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Centroids of two and three dimensional figures!

  1. Oct 16, 2010 #1
    In my Questions I usually have to calculate the Centroid of a curve whose equation is given!

    I don't know the formula which are used it that, and I have searched on google too and couldn't find anything useful. Can someone provide me with formulas for how to calculate centroid of 2 and 3 dimensional figures fr x and y axis?

    Any help will be appreciated, :smile:
  2. jcsd
  3. Oct 16, 2010 #2
    Please help me someone :D
  4. Oct 16, 2010 #3


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    All I can give you is the usual formulas:

    In two dimensions, the centroid is given by
    [tex]\overline{x}= \frac{\int\int x dydx}{\int \int dydx}[/tex]
    [tex]\overline{y}= \frac{\int\int y dydx}{\int \int dydx}[/itex]
    where the integrals are taken over the two dimensional figure.

    In three dimensions, just exend those formulas:
    [tex]\overline{x}= \frac{\int\int\int x dzdydx}{\int\int \int dzdydx}[/tex]
    [tex]\overline{y}= \frac{\int\int\int y dzdydx}{\int \int\int dzdydx}[/tex]
    [tex]\overline{z}= \frac{\int\int\int z dzdydx}{\int\int\int dzdydx}[/tex]

    where the integrations are not over the three dimensional body. The denominators of those fractions are the area of the two dimensional figure and volume of the three dimensional figure. If they have a reasonable geometry, you might be able to find it without integrating.

    For example, to find the centroid of the triangle with vertices at (0, 0), (a, h), and (b, 0) we take the area to be "one half height times base" or (1/2)bh.

    But we still have to integrate to find the centroid:
    [tex]\overline{x}= \frac{\int_{y= 0}^h\int_{x= ay/h}^{b-\frac{b-a}{b}y} x dxdy}{bh/2}[/tex]
    [tex]\overline{y}= \frac{\int_{y=0}^h\int_{x= ay/h}^{b- \frac{b-a}{b}y} y dx dy}{by/2}[/tex]
    Where the limits on the inner integral are the equations of the lines from (0, 0) to (a, h) and from (a, h) to (b, 0).

    After some tedious computation one can show that centroid is at
    [tex]\left(\frac{a+ b}{3}, \frac{h}{3}\right)[/tex]
    That is, the coordinates of the centoid are the arthmetic averages of the x and y coordinates of the vertices. That is NOT in general true of figures with more than three sides. You could divide a polygon into triangles but then the centroid of the polygon would be a weighted average of the centroids of the triangles, weighted by their area.
  5. Oct 16, 2010 #4
    [tex]\overline{x}= \frac{\int\int x dydx}{\int \int dydx}[/tex]

    In this expression, what do u mean by xdydx.

    And by 3d I meant when in xy-plane you rotate the curve about x or y axis to form a 3d shape. like if you rotate a line y=kx, about the x axis, yo will get a cone so how will you find its centroid?

    Thanks for the help.
    Last edited: Oct 16, 2010
  6. Oct 17, 2010 #5


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    I mean to take the integral of the function f(x,y)= x with respect to x and y, over the region. That is, I think, the usual meaning.

    Well, that's easier! If you have a curve rotated around the x-axis, say, then by symmetry, you know the centroid, of either the surface swept out by the curve or the solid swept out by the region contained in the curve, is on the x-axis. That is, the y and z coordinates of the centroid are 0.

    For this particular problem, I would make use of the symmetry by setting up a polar coordinate system in the yz- axis. Since the equation of the line in the xy-plane is y= kx or x= y/k, rotating around the x-axis makes y= r so the cone is given by x= r/k. Now, I don't know what method you use for getting a "differential of surface area" but for this cone it is
    [tex]dS= \frac{\sqrt{k^2+ 1}}{k} r dr d\theta[/itex].

    You did not specify an "ending x value" or height for the cone. Calling that h, we have for the surface area of the cone
    [tex]\frac{\sqrt{k^2+ 1}}{k}\int_{\theta= 0}^{2\pi}\int_{r= 0}^{kh} rdr d\theta= \pi k\sqrt{k^2+ 1}h^2[/tex]

    The "moment about the x-axis" is given by
    [tex]\frac{\sqrt{k^+ 1}}{k}\int_{\theta= 0}^{2\pi}\int_{r= 0}^{kn}x rdrd\theta= \frac{\sqrt{k^+ 1}}{k^2}\int_{\theta= 0}^{2\pi}\int_{r= 0}^{kn} r^2 drd\theta= \frac{2}{3}\pi k\sqrt{k^2+ 1}h^3[/tex]

    The x-coordinate of the centroid is the moment divided by the area:
    [tex]\frac{\frac{2}{3}\pi k\sqrt{k^2+ 1}h^3}{\pi k\sqrt{k^2+ 1}h^2}= \frac{2}{3}h[/tex].

    The centroid of the (slant surface) of the cone is at
    [tex]\left(\frac{2}{3}h, 0, 0\right)[/tex]

  7. Oct 17, 2010 #6
    Can you tell me a generalized method to find the centroid when a curve in x-y plane is rotated about the x or[b/] y axis.

    And one last thing is that how can I use [tex]
    \overline{y}= \frac{\int\int y dydx}{\int \int dydx}
    [/tex] and [tex]
    \overline{x}= \frac{\int\int x dydx}{\int \int dydx}
    [/tex] to find the centroid of the curve with equation y=(sin x)^4 from 0 to 3

    Thanks again.
  8. Oct 18, 2010 #7


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    Suppose the curve is given by y= f(x) and it is rotated around the x-axis. Then the y and z coordinates of its centroid are 0.

    Take x and [itex]\theta[/itex], the angle of rotation, as parameters. Then we can write [itex]y= f(x)cos(\theta)[/itex] and [itex]z= f(x)sin(\theta)[/itex] so that the "position vector" of a point on the surface is [itex]\vec{r}(x, \theta)= x\vec{i}+ y\vec{j}+ z\vec{k}= x\vec{i}+ f(x)cos(\theta)\vec{j}+ f(x)sin(\theta)\vec{k}[/itex].

    The derivatives of that position vector, with respect to x and [itex]\theta[/itex] are:
    [tex]\vec{r}_x= \vec{j}+ f'(x)cos(\theta)\vec{j}+ f'(x)sin(\theta)\vec{k}[/tex]
    [tex]\vec{r}_\theta= -f(x) sin(\theta)\vec{j}+ f(x)cos(\theta)\vec{k}[/tex]

    The "fundamental vector product" for this surface is the cross product of those:
    [tex]\left|\begin{array}\vec{i} & \vec{j} & \vec{k} \\ 1 & f'(x)cos(\theta) & f'(x)sin(\theta) \\ 0 & -f(x)sin(\theta) & f(x) cos(\theta)\end{array}\right|[/tex]
    [tex]= f'(x)f(x)\vec{i}- f(x)cos(\theta)\vec{j}- f(x)sin(\theta)\vec{k}[/tex]

    It's magnitude,
    [tex]\sqrt{f'^2(x)f^2(x)+ f^2(x)}= f(x)\sqrt{f'(x)^2+ 1}[/tex]
    gives the "differential of surface area"
    [tex]f(x)\sqrt{f'^2(x)+ 1}dxd\theta[/itex]

    Assuming our original curve extends from x= a to x= b, the surface area is given by
    [tex]\int_{\theta}= 0^\2\pi \int_{x= a}^b f(x)\sqrt{f'^2(x)+1}dxd\theta= 2\pi\int_a^b f(x)\sqrt{f'^2(x)+ 1} dx[/tex]

    The x coordinate of the centroid is
    [tex]\e\pi\int_a^b x f(x)\sqrt{f'^2(x)+ 1}dx[/tex]
    divided by that area:

    The centroid is at [itex]\left(\overline{x}, 0, 0\right)[/itex] where
    [tex]\overline{x}= \frac{\int_a^b x f(x)\sqrt{f'^2(x)+ 1}dx}{\int_a^b f(x)\sqrt{f'^3(x)+ 1} dx}[/tex]

    For the last problem, are you talking about the curve or the surface created by rotating the curve around the x-axis.

    If you mean the curve itself, You don't use those formulas. They are for the centroid of a surface. For the curve, you use
    [tex]\overline{x}= \frac{\int xd\sigma}{\int d\sigma}[/tex]
    [tex]\overline{y}= \frac{\int yd\sigma}{\int d\sigma}[/tex]

    where "[itex]d\sigma[/itex]" is the differential of arc length.

    For [itex]y= sin^4(x)[/itex], [itex] dy/dx= 4sin^3(x) cos(x)[itex] so
    [tex]d\sigma= \sqrt{1+ \left(\frac{dy}{dx}\right)^2} dx= \sqrt{1+ 16sin^6(x)cos^2(x)}dx[/tex]

    If you are talking about the centroid of the surface of rotation, use the formulas above with [itex]f(x)= sin^4(x)[/itex], [itex]f'(x)= 4 sin^3(x)cos(x)[/itex]:

    [tex]\overline{x}= \frac{\int_0^3 x sin^3(x)\sqrt{16 sin^6(x)cos^2(x)+ 1}dx}{\int_0^3 sin^4(x)\sqrt{16sin^6(x)cos^2(x)+ 1} dx}[/tex]
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