How to Determine the Centroid of a Complex Plane Area in Calculus?

[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

[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!

 

[AKG]

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

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

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).

 

[M98Ranger]

First of all, good luck on your final exams! Calculus can be challenging, but with practice and determination, you can definitely do well on your exams.

To solve this problem, you can use the formula for finding the centroid of a plane area:

x_{c} = \frac{\int x\cdot f(x)dx}{\int f(x)dx}

y_{c} = \frac{\int y\cdot f(y)dy}{\int f(y)dy}

Where f(x) and f(y) represent the functions that define the boundaries of the plane area. In this case, f(x) = \sqrt{4-\frac{4}{9}x^2} and f(y) = \left({2^{2/3}-x^{2/3}}\right)^{3/2}.

Using this formula, you can find the centroid of the entire plane area without having to split it into two separate planes. The integrals may still be complex, but it will be easier than trying to solve them separately for each plane.

Another alternative is to use polar coordinates to solve this problem. Converting the equations from Cartesian to polar coordinates can make the integration simpler. You can then use the formula for finding the centroid in polar coordinates:

x_{c} = \frac{\int_{\theta_{1}}^{\theta_{2}} \frac{1}{2}r^{2}\cdot\cos\theta\cdot f(r,\theta)d\theta}{\int_{\theta_{1}}^{\theta_{2}} \frac{1}{2}r^{2}\cdot f(r,\theta)d\theta}

y_{c} = \frac{\int_{\theta_{1}}^{\theta_{2}} \frac{1}{2}r^{2}\cdot\sin\theta\cdot f(r,\theta)d\theta}{\int_{\theta_{1}}^{\theta_{2}} \frac{1}{2}r^{2}\cdot f(r,\theta)d\theta}

Where f(r,\theta) represents the polar form of the equations defining the boundaries of the plane area. This method may be easier and more efficient, but it requires a good understanding of polar coordinates.

I hope these suggestions help you in solving the problem. Keep practicing and don't give up, you got this!