MHB Integral of a portion of spacetime

[Sandra Conor]

Hello, I have difficulty in evaluating this integral. Can anyone assists?

$\frac{1}{a_0^2}\int_\Sigma\frac{dy'dz'}{\bigg(y'^2+z'^2+\tfrac{1}{(2a_0)^2}\bigg)^2}$

 

[I like Serena]

Hello, I have difficulty in evaluating this integral. Can anyone assists?

$\frac{1}{a_0^2}\int_\Sigma\frac{dy'dz'}{\bigg(y'^2+z'^2+\tfrac{1}{(2a_0)^2}\bigg)^2}$

Hello Symbian and welcome to MHB! ;)

Switch to polar coordinates:
$$\frac{1}{a_0^2}\int_\Sigma\frac{dy'dz'}{\bigg(y'^2+z'^2+\tfrac{1}{(2a_0)^2}\bigg)^2} =
\frac{1}{a_0^2}\int_\Sigma\frac{rd\phi dr}{\bigg(r^2+\tfrac{1}{(2a_0)^2}\bigg)^2}$$

On a disk with radius $R$ this becomes:
$$\frac{1}{a_0^2}\int_\Sigma\frac{rd\phi dr}{\bigg(r^2+\tfrac{1}{(2a_0)^2}\bigg)^2} =
\frac{1}{a_0^2}\int_0^{2\pi}d\phi \int_0^R \frac{r dr}{\bigg(r^2+\tfrac{1}{(2a_0)^2}\bigg)^2} =
\frac{1}{a_0^2}\cdot 2\pi\cdot \left[-\frac{1}{2\bigg(r^2+\tfrac{1}{(2a_0)^2}\bigg)}\right]_0^R
$$

 

[Sandra Conor]

Thank you, Klaas for your assistance.

I would like to seek your view. Initially, I want to evaluate this integral in spacetime?

$$\int_{\Sigma} \frac{dydz}{[a_{o}(y^{2}+z^{2})+2f_{o}y+2g_{o}z+c_{o}]^{2}}$$ where $$a_{o}c_{o}-f_{o}^{2}-g_{0}^{2}=\frac{1}{4}.$$

My way is to define $y':=y+\frac{f_0}{a_0},\,z':=z+\frac{g_0}{a_0}$. Then the quadratic becomes $$a_0(y'^2+z'^2)+c_0-\frac{f_0^2+g_0^2}{a_0}=a_0\bigg(y'^2+z'^2+\tfrac{1}{(2a_0)^2}\bigg).$$So now I will need to evaluate$$\frac{1}{a_0^2}\int_\Sigma\frac{dy'dz'}{\bigg(y'^2+z'^2+\tfrac{1}{(2a_0)^2}\bigg)^2}\cdot$$

Your way is easier with the substitution with the $r^{2}$. Do you think I am correct to say the final answer is as you wrote?

 

[I like Serena]

What I wrote is the next logical step after what you did.
However, the final answer depends on the shape of $\Sigma$.
What is known about $\Sigma$?
Can we assume that it is a disk with some fixed radius?
Or is it something else?

 

[Sandra Conor]

What I wrote is the next logical step after what you did.
However, the final answer depends on the shape of $\Sigma$.
What is known about $\Sigma$?
Can we assume that it is a disk with some fixed radius?
Or is it something else?

Oops. Yes, I forgot to define that. $\Sigma$ is a spacelike 2-surface in spacetime.

 

[Sandra Conor]

Dear Klaas, may I know if you have any idea how can I continue now that it is known that $\Sigma$ is a spacelike 2-surface in spacetime? Thank you.

 

[topsquark]

Just a question. What does "space-time" have to do with this? Are you in x, y, t or is it x, y where one of these is t? I don't see what would be different with your integral.

-Dan

 

[Sandra Conor]

Just a question. What does "space-time" have to do with this? Are you in x, y, t or is it x, y where one of these is t? I don't see what would be different with your integral.

-Dan

Hello Dan. Thank you for your reply. In this spacetime, I would like to find the area of a portion of the spacetime. Hence the integral. The sigma to be precise can also refer to the boundary of the chosen portion of spacetime. Currently, I am trying another way to solve this and that is by completing the square method. Not sure will be a success or not. Calculus sure is my weak spot. How I hope mathematica, maple can do it.

 

[Sandra Conor]

By using completing the square method, I am stuck with this part:

$$\int \frac{dy'+dz'}{((y')^{2}+(z')^{2}+1)^{2}}$$

I would like to intergrate this leaving the answer in equation form. Any ideas how I can do that?