Double integral of arctan in polar coordinates

[LunaFly]

Homework Statement

Evaluate the integral using polar coordinates:

∫∫arctan(y/x) dA

Where R={ (x,y) | 1≤ x² + y² ≤ 4, 0≤y≤x

Homework Equations

X=rcos(T)
Y=rsin(T)
r²=x² +y²

The Attempt at a Solution

First thing was drawing a picture of R, which I think looks like a ring 1 unit thick centered at the origin and with a radius between 1 and 2. Since y≤x, the ring is sliced through the middle by the line y=x, and only the lower half in quadrants 1, 3, and 4 is a part of R.

I rewrote the region as R= {(r, T) | -3pi/4 ≤ T ≤ pi/4, 1 ≤ r ≤ 2}.

After evaluating r*arctan(tan(T))drdT at the above limits, I end up with 3, while the book says the answet is 3pi²/64. I think I am translating the region incorrectly, but don't see how else to describe it.

Thanks for the help!

 

[SammyS]

Homework Statement

Evaluate the integral using polar coordinates:

∫∫arctan(y/x) dA

Where R={ (x,y) | 1≤ x² + y² ≤ 4, 0≤y≤x

Homework Equations

X=rcos(T)
Y=rsin(T)
r²=x² +y²

The Attempt at a Solution

First thing was drawing a picture of R, which I think looks like a ring 1 unit thick centered at the origin and with a radius between 1 and 2. Since y≤x, the ring is sliced through the middle by the line y=x, and only the lower half in quadrants 1, 3, and 4 is a part of R.

Hello LunaFly. Welcome to PF !

Notice that 0 ≤ y ≤ x , so that not only is x ≥ y, but also y ≥ 0 and x ≥ 0.

I rewrote the region as R= {(r, T) | -3pi/4 ≤ T ≤ pi/4, 1 ≤ r ≤ 2}.

After evaluating r*arctan(tan(T))drdT at the above limits, I end up with 3, while the book says the answet is 3pi²/64. I think I am translating the region incorrectly, but don't see how else to describe it.

Thanks for the help!

 

[LunaFly]

Solved!

Ah overlooking that small detail really messed up my work!

Thanks for pointing out my mistake !