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)

$\displaystyle I_{x}=2\int_{0}^{a}(\sqrt{A^2-x^2}A^2-\sqrt{a^2-x^2}a^2)dx$

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

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

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)

$\displaystyle I_{x}=2\int_{0}^{a}(\sqrt{A^2-x^2}A^2-\sqrt{a^2-x^2}a^2)dx$

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

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

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 $\displaystyle x=Asin(\theta)$ and integrate from there. The final expression will take the form $\displaystyle g(a)-g(A)$ where $\displaystyle g$ is just the definite integral of the function:

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

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

3. Do you know trigonometric substitution?

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

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

Change the varaible by doing the trigonometric substitution $\displaystyle x=Asin(\theta)$ and integrate from there. The final expression will take the form $\displaystyle g(a)-g(A)$ where $\displaystyle g$ is just the definite integral of the function:
$\displaystyle g(x)=\int \sqrt{A^2 - x^2} dx = \frac{1}{2} x \sqrt{A^2-x^2} +\frac{1}{2}A^2\tan^{-1}\frac{x}{\sqrt{A^2-x^2}}+C$
$\displaystyle g(x)=2A^2\int \sqrt{A^2 - x^2} dx = 2A^2(\frac{1}{2} x \sqrt{A^2-x^2} +\frac{1}{2}A^2\tan^{-1}\frac{x}{\sqrt{A^2-x^2}}+C)$
If I substitute $\displaystyle x=A sin\theta$ into the equation then what is the domain of the equation
$\displaystyle I_{x}=2\int_{a}^{A}(\sqrt{A^2-x^2}A^2)dx$