(1) [図] 図の囲まれた部分 (2) ∬D(x2+y2) dydx=2∫−10∫01+x(x2+y2) dydx+2∫01∫01−x(x2+y2) dydx\iint_D (x^2+y^2)\,dydx = 2\int_{-1}^0\int_0^{1+x}(x^2+y^2)\,dydx + 2\int_0^1\int_0^{1-x}(x^2+y^2)\,dydx∬D(x2+y2)dydx=2∫−10∫01+x(x2+y2)dydx+2∫01∫01−x(x2+y2)dydx =2∫−10(x2+x3+13(1+x)3)dx+2∫01(x2−x3+13(1−x)3)dx= 2\int_{-1}^0\left(x^2+x^3+\frac{1}{3}(1+x)^3\right)dx + 2\int_0^1\left(x^2-x^3+\frac{1}{3}(1-x)^3\right)dx=2∫−10(x2+x3+31(1+x)3)dx+2∫01(x2−x3+31(1−x)3)dx =2[13x3+14x4+112(1+x)4]−10+2[13x3−14x4−112(1−x)4]01= 2\left[\frac{1}{3}x^3+\frac{1}{4}x^4+\frac{1}{12}(1+x)^4\right]_{-1}^0 + 2\left[\frac{1}{3}x^3-\frac{1}{4}x^4-\frac{1}{12}(1-x)^4\right]_0^1=2[31x3+41x4+121(1+x)4]−10+2[31x3−41x4−121(1−x)4]01 =2(13−14+112)+2(13−14+112)=23= 2\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{12}\right) + 2\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{12}\right) = \frac{2}{3}=2(31−41+121)+2(31−41+121)=32