calculus The center of mass of quarter circle y=sqrt( r^2 -x^2) is given by the point??

Question

darkigloo · Answer

$$y= \sqrt{r^2 - x^2} , x \in[0,r]$$

bobo-i-bo · Answer

Given a lamina, do you know the formula to calculate the centre of mass? (Do you know what a lamina is?)

darkigloo · Answer

i dont know what a lamina is

bobo-i-bo · Answer

A lamina is a flat shape with constant density. in this case the lamina is the quarter circle.

darkigloo · Answer

ok.

bobo-i-bo · Answer

There are one or two formulae for finding the centre of mass of a lamina, but the one you want is:
$$\frac{\int\limits_{a}^{b}xydx}{\int\limits_{a}^{b}ydx}$$
which will give you the centre of mass

bobo-i-bo · Answer

Understand what to do from here? It would be a very good idea if you tried to search up where the formula comes from though btw

darkigloo · Answer

$$\frac{ \int\limits_{0}^{r} x(\sqrt{r ^{2}-x ^{2}})dx }{ \int\limits_{0}^{r} \sqrt{r ^{2}-x ^{2}}dx }  $$

darkigloo · Answer

like that?

bobo-i-bo · Answer

yup!

darkigloo · Answer

how would i find that integral?

mathmale · Answer

the integral in the numerator can be solved using a substitution.

The one in the denom. can be solved (integrated) using a trig substitution.

darkigloo · Answer

can you help me with that?
$$\sqrt{a^2 - x^2} \rightarrow x=a \sin \theta $$
$$\sqrt{(asin \theta)^{2}-x^2} $$

darkigloo · Answer

thats for the denominator. what do i do from there?

darkigloo · Answer

@mathmale

darkigloo · Answer

$$x= r \sin \theta $$
$$dx = r \cos \theta d \theta $$ ?