Double integration - switching limits

[laser]

Homework Statement

See description

Relevant Equations

Yes

I get that the bottom answer isn't a constant - but does this physically represent anything? When I set the two answers equal to each other, I get x = +- 1/sqrt(2) and I am wondering if this represents anything significant.

I don't think (mathematically) there is anything wrong with the bottom method - it just doesn't give the desired answer. Is this correct?

 

[PeroK]

I don't think (mathematically) there is anything wrong with the bottom method - it just doesn't give the desired answer. Is this correct?

The second method is invalid. You can't take the dummy variable $x$ outside the integration by $dx$. Instead, you must change the bounds on the integral to be valid for doing the integration by $dx$ first.

 

[laser]

The second method is invalid. You can't take the dummy variable $x$ outside the integration by $dx$. Instead, you must change the bounds on the integral to be valid for doing the integration by $dx$ first.

Ya I'm aware that the proper way of doing it is by changing the bounds such that you have constants on the outer integral. I guess I was looking for an explanation for why you must have constants on the outer integral. And that's because $x$ is a dummy variable like you said if I understand correctly.

 

[PhDeezNutz]

Ya I'm aware that the proper way of doing it is by changing the bounds such that you have constants on the outer integral. I guess I was looking for an explanation for why you must have constants on the outer integral. And that's because $x$ is a dummy variable like you said if I understand correctly.

Doesn’t it make sense that if you want a hard number in the end that each subsequent integration should be more and more restrictive?

I recommend sketching the region you’re integrating over and deciding your new bounds from there.

Switching the order of integration does not always mean you can carry the bounds with you. That only works when you are given solid numbers for bounds of x and y.

 

[laser]

Doesn’t it make sense that if you want a hard number in the end that each subsequent integration should be more and more restrictive?

Yeah it does, I was just wondering.

I recommend sketching the region you’re integrating over and deciding your new bounds from there.

That is what I usually do!

Switching the order of integration does not always mean you can carry the bounds with you. That only works when you are given solid numbers for bounds of x and y.

Fair enough.

 

[PhDeezNutz]

Also for this specific problem, you’re going to have to break the Region into two parts to do the integration.

 

[FactChecker]

Ya I'm aware that the proper way of doing it is by changing the bounds such that you have constants on the outer integral. I guess I was looking for an explanation for why you must have constants on the outer integral. And that's because $x$ is a dummy variable like you said if I understand correctly.

To expand the problem with the second approach a little more:
In the first approach, the inner integral is evaluated for the particular 'x' value in the lower limit, so that x value makes the outer integral make sense.
In the second approach, the inner integral is a constant and 'x' is no longer defined, so the outer integral using 'x' does not make sense. 'x' does not have a defined value for the outer integral.

 

[WWGD]

To reinforce the point on the outer limits of integration, notice the integral spits out a number. If you instead include a formula, no such number will result. Not too deep, but arguably a good heuristic.