# Thread: Centre of mass between 2 curves

1. ## 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
$

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
$

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?

2. Originally Posted by Karl Harder
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
$

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
$

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:

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

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

3. Do you know trigonometric substitution?

When an integral has a form of $\int \sqrt{a^2-x^2}dx$, let $x=a\sin \theta$

4. opps, that should be $g(A)-g(a)$. I had it backwards.

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:

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

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

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

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