- #1

Sandra Conor

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

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In summary: Sigma\frac{dy'+dz'}{(y')^{2}+(z')^{2}+1}$$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?In summary, according to Dan, the integral can be solved in either spacetime or in polar coordinates. The final answer depends on the shape of $\Sigma$.

- #1

Sandra Conor

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- 0

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

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- #2

I like Serena

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Symbian said:

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

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

$$

- #3

Sandra Conor

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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?

- #4

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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?

- #5

Sandra Conor

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

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?

- #6

Sandra Conor

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- #7

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-Dan

- #8

Sandra Conor

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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.topsquark said:

-Dan

- #9

Sandra Conor

- 9

- 0

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

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

An integral of a portion of spacetime is a mathematical concept used in the study of general relativity. It represents the total amount of energy, matter, or other physical quantities within a specific region of spacetime.

The integral of a portion of spacetime is calculated by integrating the relevant physical quantity over the region of spacetime in consideration. This can be done using mathematical tools such as the Riemann sum or the Lebesgue integral.

The integral of a portion of spacetime plays a crucial role in understanding the distribution of energy and matter in the universe. It is also used in the study of gravitational fields and the curvature of spacetime.

Yes, the integral of a portion of spacetime can be negative. This indicates that there is a deficit of the physical quantity being measured within the region of spacetime. For example, a negative integral of energy density would suggest a deficit of energy within the given region.

The integral of a portion of spacetime is directly related to the curvature of spacetime. In general relativity, the energy and matter distribution within a region of spacetime determine the curvature of that region. Therefore, by calculating the integral of a portion of spacetime, we can gain insights into the curvature of that region.

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