I am having some confusion about the following. As I had never done much physics, this is the first time I have came across the topic.
I am asked to find the centroid of the disk segment $x^{2}+y^{2} \le 2$, $x \ge 1$
My thoughts:
All thoughts below are very possibly wrong, I will say what I think but please dont hesitate to correct me as I do not know much about this topic.
From my limited knowledge on the actual physical meaning of such, it seems to be similar to finding the centre of mass of an object etc. So it seems as my answer will be in the form $(X_{c},Y_{c})$. I have the region graphed as well to better see the region.
From what I have obtained on-line/Wikipedia, the formulas I come up with are either
$$X_{c}= \frac{\iint_{R} x dA}{A}$$
$$Y_{c}= \frac{\iint_{R} y dA}{A}$$
or put another way,
since we are dealing with a bounded region, if we considered $f(x)=\sqrt{2-x^{2}}$ and the line $x=1$, that is our bounded region , over the interval $[1,\sqrt{2}]$ ,
then via formulas obtained in the wiki article it could be evaluated.
as
if the region R is bounded by continuous function $f$ , $g$ such that $f(x) \ge g(x)$ on $[a,b]$ then $$X_{c}=\frac{1}{A} \int_{a}^{b} x[f(x)-g(x)]dx $$
$$Y_{c}=\frac{1}{A} \int_{a}^{b} \frac{f(x)+g(x)}{2}[f(x)-g(x)]dx$$
where $$A=\int_{a}^{b} [f(x)-g(x)]dx$$
Overall,
I am looking to truly understand what is being asked here. I want to know what is happening and how it should be approached. In terms of numerical solution, that is only a very small part. ( I believe the final answer is $(\frac{4}{3\pi},0)$)
But I am more interested in understand how to get the solution and which methods work best. I am confused on how to actually apply the formulas ( if they are indeed the right ones). For example, if we use the bounded region way, with $f(x)=\sqrt{2-x^{2}}$ , then what is our $g(x)?.$
Anyway, thanks all in advance for any help and or advice!
Best Answer
The centroid of a planar region is (more-or-less) the same as the center of mass of a very thin sheet of material having the shape of the region. So, on that topic, your thinking is correct.
So, if the centroid is denoted by $(X_c,Y_c)$, it's immediately clear from a picture that $Y_c = 0$. What remains is to calculate $X_c$, which we can do by computing various integrals.
The easiest approach, I think, is to view the shape as a collection of very thin vertical strips each having width $\delta x$. For a constant value of $x$, the corresponding vertical strip extends from $g(x) = -\sqrt{2-x^2}$ up to $f(x) =\sqrt{2-x^2}$, so its total overall height is $f(x)-g(x)=2\sqrt{2-x^2}$, its area is $2\sqrt{2-x^2}\delta x$, and its contribution to the centroid calculation is $2x\sqrt{2-x^2}\delta x$.
We have to "add up" all these vertical strips, which means calculating integrals. We have to consider values of $x$ between $1$ and $\sqrt2$. So, the relevant integrals are: $$ A = \int_1^{\sqrt2} \big[f(x)-g(x)\big] \, dx = \int_1^{\sqrt2} 2\sqrt{2-x^2} \, dx $$ $$ X_c = \frac{1}{A}\int_1^{\sqrt2} x\big[f(x)-g(x)\big] \, dx = \frac{1}{A}\int_1^{\sqrt2} 2x\sqrt{2-x^2} \, dx $$ I expect you can take it from here.