# Find Centroid of Plane Figure and Centre of Gravity

A plane figure is enclosed by the parabola $y^2 = 4x$ and the line $y=2x$. Determine (a) the position of the centroid of the figure, and (b) the centre of gravity of the solid formed when the plane figure rotates completely about the x-axis.

So the figure i have to consider is the one enclosed here (inside the loop, that looks like an airfoil) :

from $x=0$ to $x=1$ ( when i find the points of intersection).

This is from a book, and it gives the Centroid $(0.4,1)$ and Center of Mass $(0.5,0)$.

I only managed to find the x ordinate of the Centroid. Here's what i have done.

• Centroid$(\bar{x},\bar{y})$ is given by: $$\bar{x}=\frac{1}{A}\int_{0}^{1} xy\ dx, \bar{y}=\frac{1}{2A}\int_{0}^{1} y^2\ dx$$

Where A is the area of the figure i want, given by: $A=\int_{0}^{1} y\ dx$

To find the $y$ i figure that it's the $y$ of the curve minus the $y$ of the line from $0$ to $1$. So $$y=\sqrt{2}\ x - 2x$$

So with all these i calculate $$A=\frac{1}{3},\ \bar{x} = 0.4,\ \bar{y}= 0.2$$

• Center of Gravity $(\bar{x},\bar{y})$ is given by:

$$\bar{x}=\frac{1}{V} \int_{0}^{1} xy^2 dx$$ where $V$ $=\int_{0}^{1}y^2 dx$ is the volume generated.

and $\bar{y}=0$ here because the axis of rotation is the x-axis.

EDIT: The following line is wrong (i don't know how to strikethrough it)

So i calculate: $y^2= 4x+4x^2-8x^{3/2} \ , V=\frac{2}{15}$ and $\bar{x}=0.357$

So apparently the only thing i get right is the $\bar{x}$ ordinate of the centroid. What am i doing wrong? Please help. Thanks in advance.

I made some adjustments to the equations i used. There were mistakes. I have figured out the answer. My main mistake was that i took $y^2$ as the difference of the values of the two functions squared $(y_1-y_2)^2$ and not as the difference of the squares of two functions $y_1^2-y_2^2$ (they are two separate functions acting on their own to derive the final result, it's not one function). I didn't have proper intuition about what i needed to do. This video also helped me (it has a similar example).
One other mistake is that i also divided by $π$ when the definitions already had that sorted out. That was a careless mistake. I edited this in the question.
So the area $A=\frac{1}{3}$, the volume is $V= \frac{2}{3}$(it's $V/π$ to be precise, because the $π$'s are cancelled out in the definition of the center of gravity). And as i said wherever we see $y^2$ we consider the superposition of our functions, i.e. the combined effect they have on the final result. In my case it is $y_1 - y_2$ (where $y_1=2\sqrt{x}\ ,\ y_2=2x$ ). I hope this is clear.