Solve $a$ and $b$ for centre of mass in $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ Given ellipse:
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$
What length do $a$ and $b$ have to be so the centre of mass is $S(4;2)$?
I've tried steps to solve the equation to $$y=b\sqrt{1-\frac{x^2}{a^2}}$$ and integrate 
$$A=b\int_0^a{\sqrt{1-\frac{x^2}{a^2}}}$$
But I'm not achieving a satisfying result. There must be an easier way . Enlighten me please

 A: Combining the results in the previous discussion and answers, we set $x=a t$, then $y=b\sqrt{1-t^2}$ and $dx=adt$.  So:
$$\bar{x}=\frac{\int_0^a xy(x)\,\mathrm{d}x}{\int_0^a y(x)\,\mathrm{d}x}=\frac{a^2b}{ab}\frac{\int_0^1 t \sqrt{1-t^2}\,\mathrm{d}t}{\int_0^1 \sqrt{1-t^2}\,\mathrm{d}t}=a\frac{1/3}{\pi/4}=\frac{4a}{3\pi}$$
Similarly we obtain:
$$\bar{y}=\frac{4b}{3\pi}$$
Thus
$$\bar{x}=\frac{4a}{3\pi}=4 \implies a=3\pi$$
$$\bar{y}=\frac{4b}{3\pi}=2 \implies b=\frac{3\pi}{2}$$
A: By definition of the centre of mass $(\bar{x},\bar{y})$:
$$ \bar{x}=\frac{\int_0^a xy(x)\,\mathrm{d}x}{\int_0^a y(x)\,\mathrm{d}x} $$
and
$$ \bar{y}=\frac{\int_0^b yx(y)\,\mathrm{d}y}{\int_0^b x(y)\,\mathrm{d}y} $$
where
$$ y(x)=b\sqrt{1-\frac{x^2}{a^2}} $$
and
$$ x(y)=a\sqrt{1-\frac{y^2}{b^2}} $$
A: If you knew the location of the center of mass (COM) for a quarter circle, it'd be easy: you'd just find the scaling-transform that mapped that point to $(4, 2)$. By symmetry, the COM for the quarter-circle must be at some point $(s, s)$ along the line $y = x$. But I cannot see any way, other than actually doing the integration, to find it. The denominator, in this case, is easy -- it's just the area of the quarter circle, i.e., it's $\pi/4$. The numerator is 
\begin{align}
\int_0^1 x \sqrt{1 - x^2} dx &= \frac{1}{2} \int_0^1 2x \sqrt{1 - x^2} dx\\
&= \frac{1}{2} \int_1^0 -\sqrt{u} du \text{, substituting $u = 1 - x^2$}\\
&= \frac{1}{2} \int_0^1 u^{1/2} du \\
&= \frac{1}{2} \left.\frac{u^{3/2}}{3/2}\right|_0^1 \\
&= \frac{1}{2} (\frac{2}{3}) \\
&= \frac{1}{3}. 
\end{align}
That makes the $x$-coord of the centroid (for a circle) be $s = \frac{1/3}{\pi/4} = \frac{4}{3\pi}$. And the centroid is at location $(s, s)$. 
What I'm wondering, and hoping other stackexchangers might be able to suggest, is a geometric argument for this result that makes it completely obvious without integration. Don't ask much, do I? 
A: 
We can also use Pappus' (Second) Centroid Theorem, which states that the volume of a solid of revolution is equal to the area of the region revolved about the axis of symmetry times the circumference of the circular path "swept out" by the centroid of the region,
$$ V \ = \ A \ \cdot \ 2 \pi \  \overline{r} \ \ . $$
The area of the quarter-ellipse (in the diagram for the problem) for which we wish to locate its centroid is $ \ \frac{1}{4}  \cdot  \pi \ ab \ $ .  
If we revolve this region about the $ \ x-$ axis, we obtain a solid of revolution which is half of a "prolate spheroid", an ellipsoid with one of its semi-axes having length $ \ a \ $ and two with length $ \ b \ $ .  Its volume is then $ \ \frac{1}{2} \cdot \ \frac{4 \pi}{3} \ ab^2 \ $ .  The circumference of the centroid's path (in blue) is $ \ 2 \pi \cdot \ \overline{y} \ $ .  By Pappus' Theorem, we conclude
$$ \frac{2 \pi}{3} \ ab^2 \ = \ \frac{\pi}{4}  \ ab \ \cdot \ 2 \pi \cdot \ \overline{y} \ \ \Rightarrow \ \ \overline{y} \ = \ \frac{4}{3 \pi} \ b \ \ . $$
By similar reasoning, revolving the quarter-ellipse about the $ \ y-$ axis generates half of an "oblate spheroid" with one semi-axis of length $ \ b \ $ and two with length $ \ a \ $ ; the circumference of the centroid path (in green) is $ \ 2 \pi \cdot \ \overline{x} \ $ .  Hence,
$$ \frac{2 \pi}{3} \ a^2b \ = \ \frac{\pi}{4}  \ ab \ \cdot \ 2 \pi \cdot \ \overline{x} \ \ \Rightarrow \ \ \overline{x} \ = \ \frac{4}{3 \pi} \ a \ \ . $$
The Centroid Theorems have a direct connection to the moment integrals used to compute the coordinates of a centroid, although the means to describe them in this way was not available to Pappus.
Answering the question for the problem is then a matter of solving
$$ 4 \ = \ \frac{4}{3 \pi} \ a \ \ \ \text{and}  \ \ \ 2 \ = \ \frac{4}{3 \pi} \ b \ \ , $$
as also shown in other posted answers here.
