# [SOLVED]Centroid problem with y vector

#### Bmanmcfly

##### Member
[SOLVED] Centroid problem with y vector

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.

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

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

Last edited:

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
Sorry , I cannot understand what the question really is ?

#### Bmanmcfly

##### Member
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.

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
Are you sure that you have constructed the integral correctly ?

#### Jameson

Staff member
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.

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Welcome to MHB, Bmanmcfly!

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

##### Member
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 ive 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

##### Member
Welcome to MHB, Bmanmcfly!

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...

#### Klaas van Aarsen

##### MHB Seeker
Staff member
If this is the case then the book just messed me up, because the equation ive 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

##### Member
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.