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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

user1211030
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  • There are two possible meanings of centre of mass. It is clear that you are not dealing with the ellipse, but with the first quadrant part. You might be interested in the part of the first quadrant that is inside the ellipse, or (less likely) you might be interested in the centre of mass of the curve. – André Nicolas Jul 06 '14 at 19:08
  • As shown in the image, i'm interested in the centre of mass inside the elliplse area of the first quadrant. – user1211030 Jul 06 '14 at 19:10
  • If it is the two-dimensional region, the area is as you wrote. Easy change of variable shows the area is $ab$ times the area of a quarter-circle of radius $1$, so it is $\pi ab/4$. For the moment about the $y$-axis, the integration is by substitution. – André Nicolas Jul 06 '14 at 19:12
  • Would you be so kind and give me 1 or 2 equations to start with please. How exactly do you get $\pi ab/4$ – user1211030 Jul 06 '14 at 19:19
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    Can you find the center of mass of a unit circle? Multiply by $a$ respectively $b$ to find the center of mass of an ellipse. – Klaas van Aarsen Jul 06 '14 at 19:29
  • Start with the integral you wrote down, make the change of variable $x=at$. We end up with $ab\int_0^1\sqrt{1-t^2},dt$. You can recognize the integral as the area of a quarter-circle of radius $1$, which you know. Or you can let $t=\sin\theta$. – André Nicolas Jul 06 '14 at 19:33
  • But a much better approach is the "stretching" approach of @I like Serena. – André Nicolas Jul 06 '14 at 19:34

4 Answers4

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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 $$\bar{x}=\frac{4a}{3\pi}=4 \implies a=3\pi$$ $$\bar{y}=\frac{4b}{3\pi}=2 \implies b=\frac{3\pi}{2}$$

mike
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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 $$ x(y)=a\sqrt{1-\frac{y^2}{b^2}} $$

lemon
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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?

John Hughes
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enter image description here

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.

colormegone
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