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#1 Re: Help Me ! » Surface area of a sphere??? » 2015-11-14 01:17:57
Hi,
Basically, you would like to understand why the surface area for sphere is 4*pi*r^2 right.
As far we know, sphere perfectly round geometrical object in three-dimensional space that is the surface of a completely round ball.
So basically, it like super-circle but in way of three dimensional.
As we recall, the circumference of circle is given by 2*pi*r where it define the distance all around the circle.
Given that sphere is super-circle that shaped like round ball, we can assume that the formula of circumference multiplied with 2*r (due to the fact that the center of ball is connected with many line of radius intersect with all the point along the distance around the circle).
As conclusion, why the formula for surface area of sphere is given by 4*pi*r^2 is due to the fact of circumference of circle times with diameter of the circle or 2*r .
*** This is only my point of view. Do check with the website provided by bobbym. Thanks
#2 Re: Help Me ! » help please to understand a step in partial diff. equations » 2015-11-14 00:33:34
Hai.
Maybe you can check this domain to understands the Partial Differentials Equation --> (University of Texas)
https://www.ma.utexas.edu/users/rrother/f2011/m427k/pde.pdf
#3 Help Me ! » Singularities and Residue » 2015-11-14 00:28:29
- Maximoff
- Replies: 1
Can anyone help me on this matter?
The function f(z) has a double pole at z=0 with residue=2 and a simple pole at z=1 also with residue=2.
It is also analytic at all other finite points of the plane and is bounded as |z| -> infinity.
Also f(z=2)=5 and f(z=-1)=2.
Determine the f(z).
______________________________________________________________
I tried by using the Taylor and Laurent Series but when I solve the function back, it given me not the exact value from the above statement.
Thanks.
#4 Re: Help Me ! » The Residue Theorem » 2015-10-26 22:09:52
Hi bobbym..
Could you show the working step on how you get it? I got a less bit of knowledge on this theorem.
Thanks.
#5 Help Me ! » The Residue Theorem » 2015-10-25 20:05:37
- Maximoff
- Replies: 5
Hello everyone. Could anybody help me with this? Its about residue theorem.
Find the residue at z=i of the function
Thanks in advance.
#6 Re: Help Me ! » Confusing math problem. » 2015-10-24 22:45:51
Thanks, zetafunc.
Quite easy it is but to think it outside the box maybe take quite a time.
If I had some idea, i will bring it up. Thank you. 
#7 Re: Help Me ! » Confusing math problem. » 2015-10-24 21:07:14
Thank you for the help, mr Zetafunc. darn, this forum is great. Math sure is fun
The equation goes like these.
I just need to understand how the LHs can change to RHS.
From the basic concept that you've shown, if I'm using the complex conjugate to solve it, it will give a much longer equation.
Hope you willing to teach me again.
*I'm been following your Youtube channel and had been learn enormously new thing. Thanks.
#8 Re: Help Me ! » Confusing math problem. » 2015-10-23 23:06:33
Mr Zetafunc.
Sorry, but one more question. This time about the circuit but in parallel.
This time, I get a little bit confuse because when i try to use the complex conjugate is get a longer equation.
I like to understand how it change from LHS to RHS
\frac{1}{R - j\omega C R^2}{1+ (\omega C R)^2} = \frac{1}{R} + j \omega C
Thank you.
#9 Re: Help Me ! » Confusing math problem. » 2015-10-23 22:29:25
Thank you, zetafunc.
May the Force with you.
#10 Re: Help Me ! » Confusing math problem. » 2015-10-23 22:09:48
I'm trying to solve on how from the equation at LHS can changed to equation at RHS.
From this "1/(R-j/ωC)" into this "(R+j/ωC)/(R^2+1/(ωC)^2)" .
** Y = Z^{-1}= \frac{1}{R + jX} = \left( \frac{1}{R^2 + X^2} \right) \left(R - jX\right) ---> This what I get from the internet. But somehow i don't understand it.
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