MHB Where Did I Go Wrong With the Centroid Problem?

[bmanmcfly]

[SOLVED] Centroid problem with y vector

View attachment 720

So, I've been having a problem with this part here, the equations are y=4x^2 and y=2x3.

So, the intersections are (0,0) and (2, 16)
The area between the two is 8/3.

I know I'm doing something correct because I get the x vector correct with ease, but no matter what I can't get the y vector anywhere near the answer.View attachment 720

Can someone please point out where I'm going wrong here??

(not the answer, just where I'm going wrong)

 

[alyafey22]

Sorry , I cannot understand what the question really is ?

 

[bmanmcfly]

Sorry , I cannot understand what the question really is ?

The given equation is y vector= 1/area( int from 0-16 for y( x2-x1) dy

(sorry not good with latex)

The x2 and x1 are found with y=4x^2 and y=2x^3.

At which point am I going wrong?

Or did the image linked not appear?

Edit: the answer should by a y value between 0-16 to reflect the point of balance between the two curves.

 

[alyafey22]

Are you sure that you have constructed the integral correctly ?

 

[Jameson]

Before we try to follow your work and check for an error, what method are you using? Are you using the method outlined here? I see the 1/A term but the y moment should be integrating f(x)^2-g(x)^2. Just checking we're talking about the same thing. :)

 

[I like Serena]

Welcome to MHB, Bmanmcfly! :)

Your calculation is fine!
But as ZaidAlyafey already suggested, the expression for your initial integral contains a small mistake.

Btw, what you refer to as "y vector" is actually the "mean of y".

 

[bmanmcfly]

Before we try to follow your work and check for an error, what method are you using? Are you using the method outlined here? I see the 1/A term but the y moment should be integrating f(x)^2-g(x)^2. Just checking we're talking about the same thing. :)

If this is the case then the book just messed me up, because the equation I've been given in the book is the
X vector uses x(y2-y1) then divide by the area equation. Then the y vector is y(x2-x1)...

But the square method in that link might prove useful... I used the 1/a because I calculated that separately to use less page room from calculating all...

I think you got me going in a good direction here, I'll try and report.

I did notice that I chose the wrong x2 and x1 lines, I think.

 

[bmanmcfly]

Welcome to MHB, Bmanmcfly! :)

Your calculation is fine!
But as ZaidAlyafey already suggested, the expression for your initial integral contains a small mistake.

Btw, what you refer to as "y vector" is actually the "mean of y".

Oh, oops, I thought the bar over the y meant vector...

 

[I like Serena]

If this is the case then the book just messed me up, because the equation I've been given in the book is the
X vector uses x(y2-y1) then divide by the area equation. Then the y vector is y(x2-x1)...

But the square method in that link might prove useful... I used the 1/a because I calculated that separately to use less page room from calculating all...

I think you got me going in a good direction here, I'll try and report.

I did notice that I chose the wrong x2 and x1 lines, I think.

Both methods work.
Either you calculate the inverse of f(x) and g(x) and integrate y(x2-x1) with respect to y as you did.
Or you use the integral of (1/2)(x2^2 - x1^2) and integrate with respect to x as Jameson suggests.

 

[bmanmcfly]

Both methods work.
Either you calculate the inverse of f(x) and g(x) and integrate y(x2-x1) with respect to y as you did.
Or you use the integral of (1/2)(x2^2 - x1^2) and integrate with respect to x as Jameson suggests.

I realized the mistakes I was making... First, x1 should have been closest to the axis.

There were other issues also in the 20 or so times I tried to solve the problem before posting, but I finally got the answer.

It may not feel like you all helped that much, but it definitely worked.

Thanks.