Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Volume of a cylinder by integration

  1. Apr 15, 2004 #1
    please forgive me, because i haven't done this in quite some time. :confused: any help is appreciated.

    i'm trying to calculate the volume of material that was shaved off a solid cylindrical temperature thermowell. looking from the side, a triangular portion was cut off at an angle from the top down.

    the diameter of the cylinder is 1.1 cm
    the height of the cylinder is 0.8 cm

    looking top down at the cylinder, perpendicular to the measured diameter, the circle is cut 0.4 cm from the side of the cylinder, leaving the remaining 0.7 cm intact.

    the cut was at such an angle (approx 63 deg by my calcs) that it exited the thermowell 0.8 cm down from the top.

    what is the missing volume? as mentioned before, any help is appreciated...

  2. jcsd
  3. Apr 15, 2004 #2
    Look at the picture I attached. I think you need to find the equation of the right line (the cut), then revolve that line around the Y axis and find the created volume. Divide that volume by half, and substract it from half of the volume of the original cylinder.

    Actually, ignore this. I'm wrong... :rolleyes: I'll try to think of something else.

    Attached Files:

  4. Apr 15, 2004 #3
    Ok, what about this? I've rotated the axes sytem so that the line I was previously talking about now lies on the Y axis. You need to find the equations of the two dashed lines, and revolve them around the new Y axis. The volume you get divided by half should be the volume that was removed from the cylinder.

    Attached Files:

    Last edited: Apr 15, 2004
  5. Apr 15, 2004 #4
    that wont work Chen.. if you do that, you'll be revolving a different shape than what was removed

    its hard to explain, but think about it, the shape will be different... you have to revolve it around an axis which is parallel to the long side of the cutout inorder for the shape to be the same

    this is certainly an interesting problem tho.. i'll think about it a bit
  6. Apr 15, 2004 #5
    Isn't that what I did? By rotating the axes system, I made the Y axis lie on the long side of the cutout... or doesn't it? :confused:
  7. Apr 15, 2004 #6
    Try this.. rotate sections A about the GREY axis

    Attached Files:

  8. Apr 15, 2004 #7
    OK.. i got it now.. let me draw it out so it makes some sense
  9. Apr 15, 2004 #8
    you cant simply revolve what you see in the red, since it wont fill the entire cylinder

    rotate the section you see in the yellow, then subtract B (which will be a triangular prism, plus an erregular shape that is somewhat like a rectangular prism.. which i cant remember the name of.. lol)

    EDIT: is what I'm saying making any sense? or am i just sounding crazy?

    Attached Files:

    Last edited: Apr 15, 2004
  10. Apr 16, 2004 #9
    At any rate, I'd be interested to see the correct answer once you find it out, johnnyv.
  11. Apr 16, 2004 #10
    me, too!

    definitely doesn't sound like a simple "plug and chug" type problem...
  12. Apr 16, 2004 #11

    Try this. On an x-y cooordinate axis, draw a circle of radius R whose center is below the x -axis by some distance L<R. So part of the circle is above the axis and part is below. Let the height of the part that's above be h = R-L. Now figure out the area A(h) of this part of the circle. The volume you're trying to find will be smiply related to the integral of this function.
  13. Apr 16, 2004 #12

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    It is if you konw what to plug in to where.

    You need to evaluate the intergral of 1 over the volume, it's just finding the equations that parametrize the surface that is tedious. You may assume that the cylinder is centred on the origin and sat on the x-y plane, and that the planar cut's normal lies in the plane y=0 if it helps.

    Once you've worked out the surface equations it seems like a plug and chug to do the integral. It won't be very pleasant but, seeing as you've used the phrase 'about 63 degrees' then the answer presumably isn't required to any unreasonable degree of accuracy either.

    Of course you could also figure it out using a shell argument: work out the volume of a disc of thickness dz at the point z and integrate the z from 0 to 0.8
  14. Apr 16, 2004 #13


    User Avatar

    Here's what I did, but I think I made a mistake somewhere.

    I put a circle into the coordinate system in such a way that the cutting line is parallel with the y axis. By integration I found the formula for the "cut off" area of the circle in relation to c (where c is the x coordinate of the rightmost point where the "cutting" line crosses the circle).

    I got this for the area of the "cut off" surface (I don't know the correct mathematical term in English):

    S(c) = rc*sqrt(1-c^2/r^2)+r^2*arcsin(c/r) - 2c*sqrt(r^2-c^2)

    I think this is correct so far. I then found the relation of c to z so I can integrate the surfaces over z. I positioned the cyllinder in such a way that the top circle of the cyllinder is parallel with xy plane and its z=0.8. That way, when z is 0, c is 0, when z is 0.8, c is cmax. Since c and z are proportional (I think they are):

    c(z) = z*cmax/h

    So S(z) is the S(c) formula with c(z) instead of c everywhere, I am too lazy to type that. So the volume is the integral from 0 to 0.8 of S(z)dz.

    I typed that into the wolfram integrator (replacing z with x because of the program) and got a huge formula (the solution would be http://img5.imageshack.us/img5/6504/integral2.gif [Broken] from 0 to 0.8, but it's probably wrong).

    Since all those r, v and h are given numbers it could be simplified, but it is too big and ugly for me to even look at, so I guess I just wasted an hour :( I hope at least I figured out correctly what the question was and what the cyllinder looked like :)

    [ r = 0.55, h = 0.8, v = cmax = sqrt(7)/5. ]
    Last edited by a moderator: May 1, 2017
  15. Apr 16, 2004 #14

    if you actually have the part, and you dont have to be extremely accurate, the easiest way to measure the amount cut off, may be to get another uncut part

    then see the difference in the amount of water they displace
  16. Apr 16, 2004 #15
    good idea. i thought about doing just that, but the replacement part isn't identical to the damaged part, therefore weight and volume comparisons are out the door. :frown:
  17. Apr 16, 2004 #16


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member


    The replacement need not be identical in volume, just in proportion to the dimensions of the original. And the density doesn't matter at all.
  18. Apr 16, 2004 #17


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    What is your question?
  19. Apr 16, 2004 #18
    Do you know what the answer is, either as an expression or a numeric answer?

    Ok, here is what I did. I defined an XYZ space as described in the attached image. The equation of the circle in the XY plane, with a radius of 0.55 and the center located at (-0.15, 0) is:

    [tex](x + 0.15)^2 + y^2 = 0.55^2[/tex]

    The value of Y for every X is then:

    [tex]y = \pm \sqrt{-x^2 - 0.3x + 0.28}[/tex]

    From now on I will refer to that expression as "sqrt" for sake of simplicity. So now every point on the circle can be described as (x, sqrt, 0.8). I've defined another point (0.4, 0, 0) which is located at the bottom of the cylinder, on the circle's perimeter.

    Let's consider two points on the top circle perimeter, A (x, +sqrt, 0.8) and B (x, -sqrt, 0.8), and the bottom point C (0.4, 0, 0). If we connect these points we create a triangle. The missing volume from the clyinder is actually the sum of areas of all these triangles, for x running from 0 (beginning of cut) to 0.4 (end of it).

    So first we need to find the expression for the area of those triangles. If we create a vector AC and BC, then the area of the ABC triangle is:

    [tex]S_x = \frac{1}{2}|\vec{AC} \times \vec{BC}|[/tex]

    AC = (x - 0.4, +sqrt, 0.8)
    BC = (x - 0.4, -sqrt, 0.8)

    I don't know how to write the calculation here, so you will just have to trust me that:
    AC x BC = (0.8*sqrt, 0, 0.4*sqrt - x*sqrt)
    Where x is the X coordinate of the points A and B. So now the magnitude of that vector is:

    [tex]|\vec{AC} \times \vec{BC}| = sqrt\sqrt{x^2 - 0.8x + 0.8}[/tex]

    And thus the area of every triangle is:

    [tex]S_x = \frac{1}{2}|\vec{AC} \times \vec{BC}| = \frac{sqrt}{2}\sqrt{x^2 - 0.8x + 0.8}[/tex]

    And if we use the real expression of "sqrt" we get:

    [tex]S_x = \frac{1}{2}\sqrt{(-x^2 - 0.3x + 0.28)(x^2 - 0.8x + 0.8)}[/tex]

    [tex]S_x = \frac{1}{2}\sqrt{-x^4 + 0.5x^3 - 0.28x^2 - 0.464x + 0.224}[/tex]

    And finally, we need to find the sum of all these areas for x from 0 to 0.4:

    [tex]V = \frac{1}{2}\sum_{x = 0}^{0.4}S_x = \frac{1}{2}\sum_{x = 0}^{0.4} \sqrt{-x^4 + 0.5x^3 - 0.28x^2 - 0.464x + 0.224}[/tex]

    And this is where I'm stuck because I don't know the formulas for the sums of x4 and x3. :frown:

    Anyway, I hope this has helped... and not caused too much confusion. I may have gone about this in the wrong way, but I believe it's correct. And at least with this answer you got a definite expression. :smile: Let me know how it works out, and if anyone can spot an error in my logic/calculations please do let me know.

    Attached Files:

  20. Apr 17, 2004 #19

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    I think the sqrt buggers up any hope of summing that, Chen.
  21. Apr 20, 2004 #20

    Just take few moments to analyze the two cases (images attached)

    if it is the first case you are wondering about, well, then it does need some thought but if it is the second one then it's easy.
    1) take a cylinder with the same height and radius 'r1' (base radius being 'r2' and r2 > r1) and find the volume.
    2) take a right triangle of the same height and base equal to r2-r1, multiply the area of this rectangle with '2 * pi * (r2- (2/3)(r2-r1))' to get a volume of revolution.

    add the volumes calculated in the above steps, to get the volume left in your part

    Attached Files:

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook