What does a volume running between fixed bounds in the x, y and z directions look like?physicss said:∫0h ∫0r ∫0r z(x^2 + y^2) dx dy dz
would that be right?
like a rectangular prismharuspex said:What does a volume running between fixed bounds in the x, y and z directions look like?
Right, so not a cone.physicss said:like a rectangular prism
yes, I more or less know how to convert Cartesian coordinates into cylindrical coordinatesharuspex said:Right, so not a cone.
Are you familiar with cylindrical coordinates?
To post math equations, it's best to use the LaTeX engine that PF provides. There is a helpful "LaTeX Guide" link below the Edit window to get you started. Note that you put double-$ delimiters at the start and end of each stand-alone line of LaTeX, and double-# delimiters at the start and end of in-line LaTeX that does not need to be on its own line. Also, if you right-click on a LaTeX equation in a post, you get a pop-up menu to let you view the LaTeX source or view it in other formats.physicss said:∫0h ∫0r ∫0r z(x^2 + y^2) dx dy dz
It should be obvious that the bounds will be more convenient in cylindrical coordinates, so try that.physicss said:yes, I more or less know how to convert Cartesian coordinates into cylindrical coordinates
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)haruspex said:It should be obvious that the bounds will be more convenient in cylindrical coordinates, so try that.
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.physicss said: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?physicss said: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
wouldn´t the general limits be (0,h), (0,2pi) and (0, r-rz/h) rdrdφdz?haruspex said: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.
r cannot feature in its own bounds. Use different symbols for the base radius and the variable radius.physicss said:wouldn´t the general limits be (0,h), (0,2pi) and (0, r-rz/h) rdrdφdz?
May I send you my solution in spherical via PM?kuruman said: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##.
I will be happy to look at it.PhDeezNutz said:May I send you my solution in spherical via PM?
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 said:I will be happy to look at it.
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 said: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)
I give up but still thank you for your time and helpkuruman said: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.
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.physicss said:I give up but still thank you for your time and help
physicss said: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)
thank you for your answer. could H(1−r/R) be right ?PhDeezNutz said: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 said:thank you for your answer. could H(1−r/R) be right ?
PhDeezNutz said: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
R/hkuruman said: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##.
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 said: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
I think I found my mistake, thanks for the helpkuruman said: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.
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.physicss said:I think I found my mistake, thanks for the help
The formula for the volume of a cone is \( V = \frac{1}{3} \pi r^2 h \), where \( r \) is the radius of the base and \( h \) is the height of the cone.
To set up the volume integral in Cartesian coordinates, you would integrate over the region bounded by the cone. The limits of integration would be determined by the equation of the cone and the base. The integral would typically be expressed as a triple integral in the form \( \int \int \int dV \), where \( dV \) is the differential volume element.
In cylindrical coordinates, the volume integral for a cone can be set up as \( \int_{0}^{2\pi} \int_{0}^{r} \int_{0}^{h - \frac{h}{r}r'} r' \, dz \, dr' \, d\theta \), where \( r' \) is the radial coordinate, \( \theta \) is the angular coordinate, and \( z \) is the height coordinate. This simplifies the integration process.
The limits of integration in cylindrical coordinates are: \( \theta \) ranges from 0 to \( 2\pi \), \( r' \) ranges from 0 to \( r \), and \( z \) ranges from 0 to \( h - \frac{h}{r}r' \). These limits cover the entire volume of the cone.
While it is possible to set up the volume integral of a cone in spherical coordinates, it is generally more complicated than using cylindrical coordinates. Spherical coordinates are more suitable for objects with radial symmetry from a central point, such as spheres, rather than cones.