x？ +y？ ≤x+y

Let x=rcosθ and y=rsinθ.

And then r? ≤r(sinθ+cosθ)

r≤sinθ+cosθ

I=∫∫ xdxdy

=∫(-π/4，3π/4) dθ ∫(0，sinθ+cosθ) r？ cosθdr

= 1/3 ∫(-π/4，3π/4) (sinθ+cosθ)？ cosθdθ

= 1/3 ∫(-π/4，3π/4) (sinθcosθ+2sin？ θcos？ θ+cos？ θ+2sinθcos？ θ)dθ

= 1/6 ∫(-π/4，3π/4) sin2θdθ+ 1/6 ∫(-π/4，3π/4) sin？ 2θdθ+ 1/3 ∫(-π/4，3π/4) cos？ θdθ+2/3 ∫(-π/4，3π/4) sinθcos？ θdθ

= 1/ 12 ∫(-π/4，3π/4)sin 2θd(2θ)+ 1/ 12∫(-π/4，3π/4)( 1-cos 4θ)dθ+ 1/6∫(-π/4，3π/4) ( 1+cos2θ)dθ -θdcosθ

=[- 1/ 12 cos 2θ+θ/ 12- 1/48 sin 4θ+θ/6 ++ 1/ 12 sin 2θ- 1/6(cosθ)^4)|(-π/4，3π/4)

=π/ 16+π/8- 1/ 12 - 1/24+π/48+π/24+ 1/ 12+ 1/24

=(3π+6π+π+2π)/48

= 12π/48

=π/4

Method 2:

x？ +y？ ≤x+y

(x- 1/2)？ +(y- 1/2)？ ≤ 1/2

Let x = 1/2+rcos θ, y = 1/2+rsin θ.

So |=∫∫ xdxdy

=∫(0，2π) dθ∫(0，∨2/2)？ ( 1/2 +rcosθ)rdr

=∫(0,2π)? ( 1/8+∞2/ 12 cosθ)dθ

=( 1/8θ+∞2/ 12 sinθ)|(0，2π)

= 1/8 ×2π

=π/4