Recent content by 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?

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...

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.

Thanks Svein for the ideas. I will check these out. Yes, mathman. The summation sign represent a spacelike 2 surface in spacetime. Its mentioned in the third post. Sorry about that.

I have been thinking of polar form too. The summation sign represent a spacelike 2 surface in spacetime. 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 $$a_{o}c_{o}-f_{o}^{2}-g_{0}^{2}=\frac{1}{4}.$$...

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

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...

Hello, can anyone show me if this integral can be evaluated? ##\frac{1}{a_0^2}\int_\Sigma\frac{dy'dz'}{\bigg(y'^2+z'^2+\tfrac{1}{(2a_0)^2}\bigg)^2}##

Hello, I have difficulty in evaluating this integral. Can anyone assists? $\frac{1}{a_0^2}\int_\Sigma\frac{dy'dz'}{\bigg(y'^2+z'^2+\tfrac{1}{(2a_0)^2}\bigg)^2}$