计算I=∫∫D|(x+y)/2-x^2-y^2|dxdy,其中D={(x,y):x^2+y^2≤1}

Question

计算I=∫∫D |(x+y)/2 - x^2 - y^2| dxdy,其中D={(x,y):x^2+y^2≤1}

Answer 1

f(x,y) = (x+y)/2-x^2-y^2 = 1/8-(x-1/4)^2+(y-1/4)^2,
f(x,y)=0 即圆 D1:(x-1/4)^2+(y-1/4)^2=1/8,
在 D1 内 f(x,y)≤0,在 D 内且 D1 外 f(x,y)≥0,则
I =∫∫|(x+y)/2-x^2-y^2|dxdy
=∫∫[(x+y)/2-x^2-y^2]dxdy
-∫∫[(x+y)/2-x^2-y^2]dxdy
=∫∫[(x+y)/2-x^2-y^2]dxdy
-2∫∫[(x+y)/2-x^2-y^2]dxdy
=∫dt∫[r(cost+sint)/2-r^2]rdr
-2∫dt∫[r(cost+sint)/2-r^2]rdr
=∫[(cost+sint)/6-1/4]dt
-2∫dt[r^3(cost+sint)/6-r^4/4]
=∫[(cost+sint)/6-1/4]dt
-2∫[(cost+sint)^4/192]dt
= [(sint-cost)/6-t/4]
-(1/96)∫[1+2sin2t+(sin2t)^2]dt
= -π/2-(1/96)∫[3/2+2sin2t-(1/2)cos4t]dt
= -π/2-(1/96)[3t/2-cos2t-(1/8)sin4t]
= -π/2-π/64 = -33π/64.