Please help me finding this centroid of a graph. Thank you.

[sun1234]

Homework Statement

Find the x-coordinate of the centroid of the region bounded by the graphs of

Homework Equations

y= 5/(√(25-x²))

The Attempt at a Solution

I stuck at finding Mx

 

[haruspex]

Homework Statement

Find the x-coordinate of the centroid of the region bounded by the graphs of

Homework Equations

y= 5/(√(25-x²))

The Attempt at a Solution

I stuck at finding Mx[/B]

Please post some attempt.
What you quote as a 'relevant equation' is part of the problem statement. A relevant equation here would be a general equation for finding a centroid.

 

[sun1234]

Here is what I got so far.
https://lh4.googleusercontent.com/-Hm-tHWU3uR5FHIcZIqSHVDGDjaKA1lA68L2m_DgLeU7LMUmwV2qcVIQr8_kEQ_AjkFJ5PjlGz8=w1256-h843

 

[haruspex]

In the OP you did not state the other boundaries, so I assume they are indeed x=0, y=0 and x=5.
You are only asked for the x coordinate. You found this, but called it M_Y.

 

[sun1234]

Thank you so much. How blind I am!

 

[Ray Vickson]

Here is what I got so far.
[PLAIN]https://lh4.googleusercontent.com/-Hm-tHWU3uR5FHIcZIqSHVDGDjaKA1lA68L2m_DgLeU7LMUmwV2qcVIQr8_kEQ_AjkFJ5PjlGz8=w1256-h843[/QUOTE]

The $x$-corrdinate of the centroid of the region between $y = 0$ and $y = f(x) (\geq 0)$, with $a \leq x \leq b$ is
$$C_x = \frac{1}{A} \int_a^b x f(x) \, dx, \; \text{where} \; A = \int_a^b f(x) \, dx$$
I cannot make out exactly what you have computed, but I do not see an integral like $\int x f(x) \, dx$ anywhere. Did I miss it?

BTW: such issues arise often when people submit photos of written work, instead of typing it out as they are supposed to do in this Forum.

 

[sun1234]

That's My (not my in English). I already calculated, so don't bother about it. Thank you for trying to help. By the way, do you know where I can post quick questions? I have an integral that I can't figure out, but I don't want to make a new thread about it. Thank you.

 

[Ray Vickson]

That's My (not my in English). I already calculated, so don't bother about it. Thank you for trying to help. By the way, do you know where I can post quick questions? I have an integral that I can't figure out, but I don't want to make a new thread about it. Thank you.

For the record, what you wrote for My (by which you meant Mx) is NOT CORRECT. Mx = a ratio of two quantities, and what you called My is the numerator of the ratio, not the actual ratio itself.