Centroids of two and three dimensional figures

  • Thread starter rdajunior95
  • Start date
  • Tags
    Centroids
In summary: The y and z coordinates are 0. In the particular case of [itex]y= (sin(x))^4, we have [itex]f(x)= (sin(x))^4[/itex] and [itex]f'(x)= 4sin^3(x)cos(x)[/tex] The centroid is at[tex]\left(\frac{\sqrt{1+ 16sin^6(x)}}{16sin^3(x)}dx, 0, 0\right) I don't know what you mean by "from 0 to 3" but if you mean to take the curve from x= 0 to x= 3, the centroid is at
  • #1
rdajunior95
25
0
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:
 
Mathematics news on Phys.org
  • #2
Please help me someone :D
 
  • #3
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.
 
  • #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:
  • #5
rdajunior95 said:
[tex]\overline{x}= \frac{\int\int x dydx}{\int \int dydx}[/tex]

In this expression, what do u mean by xdydx.
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.

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?
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]

Thanks for the help.
 
  • #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.
 
  • #7
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]
and
[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]
and
[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]
 

Related to Centroids of two and three dimensional figures

1. What is the definition of a centroid?

A centroid is the geometric center of a shape or figure, often referred to as the "center of mass". It is the point where all the mass of the object is evenly distributed.

2. How is the centroid calculated for a two-dimensional figure?

The centroid of a two-dimensional figure can be calculated by finding the average of the x and y coordinates of all the points that make up the figure. This can be done using the formula: x̄ = (x1 + x2 + x3 + ... + xn) / n and ȳ = (y1 + y2 + y3 + ... + yn) / n, where and represent the x and y coordinates of the centroid, and n represents the number of points.

3. How is the centroid calculated for a three-dimensional figure?

The centroid of a three-dimensional figure can be calculated by finding the average of the x, y, and z coordinates of all the points that make up the figure. This can be done using the formula: x̄ = (x1 + x2 + x3 + ... + xn) / n, ȳ = (y1 + y2 + y3 + ... + yn) / n, and z̄ = (z1 + z2 + z3 + ... + zn) / n, where , , and represent the x, y, and z coordinates of the centroid, and n represents the number of points.

4. What is the significance of a centroid in geometry and physics?

In geometry, the centroid is used to determine the balance point of a shape, and it also has applications in finding the center of gravity. In physics, the centroid is important in calculating the moment of inertia of an object, which measures its resistance to rotational motion.

5. Can the centroid lie outside of the figure?

No, the centroid always lies within the boundaries of the figure. It can be located at the center of the figure, or closer to one side depending on the distribution of mass within the figure.

Similar threads

  • Calculus and Beyond Homework Help
Replies
9
Views
828
Replies
2
Views
441
  • Introductory Physics Homework Help
Replies
2
Views
1K
Replies
10
Views
375
Replies
3
Views
2K
  • Calculus and Beyond Homework Help
Replies
2
Views
4K
  • Engineering and Comp Sci Homework Help
Replies
5
Views
3K
  • Calculus and Beyond Homework Help
Replies
1
Views
1K
Back
Top