Math question! Help needed...?

I = ∫(y = 0 to 2) ∫(x = 0 to √(4-y²) ) 4 - x - y dx dy

Change to polar coordinates, then the domain

0 ≤ x ≤ √(4-y²) and 0 ≤ y ≤ 2

in other words, the quarter circle with radius 2 in the first quadrant,

can be written:

0 ≤ r ≤ 2 and 0 ≤ t ≤ π/2

The Jacobian is r.

The integrand is r(4 - r cos t - r sin t)

I = ∫(t = 0 to π/2) ∫(r = 0 to 2) 4r - r² (cos t + sin t) dr dt =

= ∫(t = 0 to π/2) 2r² - 1/3 r³ (cos t + sin t) | (r = 0 to 2) dt =

= ∫(t = 0 to π/2) 8 - 8/3 (cos t + sin t) dt =

= 8t - 8/3( sin t - cos t) | (t = 0 to π/2) =

= 4π - 8/3 ( sin π/2 - sin 0 - cos π/2 + cos 0 ) =

= 4π - 16/3

Of course, it can be done in Cartesian coordinates as well, but this way is easier.