Using Fubini's theorem, we get:
∬D(x+y)dA=∫02dx∫04−x2(x+y)dy=∫02((xy+y22)∣04−x2)dx=\iint_D(x+y)dA=\int_0^2dx\int_{0}^{\sqrt{4-x^2}}(x+y)dy=\int_0^2((xy+\frac{y^2}{2})\biggr\rvert_{0}^{\sqrt{4-x^2}})dx=∬D(x+y)dA=∫02dx∫04−x2(x+y)dy=∫02((xy+2y2)∣∣04−x2)dx=
=∫02(x4−x2+12(4−x2)))dx=−12∫024−x2d(4−x2)+∫02(2−12x2)dx==\int_0^2(x\sqrt{4-x^2}+\frac12(4-x^2)))dx=-\frac12\int_0^2\sqrt{4-x^2}d(4-x^2)+\int_0^2(2-\frac12x^2)dx==∫02(x4−x2+21(4−x2)))dx=−21∫024−x2d(4−x2)+∫02(2−21x2)dx=
=83+4−43=163=\frac83+4-\frac43=\frac{16}3=38+4−34=316
Answer:d). 163\frac{16}3316
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