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Math Help - Centre of mass between 2 curves

  1. #1
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    Centre of mass between 2 curves

    I have a question where it asks me to find the centre of mass in the x plane between 2 hemispheres where the centre of both hemispheres passes through the origin. The larger one has a radius A and the smaller one radius a. I have broken the area into 2 parts. A1 and A2.

    For A1 the area used is between the 2 curves from origin to a (being the point where the smaller curve meets the line y=0)

     I_{x}=2\int_{0}^{a}(\sqrt{A^2-x^2}A^2-\sqrt{a^2-x^2}a^2)dx<br />

    For A2 the area used is between the curve from a to A (the area is between the larger hemisphere and y=0):

     I_{x}=2\int_{a}^{A}(\sqrt{A^2-x^2}A^2)dx<br />

    How then do you integrate the functions and then how do you substitute the limits a and A into the problem once you have integrated the functions? do you replace x with a and A?
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  2. #2
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    Quote Originally Posted by Karl Harder View Post
    I have a question where it asks me to find the centre of mass in the x plane between 2 hemispheres where the centre of both hemispheres passes through the origin. The larger one has a radius A and the smaller one radius a. I have broken the area into 2 parts. A1 and A2.

    For A1 the area used is between the 2 curves from origin to a (being the point where the smaller curve meets the line y=0)

     I_{x}=2\int_{0}^{a}(\sqrt{A^2-x^2}A^2-\sqrt{a^2-x^2}a^2)dx<br />

    For A2 the area used is between the curve from a to A (the area is between the larger hemisphere and y=0):

     I_{x}=2\int_{a}^{A}(\sqrt{A^2-x^2}A^2)dx<br />

    How then do you integrate the functions and then how do you substitute the limits a and A into the problem once you have integrated the functions? do you replace x with a and A?
    Change the varaible by doing the trigonometric substitution x=Asin(\theta) and integrate from there. The final expression will take the form g(a)-g(A) where g is just the definite integral of the function:

    g(x)=\int \sqrt{A^2 - x^2} dx = \frac{1}{2} x \sqrt{A^2-x^2}<br />
+\frac{1}{2}A^2\tan^{-1}\frac{x}{\sqrt{A^2-x^2}}+C

    In your situation this is just:
    g(x)=2A^2\int \sqrt{A^2 - x^2} dx = 2A^2(\frac{1}{2} x \sqrt{A^2-x^2}<br />
+\frac{1}{2}A^2\tan^{-1}\frac{x}{\sqrt{A^2-x^2}}+C)
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  3. #3
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    Do you know trigonometric substitution?

    When an integral has a form of \int \sqrt{a^2-x^2}dx, let x=a\sin \theta
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  4. #4
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    opps, that should be g(A)-g(a). I had it backwards.
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  5. #5
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    Quote Originally Posted by adkinsjr View Post
    Change the varaible by doing the trigonometric substitution x=Asin(\theta) and integrate from there. The final expression will take the form g(a)-g(A) where g is just the definite integral of the function:

    g(x)=\int \sqrt{A^2 - x^2} dx = \frac{1}{2} x \sqrt{A^2-x^2}<br />
+\frac{1}{2}A^2\tan^{-1}\frac{x}{\sqrt{A^2-x^2}}+C

    In your situation this is just:
    g(x)=2A^2\int \sqrt{A^2 - x^2} dx = 2A^2(\frac{1}{2} x \sqrt{A^2-x^2}<br />
+\frac{1}{2}A^2\tan^{-1}\frac{x}{\sqrt{A^2-x^2}}+C)

    If I substitute  x=A sin\theta into the equation then what is the domain of the equation

     I_{x}=2\int_{a}^{A}(\sqrt{A^2-x^2}A^2)dx

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