# Difficult Centroid Problem

1. May 31, 2004

### agro

This is a problem from a past final calculus test in my university (I'm studying for the finals which will come in around 1 week :) ):

Find the centroid of a plane area in the first quadrant bounded by

$x^{2/3}+y^{2/3}=2^{2/3}$

$\frac{x^2}{9}+\frac{y^2}{4}=1$

and the x axis.

I tried splitting the plane into 2 parts, one from x = 0 to x = 2 (call this plane 1), and the other from x = 2 to x = 3 (call this plane 2). My tactic is to find the centroids of each plane. I can then easily find the centroid of the composite plane. Then I define these equations:

$y_1 = \sqrt{4-\frac{4}{9}x^2}$
$y_2 = \left({2^{2/3}-x^{2/3}}\right)^{3/2}$

Here are the integrals I formulated:

To find the area of plane 1:

$A_1=\int_0^2(y_1-y_2)dx$

To find the area of plane 2:

$A_2=\int_2^3y_1dx$

To find the (first) moment of plane 1 with respect to the y axis:

$M_{1y}=\int_0^2x(y_1-y_2)dx$

To find the moment of plane 1 with respect to the x axis:

$M_{1x}=\frac{1}{2}\int_0^2(y_1+y_2)(y_1-y_2)dx$

And similiarly,

$M_{2y}=\int_2^3xy_1dx$
$M_{2x}=\frac{1}{2}\int_2^3y_1^2dx$

I'm stuck at finding the area. The definite integral which I must evaluate is:

$\int_0^2\sqrt{4-\frac{4}{9}x^2}dx-\int_0^2\left({2^{2/3}-x^{2/3}}\right)^{3/2}dx$

The first term evaluates to (If I had done it correctly)

$3\arcsin{\frac{2}{3}}+\frac{2\sqrt{5}}{3}$

On the other hand, I have no idea how to integrate the second term. Can anyone give me a hint?

Or maybe, the method that I have chosen (dividing it into 2 plane, etc etc) results in a complex calculation. Is there any easier alternatives?

Thanks a lot!!!

2. May 31, 2004

### AKG

I don't think you did the first integration correctly.

$$\int _0 ^2 \sqrt{4 - \frac{4}{9}x^2} dx$$
$$= 2\int _0 ^2 \sqrt{1 - {\left ( \frac{x}{3} \right ) }^2} dx$$
$$= 6\int_{x=0} ^{x=2} \cos ^2 \theta d\theta$$

You can figure out how to evaluate this one, as for the second, I have to sleep. If no one has helped you, I'll try to figure it out tomorrow (it looks tough).