problem of double integral

[fisher garrry]

Below they are integrating a infinte long sheet of charge. I am wondering about how they integrate along x as the first part of a double integral. The point they want to find the field in is an upwards distance r from the sheet.




below is a calculation on how to integrate with respect to x in the double integral above

Can someone derive why

This is substitution - there is NO WHY to it. It makes the expression simpler to integrate!!

[MATH]x=\sqrt{y^2+r^2}tan(u)[/MATH]
I will add the rest of the integral since I have an issue in the end. The rest of the integral is as follows




Factor out 'x' from the denominator and cancel out the x in the numerator - then apply limits x \(\displaystyle \to \pm \infty \)



I don't get the next step where they add the integration limits:




How do they get this definite integral [MATH]\frac{x}{(y^2+r^2)\sqrt{x^2+y^2+r^2}}+C[/MATH] from [MATH]-\infty[/MATH] to [MATH]\infty[/MATH] to become [MATH]\frac{2}{y^2+r^2}[/MATH]?

Explained above.

 

[Deleted member 4993]

Below they are integrating a infinte long sheet of charge. I am wondering about how they integrate along x as the first part of a double integral. The point they want to find the field in is a distance r from the sheet.
View attachment 15641



below is a calculation on how to integrate with respect to x in the double integral above
View attachment 15643
Can someone derive why

This is substitution - there is NO WHY to it. It makes the expression simpler to integrate!!

[MATH]x=\sqrt{y^2+r^2}tan(u)[/MATH]
I will add the rest of the integral since I have an issue in the end. The rest of the integral is as follows


View attachment 15646

Factor out 'x' from the denominator and cancel out the x in the numerator - then apply limits x \(\displaystyle \to \pm \infty \)



I don't get the next step where they add the integration limits:


View attachment 15647

How do they get this definite integral [MATH]\frac{x}{(y^2+r^2)\sqrt{x^2+y^2+r^2}}+C[/MATH] from [MATH]-\infty[/MATH] to [MATH]\infty[/MATH] to become [MATH]\frac{2}{y^2+r^2}[/MATH]?

Explained above.

Look at my comments in Red.

 

[fisher garrry]

But should not the relation [MATH]x=\sqrt{y^2+r^2}tan(u)[/MATH] obey mathematical relations? Where is the angle u? And why do we use tan and not cos for example. Since it is tan we must have the two sides of the triangle that is not hypotenus. How can I find this relation?

 

[Deleted member 4993]

But should not the relation [MATH]x=\sqrt{y^2+r^2}tan(u)[/MATH] obey mathematical relations? Where is the angle u? And why do we use tan and not cos for example. Since it is tan we must have the two sides of the triangle that is not hypotenus. How can I find this relation?

I am a bit perplexed by your question.

You ask:

But should not the relation [MATH]x=\sqrt{y^2+r^2}tan(u)[/MATH] obey mathematical relations?​

................. of course it must. What makes you believe otherwise? You also ask:

why do we use tan and not cos​

because using "tan" will simplify the expression - and using "cos" will probably complicate things! You also state:

Since it is tan we must have the two sides of the triangle that is not hypotenus.​

Yes we have. In this triangle one side = x and the other side \(\displaystyle \to \ \sqrt{y^2+r^2}\)

 

[fisher garrry]

I am a bit perplexed by your question.

You ask:

But should not the relation [MATH]x=\sqrt{y^2+r^2}tan(u)[/MATH] obey mathematical relations?​

................. of course it must. What makes you believe otherwise?

[MATH]\sqrt{y^2+r^2}[/MATH] is a constant? tan is a function where both sides change in length as function of the angle?

Here is a bit more info but i still don't understand it

 

[fisher garrry]

Sorry about the mess but here is a bit more info:

 

[Deleted member 4993]

tan is a function where both sides change in length as function of the angle?

Here is a bit more info but i still don't understand it

View attachment 15661

[MATH]\sqrt{y^2+r^2}[/MATH] is a constant?

This substitution is being done while you are integrating w.r.t. 'x'. During that action - 'y' is held constant.

You said:

Here is a bit more info but i still don't understand it....

Exactly where are you getting lost?