double integral in polar coordinates

[stone meadow]

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[sand solar]

Well, how has converting to polar coordinates worked out so far?

[stone meadow]

Got this so far, but I don't know why it's 0 to 1 instead of -1 to 1

[sand solar]

Think about the integral as a limit of sums

The original integral summed the function values over these vertical strips

The conversion to polar coordinates changes the summing to be over these radial sectors

If the radius is allowed to vary from -1 to 1, then the summation will be over sectors formed between successive diameters

which is not what we wanted in the first place

[jolly torrent]

Another way to see it is that you can compute only the part where x goes from 0 to 1

Because the function you integrate is even with respect to x

That is a perfectly legal argument to just double the integral where x goes from 0 to 1 instead

[sand solar]

That's not the point
They are asking why the change of coordinates to polar would result in integrating w.r.t. r only from 0 to 1, not -1 to 1
Your proposal only halves the domain of integration for the angular argument upon change of coordinates, not the radial

[jolly torrent]

Ah my bad, I apologize for the confusion

Then I will rephrase it in another way

There is a requirement for variable changes that impose that the coordinate change should be bijective

You can see on your board that r² = 1 for instance will yield two solutions, but in reality you must pick one of the two solutions anyway

Conventionally we only use the nonnegative radii for the reason that is explained by @sand solar