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

[itex]x^{2/3}+y^{2/3}=2^{2/3}[/itex]

[itex]\frac{x^2}{9}+\frac{y^2}{4}=1[/itex]

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:

[itex]y_1 = \sqrt{4-\frac{4}{9}x^2}[/itex]

[itex]y_2 = \left({2^{2/3}-x^{2/3}}\right)^{3/2}[/itex]

Here are the integrals I formulated:

To find the area of plane 1:

[itex]A_1=\int_0^2(y_1-y_2)dx[/itex]

To find the area of plane 2:

[itex]A_2=\int_2^3y_1dx[/itex]

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

[itex]M_{1y}=\int_0^2x(y_1-y_2)dx[/itex]

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

[itex]M_{1x}=\frac{1}{2}\int_0^2(y_1+y_2)(y_1-y_2)dx[/itex]

And similiarly,

[itex]M_{2y}=\int_2^3xy_1dx[/itex]

[itex]M_{2x}=\frac{1}{2}\int_2^3y_1^2dx[/itex]

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

[itex]\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[/itex]

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

[itex]3\arcsin{\frac{2}{3}}+\frac{2\sqrt{5}}{3}[/itex]

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!!!