Find the centroid

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  • #1
Jbreezy
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Homework Statement



Find the centroid bounded by the given curves.
y = x^3 , x+y = 2, y = 0
So if you graph this you will see that if you did it with respect to x you would have the break the graph at the point (1,1). Or at least I think you would have to. So in my book they give the equations


Homework Equations



x bar = (1/A)∫ x(f(x) - g(x)) dx and y bar = (1/A)∫ (1/2) ( f(x)^2 - g(x)^2 ) dx

The Attempt at a Solution




So I was thinking what if I just turned did it with respect to Y.


So would it be OK if I did.

y bar = (1/A)∫ y(f(y) - g(y)) dy and x bar = (1/A)∫ (1/2) ( f(x)^2 - g(x)^2 ) dx

Would this be OK to try? I just want to make sure I have it straight before I plug and chug.
Thanks,
J
 

Answers and Replies

  • #2
verty
Homework Helper
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I think this is similar to a center of mass calculation, you can read more here.
 
  • #3
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8,293

Homework Statement



Find the centroid bounded by the given curves.
y = x^3 , x+y = 2, y = 0
So if you graph this you will see that if you did it with respect to x you would have the break the graph at the point (1,1). Or at least I think you would have to. So in my book they give the equations


Homework Equations



x bar = (1/A)∫ x(f(x) - g(x)) dx and y bar = (1/A)∫ (1/2) ( f(x)^2 - g(x)^2 ) dx

The Attempt at a Solution




So I was thinking what if I just turned did it with respect to Y.
If I'm understanding what you're asking, no, you can't do this. The geometry of the two integrands is very different, so you can't just arbitrarily switch from integrating w.r.t x to integrating w.r.t. y.

For ##\bar{x}##, the typical integration element is the moment for a vertical slice, and looks like x * (yupper - ylower)Δx.

For ##\bar{y}##, the typical integration element is the moment for a horizontal slice, and looks like y * (xright - xleft)Δy.
So would it be OK if I did.

y bar = (1/A)∫ y(f(y) - g(y)) dy and x bar = (1/A)∫ (1/2) ( f(x)^2 - g(x)^2 ) dx

Would this be OK to try? I just want to make sure I have it straight before I plug and chug.
Thanks,
J
 
  • #4
HallsofIvy
Science Advisor
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The definition of "centroid" of a region is [itex]\bar{x}= \frac{1}{A}\int x dA[/itex] and [itex]\bar{y}= \frac{1}{A}\int ydA[/itex] where A is the area of the region and dA is the differential of area.

In particular, if the region lies between y= f(x) and y= g(x), x between a and b, with f(x)< g(x) for all x in [a, b], then [itex]\int xdA= \int_{x= a}^b\int_{y= (x)}^{g(x)} x dy dx= \int_a^b x(f(x)- g(x))dx[/itex] and [itex]\int_a^b \int_{f(x)}^{g(x)} y dydx= \int_a^b \frac{1}{2}\left[y^2\right]_{f(x)}^{g(x)} dx[/itex]
[itex]= \frac{1}{2}\int_a^b g^2(x)- f^2(x) dx[/itex], just as you say.

But if the region lies between x= f(y) and x= g(y), y between a and b, with f(y)< g(y) for all y in [a, b], you can also do [itex]\int dA[/itex] reversing the orders of x and y. That would give [itex]\int xdA= \int_{y= a}^b\int_{x= f(y)}^{g(y)} x dxdy= \frac{1}{2}\int_a^b g^2(y)- f^2(y) dy[/itex] and [itex]\int ydA= \int_{y= a}^b \int_{x= f(y)}^{g(y)} y dxdy= \int_a^b y(f(y)- g(y))dy[/itex].

In other words, all you are doing is "swapping" x and y. And, of course, it really doesn't matter which axis you call "x" and which "y", as long as you don't get them confused.
 
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  • #5
Jbreezy
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Wait I'm confused. Mark44 said no and Hallsofivy yes? Or is HallsofIvy saying something different?
 
  • #6
Jbreezy
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Different I should of had dy
 
  • #7
Jbreezy
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Going back to my original post then.
Find the centroid bounded by the given curves.
y = x^3 , x+y = 2, y = 0
A = 3/4
So I did x = 1/A (1/2) ∫ (2-y)^2 -(y)^1/3 dy

= 2/3 ∫ 4 - 4y +y^2 - y^(2/3) dy
= (2/3) [ 4y -2y^2 + y^3/3 - (3/5)y^5/3]
evaluate from [0,1] and you get 52/45 = 1.155


And y = (4/3) ∫ y(2-y - y^1/3) dy = (4/3) ∫ 2y -y^2 -^4/3 dy
= (4/3)[y^2 - y^3 /3 - 93/7)^7/3 from [0,1]

= 20/ 63 = .317460

So centroid is (52/45 , 20/ 63)

Yes no ?
 
  • #8
Jbreezy
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The definition of "centroid" of a region is [itex]\bar{x}= \frac{1}{A}\int x dA[/itex] and [itex]\bar{y}= \frac{1}{A}\int ydA[/itex] where A is the area of the region and dA is the differential of area.

In particular, if the region lies between y= f(x) and y= g(x), x between a and b, with f(x)< g(x) for all x in [a, b], then [itex]\int xdA= \int_{x= a}^b\int_{y= (x)}^{g(x)} x dy dx= \int_a^b x(f(x)- g(x))dx[/itex] and [itex]\int_a^b \int_{f(x)}^{g(x)} y dydx= \int_a^b \frac{1}{2}\left[y^2\right]_{f(x)}^{g(x)} dx[/itex]
[itex]= \frac{1}{2}\int_a^b g^2(x)- f^2(x) dx[/itex], just as you say.

But if the region lies between x= f(y) and x= g(y), y between a and b, with f(y)< g(y) for all y in [a, b], you can also do [itex]\int dA[/itex] reversing the orders of x and y. That would give [itex]\int xdA= \int_{y= a}^b\int_{x= f(y)}^{g(y)} x dxdy= \frac{1}{2}\int_a^b g^2(y)- f^2(y) dy[/itex] and [itex]\int ydA= \int_{y= a}^b \int_{x= f(y)}^{g(y)} y dxdy= \int_a^b y(f(y)- g(y))dy[/itex].

In other words, all you are doing is "swapping" x and y. And, of course, it really doesn't matter which axis you call "x" and which "y", as long as you don't get them confused.
I think you made a mistake

[itex]\int ydA= \int_{y= a}^b \int_{x= f(y)}^{g(y)} y dxdy= \int_a^b y(f(y)- g(y))dy[/itex].

Should if be g(y) - f(y) ? Since f(y) <g(y) ?
 
  • #9
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I think you made a mistake

[itex]\int ydA= \int_{y= a}^b \int_{x= f(y)}^{g(y)} y dxdy= \int_a^b y(f(y)- g(y))dy[/itex].

Should if be g(y) - f(y) ? Since f(y) <g(y) ?
Yes. The last integral should have an integrand of y(g(y) - f(y)). The upper limit of integration is g(y) and the lower limit is f(y).
 
  • #10
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Wait I'm confused. Mark44 said no and Hallsofivy yes? Or is HallsofIvy saying something different?

Different I should of had dy
That makes a difference. Also, I'm not sure I understood what you were asking, so we might have interpreted what you wrote in different ways.
 
  • #11
Jbreezy
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Ah. It;s Ok. I usually write like a 6 year old. Doesn't help ...ha!
 

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