, y, z) \mid \sqrt{x^2+y^2}\leq z\leq 1\}$ represent?

The surface $z= \sqrt{x^2+y^2}$ is a cone, or not? But why when we want to calculate an integral over $D$ we use cylindrical coordinates? (Wondering)

Hey mathmari!

Yep. It's a cone up to z=1. (Nod)

It has rotational symmetry around the z-axis, but not around any other axis.
That makes it a prime candidate for cylindrical coordinates, and not for spherical coordinates.
More specifically, if we substitute $\rho=\sqrt{x^2+y^2}$, we have one variable less to deal with, which is only the case in cylindrical coordinates. (Nerd)

 

[mathmari]

It has rotational symmetry around the z-axis, but not around any other axis.

What do you mean? I got stuck right now. (Wondering)

 

[I like Serena]

What do you mean? I got stuck right now. (Wondering)

Isn't a cone round?
Doesn't it map to itself when we rotate it around the z-axis?

 

[mathmari]

Isn't a cone round?
Doesn't it map to itself when we rotate it around the z-axis?

Yes.

We have the following:

View attachment 7420

 

[I like Serena]

Yes.

We have the following:

Yep. (Nod)

It means that when we integrate over $D$, the integrals will generally be easier.
We should probably look at an example to see why that is exactly. (Thinking)

 

[mathmari]

Yep. (Nod)

It means that when we integrate over $D$, the integrals will generally be easier.

So, to use the cylindrical coordinates it is not necessary to have a cylinder as the region $D$ ? (Wondering)

 

[I like Serena]

So, to use the spherical coordinates it is not necessary to have a cylinder as the region $D$ ? (Wondering)

Indeed.
Just like we don't need a cuboid to integrate in cartesian coordinates. (Thinking)

 

[mathmari]

Indeed.
Just like we don't need a cuboid to integrate in cartesian coordinates. (Thinking)

Ah ok. Do we use the cylindrical coordinates when we have an inequality for $x^2+y^2$ and the spherical coordinates when we have an inequality for $x^2+y^2+z^2$ ? (Wondering)

 

[I like Serena]

Ah ok. Do we use the cylindrical coordinates when we have an inequality for $x^2+y^2$ and the spherical coordinates when we have an inequality for $x^2+y^2+z^2$ ? (Wondering)

Yes.
It usually makes the integration a lot easier. (Mmm)

 

[mathmari]

Yes.
It usually makes the integration a lot easier. (Mmm)

Ah ok! Thank you so much! (Yes)