How do I solve for the centroid of a function with a given range?

[Guillem_dlc]

Summary:: I'm solving an exercise.

I have the following center of gravity problem:

Having the function Y(x)=96,4*x(100-x) cm, where X is the horizontal axis and Y is the vertical axis, ranged between the interval (0, 93,7) cm. Determine:
a) Area bounded by this function, axis X and the line X=93,7 (in cm²)
b) The bar{x} coordinate of its centroid (in cm)
c) The bar{y} coordinate of its centroid (in cm)

My attempt at resolution:

How can I calculate the last section if I can't clear the x?

Thanks!

 

[Mark44]

How can I calculate the last section if I can't clear the x?

You are given y as a function of x, $y = 96.4x(100 - x)$. Solve this equation for x in terms of y.
IOW, you have y = f(x), Solving for x gives you $x = f^{-1}(y)$.

 

[Guillem_dlc]

You are given y as a function of x, $y = 96.4x(100 - x)$. Solve this equation for x in terms of y.
IOW, you have y = f(x), Solving for x gives you $x = f^{-1}(y)$.

Okay perfect thank you!

 

[etotheipi]

You can do it with the inverse that can sometimes be unwieldy if the function is complicated. You can also find $\bar{y}$ by integrating wrt $x$. Consider a vertical strip of width $\delta x$ and height $f(x)$, then the incremental change in the first moment of area in the $y$ direction is

$\delta S_y = \frac{1}{2} {f(x)}^2 \delta x$

and then you can just integrate up over your limits and divide by the total area.

 

[Guillem_dlc]

You can do it with the inverse but as an additional point that might sometimes be a little unwieldy if the function is complicated. You can also find $\bar{y}$ by integrating wrt $x$. Consider a vertical strip of width $\delta x$ and height $f(x)$, then the incremental change in the first moment of area in the $y$ direction is

$\delta S_y = \frac{1}{2} {f(x)}^2 \delta x$

and then you can just integrate up over your limits and divide by the total area.

Yes, I applied this second method in the end, i.e. integrating with respect to $x$ and considering that $y$ was in the middle of the differential.