Force calculation on a spherical body

Morteza

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Jul 20, 2021
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Please find the integration of:

[math]\int \frac{(cos^2(a+b)cos(a)}{x} dx[/math]
Where,
[math]cos(a) = \frac{r^2+R^2-x^2}{2rR}[/math]and
[math]cos(b) = \frac{r^2+x^2-R^2}{2rx}[/math]
Thank you for your help.
 

Zermelo

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Jan 7, 2021
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Hi there, you should read the guidelines and understand that we’re not allowed to solve your problems, or help you before seeing your work. You need to show us what you did first, and where you’re stuck.
I personally would look up the formula for cos(x + y) if I were you, then the numerator should (eventually) simplify to a polynomial.
 

Subhotosh Khan

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Jun 18, 2007
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25,498
Please find the integration of:

[math]\int \frac{(cos^2(a+b)cos(a)}{x} dx[/math]
Where,
[math]cos(a) = \frac{r^2+R^2-x^2}{2rR}[/math]and
[math]cos(b) = \frac{r^2+x^2-R^2}{2rx}[/math]
Thank you for your help.
The integrand is an expression with unbalanced parentheses. Please correct it post the correct expression to be integrated.

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:


Please share your work/thoughts about this problem
 

Morteza

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Jul 20, 2021
Messages
15
Hi there, you should read the guidelines and understand that we’re not allowed to solve your problems, or help you before seeing your work. You need to show us what you did first, and where you’re stuck.
I personally would look up the formula for cos(x + y) if I were you, then the numerator should (eventually) simplify to a polynomial.
Thank you Zermelo for your suggestion

Yes, I did it the same way, but after that, it is a long calculation to get the result and difficult to do that. Actually, I am stuck from the beginning of my work.
 

Subhotosh Khan

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Jun 18, 2007
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Thank you Zermelo for your suggestion

Yes, I did it the same way, but after that, it is a long calculation to get the result and difficult to do that. Actually, I am stuck from the beginning of my work.
You have to show us what you did and EXACTLY where you are stuck.

First thing you need to do is to correct your original post.

Then show us how you had used expansion of cos(x + y) and what did you do after that - even if you know that it is incorrect.
 

Morteza

New member
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Jul 20, 2021
Messages
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You have to show us what you did and EXACTLY where you are stuck.

First thing you need to do is to correct your original post.

Then show us how you had used expansion of cos(x + y) and what did you do after that - even if you know that it is incorrect.
 

Morteza

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Jul 20, 2021
Messages
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[math]= \int \frac{cos^3(a)cos^2(b)+cos(a)sin^2(a)sin^2(b)-2cos^2(a)cos(b)sin(a)sin(b)} {x} dx[/math]and what would be the next step?
 

topsquark

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Aug 27, 2012
Messages
1,267
This is the direct way. I usually find it's the more difficult way but looking at this problem, from experience I think (thought, I checked), that there are a lot of cancellations here. But it's going to take some time to work it out.

First, if
[imath]cos(a) = \dfrac{r^2 + R^2 - x^2}{2rR}[/imath] then [imath]sin(a) = \dfrac{\sqrt{(2rR)^2 - (r^2 + R^2 - x^2)^2}}{2rR}[/imath]
and if
[imath]cos(b) = \dfrac{r^2 + x^2 - R^2}{2rx}[/imath] then [imath]sin(b) = \dfrac{\sqrt{(2rx)^2 - (r^2 + x^2 - R^2)^2}}{2rx}[/imath]

Then
[imath]cos(a + b) = cos(a) ~ cos(b) - sin(a) ~ sin(b)[/imath]
[math] = \left ( \dfrac{r^2 + R^2 - x^2}{2rR} \right ) \cdot \left ( \dfrac{r^2 + x^2 - R^2}{2rx} \right ) - \left ( \dfrac{\sqrt{(2rR)^2 - (r^2 + R^2 - x^2)^2}}{2rR} \right ) \cdot \left ( \dfrac{\sqrt{(2rx)^2 - (r^2 + x^2 - R^2)^2}}{2rx} \right )[/math]
It's a serious mess but the end result of multiplying those square roots is a simple expression: [imath]r^4- 2r^2 (R^2 + x^2) + (R^2 - x^2)^2[/imath].

It's a lot easier from here. Finish out the cos(a + b) calculation, square and multiply by cos(a), then separate everything out in terms of factors of x. Then integrate. It's algebraic busywork but the numerator is just a polynomial in x.

I'll let you fill in the details.

-Dan
 
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