17、已知平面曲线 = (x,y)|sqrt {1-{y)^2}leqslant xleqslant 1,-1leqslant yleqslant 1} , 计算 iint dfrac (x)(sqrt {{x)^2+(y)^2}}dxdy-|||-[答案] sqrt (2)+ln (1+sqrt (2))-2

解析:积分区域D如图所示, $sqrt {(1-y^{2})}$≤x≤1, 0≤y≤1 $sqrt {(1-y^{2})}$≤x≤1, -1≤y≤0 $sqrt {(1-y^{2})}$≤x≤1, 0≤y≤1 $sqrt {(1-y^{2})}$≤x≤1, -1≤y≤0 $sqrt {(1-y^{2})}$≤x≤1, 0≤y≤1 $sqrt {(1-y^{2})}$≤x≤1, -1≤y≤0 $sqrt {(1-y^{2})}$≤x≤1, 0≤y≤1 $sqrt {(1-y^{2})}$≤x≤1, -1≤y≤0 $sqrt {(1-y^{2})}$≤x≤1, 0≤y≤1 $sqrt {(1-y^{2})}$≤x≤1, -1≤y≤0 $sqrt {(1-y^{2})}$≤x≤1, 0≤y≤1 $sqrt {(1-y^{2})}$≤x≤1, -1≤y≤0 $sqrt {(1-y^{2})}$≤x≤1, 0≤y≤1 $sqrt {(1-y^{2})}$≤x≤1, -1≤y≤0 $sqrt {(1-y^{2})}$≤x≤1, 0≤y≤1 $sqrt {(1-y^{2})}$≤x≤1, -1≤y≤0 $sqrt {(1-y^{2})}$≤x≤1, 0≤y≤1 $sqrt {(1-y^{2})}$≤x≤1, -1≤y≤0 $sqrt {(1-y^{2})}$≤x≤1, 0≤y≤1 $sqrt {(1-y^{2})}$≤x≤1, -1≤y≤0 $sqrt {(1-y^{2})}$≤x≤1, 0≤y≤1 $sqrt {(1-y^{2})}$≤x≤1, -1≤y≤0 $sqrt {(1-y^{2})}$≤x≤1, 0≤y≤1 $sqrt {(1-y^{2})}$≤x≤1, -1≤y≤0 $sqrt {(1-y^{2})}$≤x≤1, 0≤y≤1 $sqrt {(1-y^{2})}$≤x≤1, -1≤y≤0 $sqrt {(1-y^{2})}$≤x≤1, 0≤y≤1 $sqrt {(1-y^{2})}$≤x≤1, -1≤y≤0 $sqrt {(1-y^{2})}$≤x≤1, 0≤y≤1 $sqrt {(1-y^{2})}$≤x≤1, -1≤y≤0 $sqrt {(1-y^{2})}$≤x≤1, 0≤y≤1 $sqrt {(1-y^{2})}$≤x≤1, -1≤y≤0 $sqrt {(1-y^{2})}$≤x≤1, 0≤y≤1 $sqrt {(1-y^{2})}$≤x≤1, -1≤y≤0 $sqrt {(1-y^{2})}$≤x≤1, 0≤y≤1 $sqrt {(1-y^{2})}$≤x≤1, -1≤y≤0 $sqrt {(1-y^{2})}$≤x≤1, 0≤y≤1 $sqrt {(1-y^{2})}$≤x≤1, -1≤y≤0 $sqrt {(1-y^{2})}$≤x≤1, 0≤y≤1 $sqrt {(1-y^{2})}$≤x≤1, -1≤y≤0 $sqrt {(1-y^{2})}$≤x≤1, 0≤y≤1 $sqrt {(1-y^{2})}$≤x≤1, -1≤y≤0 $sqrt {(1-y^{2})}$≤x≤1, 0≤y≤1 $sqrt {(1-y^{2})}$≤x≤1, -1≤y≤0 $sqrt {(1-y^{2})}$≤x≤1, 0≤y≤1 $sqrt {(1-y^{2})}$≤x≤1, -1≤y≤0 √