Centroid Problem: Find xbar, ybar in Quadrant 1 of xy-plane

[3soteric]

Homework Statement

Find the centroid of the region in the first quadrant of the xy-plane bounded by the graphs of the equations:
x=0, x=1, y=x-x^2, and y^2=2x

Homework Equations

xbar: integral of x dA/ integral of dA
ybar: integral y dA/ integral of dA

The Attempt at a Solution

my attempt at it was not so great, i determined the limit for the x values were from 0 to 1, the y limits were from x-x^2 to square root of 2x. I am not sure if the limits I chose are even valid, but that is what I could grasp. Then integrated dydx respectively with the limits, and ended up with 1/2 as the dA. I tried to find ybar and ended up with 29/15, which is clearly wrong if I have the concept of centroids correct. I attempted xbar but I can't seem to get too far.

Can I get some help please? this is a test review and need to understand this concept

 

[SammyS]

Homework Statement

Find the centroid of the region in the first quadrant of the xy-plane bounded by the graphs of the equations:
x=0, x=1, y=x-x^2, and y^2=2x

Homework Equations

xbar: integral of x dA/ integral of dA
ybar: integral y dA/ integral of dA

The Attempt at a Solution

my attempt at it was not so great, i determined the limit for the x values were from 0 to 1, the y limits were from x-x^2 to square root of 2x. I am not sure if the limits I chose are even valid, but that is what I could grasp. Then integrated dydx respectively with the limits, and ended up with 1/2 as the dA. I tried to find ybar and ended up with 29/15, which is clearly wrong if I have the concept of centroids correct. I attempted xbar but I can't seem to get too far.

Can I get some help please? this is a test review and need to understand this concept

Hello 3soteric. Welcome to PF !

First of all, sketch the region of interest.

The area is not 1/2. Is that what you had?

What Is it that you integrated for the area, A, for xbar for ybar?

 

[3soteric]

thanks for the welcome sam!

do i sketch it and upload the image? the area seems to be limited vertically in y sense by x-x^2 and sqrt2x, horizontally by x=0, and x=1, for area yes i ended up with 1/2 but can you show me why is it wrong or the process? for ybar i got 29/15 and i couldn't calculate xbar by the nature of integration but i ended up with 10.8 as the final value if i remember which seems wrong as well

 

[SammyS]

thanks for the welcome sam!

do i sketch it and upload the image? the area seems to be limited vertically in y sense by x-x^2 and sqrt2x, horizontally by x=0, and x=1, for area yes i ended up with 1/2 but can you show me why is it wrong or the process? for ybar i got 29/15 and i couldn't calculate xbar by the nature of integration but i ended up with 10.8 as the final value if i remember which seems wrong as well

It seems that you do understand the region correctly.

The area is simply $\displaystyle \text{A}=\int_{0}^{1} (\sqrt{2x}-(x-x^2))\,dx\ .$

The integral for $\bar{x}$ should be less complicated than the integral for $\bar{y}\ .$

 

[3soteric]

yo sam thanks ! i ended up getting the values for x bar, y bar respectively as .5141, .408 which make much more sense than the values i derived at the beginning of the problem.

 

[SammyS]

yo sam thanks ! i ended up getting the values for x bar, y bar respectively as .5141, .408 which make much more sense than the values i derived at the beginning of the problem.

What did you get for the area?

 

[3soteric]

What did you get for the area?

i ended up getting .776!

 

[SammyS]

i ended up getting .776!

That's correct, (if it's not a factorial LOL).

I got something a bit different for $\bar{x}\,,$ but I wasn't all that careful in getting it.

 

[3soteric]

That's correct, (if it's not a factorial LOL).

I got something a bit different for $\bar{x}\,,$ but I wasn't all that careful in getting it.

LOL its not a factorial ! phew!

im sure i have the right idea regardless right?

by the way you think you can help me with the new problem i posted in this same section about cylindrical coordinates LOL, if you have time of course. I've been trying for the longest and don't seem to understand the concept