Difficult Centroid Problem

[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!!!

[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).