Calculate the volume integral over a cone of height h and radius r

[physicss]

Homework Statement

hello; I have to calculate the volume integral over a cone of height h and radius r located on the peak in the
origin stands of the function z(x^2+y^2)

Relevant Equations

z(x^2+y^2)

• x from 0 to r
• y from 0 to r
• z from 0 to h

∫0h ∫0r ∫0r z(x^2 + y^2) dx dy dz

would that be right?

 

[haruspex]

∫0h ∫0r ∫0r z(x^2 + y^2) dx dy dz

would that be right?

What does a volume running between fixed bounds in the x, y and z directions look like?

 

[physicss]

What does a volume running between fixed bounds in the x, y and z directions look like?

like a rectangular prism

 

[haruspex]

like a rectangular prism

Right, so not a cone.
Are you familiar with cylindrical coordinates?

 

[physicss]

Right, so not a cone.
Are you familiar with cylindrical coordinates?

yes, I more or less know how to convert Cartesian coordinates into cylindrical coordinates

 

[berkeman]

∫0h ∫0r ∫0r z(x^2 + y^2) dx dy dz

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[haruspex]

yes, I more or less know how to convert Cartesian coordinates into cylindrical coordinates

It should be obvious that the bounds will be more convenient in cylindrical coordinates, so try that.

 

[physicss]

It should be obvious that the bounds will be more convenient in cylindrical coordinates, so try that.

could I also use the limits from 0 to r, from 0 to h and from 0 r/h (h - √(x^2 + y^2))? (in cartesian)

and would 0 to r, 0 to 2pi and 0 to z = h - (h/r) * sqrt(x^2 + y^2) in cylindrical coordinates be right for the limits?

thank you in advance

 

[PeroK]

could I also use the limits from 0 to r, from 0 to h and from 0 r/h (h - √(x^2 + y^2))? (in cartesian)

and would 0 to r, 0 to 2pi and 0 to z = h - (h/r) * sqrt(x^2 + y^2) in cylindrical coordinates be right for the limits?

thank you in advance

That looks right, but I would take $z$ from $0$ to $h$ and calculate the radius of the disk at each height as a function of $z$. That seems to me a geometrically more logical approach. A cone is effectively a set of smaller and smaller concentric disks/circles stacked on top of each other.

Or, larger and larger if the apex is at the bottom.

 

[haruspex]

could I also use the limits from 0 to r, from 0 to h and from 0 r/h (h - √(x^2 + y^2))? (in cartesian)

and would 0 to r, 0 to 2pi and 0 to z = h - (h/r) * sqrt(x^2 + y^2) in cylindrical coordinates be right for the limits?

thank you in advance

That's rather unclear. What variables are going between those bounds?
E.g. "r from 0 to r" would make no sense.

You are taking the cone as made of cylindrical elements, but expressing in terms of x and y. That is not using cylindrical coordinates. Cylindrical coordinates means $(r, \theta, z)$, no references to x or y.
And using cylindrical coordinates does not mean you have to use cylindrical elements. As @PeroK notes, it would be more convenient in disc elements.

 

[physicss]

That's rather unclear. What variables are going between those bounds?
E.g. "r from 0 to r" would make no sense.

You are taking the cone as made of cylindrical elements, but expressing in terms of x and y. That is not using cylindrical coordinates. Cylindrical coordinates means $(r, \theta, z)$, no references to x or y.
And using cylindrical coordinates does not mean you have to use cylindrical elements. As @PeroK notes, it would be more convenient in disc elements.

wouldn´t the general limits be (0,h), (0,2pi) and (0, r-rz/h) rdrdφdz?

 

[haruspex]

wouldn´t the general limits be (0,h), (0,2pi) and (0, r-rz/h) rdrdφdz?

r cannot feature in its own bounds. Use different symbols for the base radius and the variable radius.

 

[PhDeezNutz]

Is it me or is this problem more suitable for spherical coordinates?

 

[kuruman]

I think it is you. Try following @haruspex's advice in post #12. There is a constraint that relates cylindrical variable $r$ to cylindrical variable $z$.

 

[PhDeezNutz]

I think it is you. Try following @haruspex's advice in post #12. There is a constraint that relates cylindrical variable $r$ to cylindrical variable $z$.

May I send you my solution in spherical via PM?

 

[kuruman]

May I send you my solution in spherical via PM?

I will be happy to look at it.

 

[physicss]

I will be happy to look at it.

Hello, I used (0,R), (0,2pi), (0,H/R*z) as the limits and I get pi*H^4*r^4/4R^2. is my solution correct? thanks in advance ( the function I used: r^3 z)

 

[kuruman]

Hello, I used (0,R), (0,2pi), (0,H/R*z) as the limits and I get pi*H^4*r^4/4R^2. is my solution correct? thanks in advance ( the function I used: r^3 z)

Your answer is not correct. It should in terms of constant numbers and the fixed quantities $h$ and $R$. Specifically, variable $r$ "integrates out" and does not belong. Please learn to use LaTeX when you post algebraic expressions. It's easy to learn and a useful skill to have.

 

[physicss]

Your answer is not correct. It should in terms of constant numbers and the fixed quantities $h$ and $R$. Specifically, variable $r$ "integrates out" and does not belong. Please learn to use LaTeX when you post algebraic expressions. It's easy to learn and a useful skill to have.

I give up but still thank you for your time and help

 

[kuruman]

I give up but still thank you for your time and help

Please don't give up because you are almost there. Your method is correct. If you post the details of your solution step by step, we can help you reach the correct answer.

 

[PhDeezNutz]

Hello, I used (0,R), (0,2pi), (0,H/R*z) as the limits and I get pi*H^4*r^4/4R^2. is my solution correct? thanks in advance ( the function I used: r^3 z)

You are really really close as @kuruman pointed out.

That H/R is close but not quite right. Think about it; at the very bottom you want to be integrating over a radius of 0. At the very top when z = H what is the radius you want to be integrating over? And after you answer that question how would you change H/R?

 

[physicss]

You are really really close as @kuruman pointed out.

That H/R is close but not quite right. Think about it; at the very bottom you want to be integrating over a radius of 0. At the very top when z = H what is the radius you want to be integrating over? And after you answer that question how would you change H/R?

thank you for your answer. could H(1−r/R) be right ?

 

[PhDeezNutz]

thank you for your answer. could H(1−r/R) be right ?

The entire point of integrating over $r$ and then evaluating it at the bounds is to get rid of $r$. As you do each successive integral you want to get rid of changing variables not keep them so in the end you can get a hard number. As others have said $r$ cannot be featured in its own bounds.

When you are integrating over $r$ you need it as a function of OTHER VARIABLES (not itself)

Let's break it down$r\left(z\right) =$ ?

When $z = 0$ we have $r = 0$

When $z = H$ we have $r=R$

think of a $r$ as a linear function of $z$ and tell me what you get. Your response before was $\frac{H}{R} z$ and that was painfully close. You're like an inch away.

 

[kuruman]

Write $r=\kappa~z$ where $\kappa$ is some constant.
Clearly, when $z =0,$ $r=0$.
What must $\kappa$ be so that $r=R$ when $z=h$?
Substitute the value of $\kappa$ in the equation for $r$.

 

[physicss]

The entire point of integrating over $r$ and then evaluating it at the bounds is to get rid of $r$. As you do each successive integral you want to get rid of changing variables not keep them so in the end you can get a hard number. As others have said $r$ cannot be featured in its own bounds.

When you are integrating over $r$ you need it as a function of OTHER VARIABLES (not itself)

Let's break it down$r\left(z\right) =$ ?

When $z = 0$ we have $r = 0$

When $z = H$ we have $r=R$

think of a $r$ as a linear function of $z$ and tell me what you get. Your response before was $\frac{H}{R} z$ and that was painfully close. You're like an

Write $r=\kappa~z$ where $\kappa$ is some constant.
Clearly, when $z =0,$ $r=0$.
What must $\kappa$ be so that $r=R$ when $z=h$?
Substitute the value of $\kappa$ in the equation for $r$.

R/h

 

[kuruman]

Very good. So what should the limits of the integral over $r$ be? You should be able to finish this now.

 

[kuruman]

thank you for the help. my last question is if (1/3)πr^2h is the right result? If not I swapped the order of the cylindrical coordinates while integrating. again thanks for the help and time

The right result for what? It looks like the volume of a cone of radius $r$ and height $h$. That's not what you have here.

 

[physicss]

The right result for what? It looks like the volume of a cone of radius $r$ and height $h$. That's not what you have here.

I think I found my mistake, thanks for the help

 

[kuruman]

I think I found my mistake, thanks for the help

Maybe you found your mistake maybe not. Posting a wrong answer, without showing how you got it, and asking if it is correct doesn't work very well when you need an extra pair of eyes looking over you shoulder.