题目
已知平面区域 = (x.y)|sqrt {1-{y)^2}leqslant xleqslant 1,-1leqslant yleqslant 1} .计算 (iint )_(D)dfrac (x)(sqrt {{x)^2+(y)^2}}dxdy
题目解答
答案
本题考查了二重积分的计算，属于基础题.
：
$∫∫_D {x \over {x^2 + y^2}} dxdy = \int_{-1}^1 {dy} \int_{\sqrt {1 - y^2}}^1 {x \over {x^2 + y^2}} dx$
$= \int_{-1}^1 {dy} \left( {1 \over y} - {1 \over {\sqrt {1 - y^2}}} \right)$
$= 2\int_0^1 {dy \over y} - 2\int_0^1 {dy \over {\sqrt {1 - y^2}}}$
$= 2\left( \ln \left| {y} \right| - \ln \left| {\sqrt {1 - y^2}} \right| \right) \left| \begin{matrix} {1 \\ 0} \end{matrix} \right.$
$= 2\ln 2$
：
$∫∫_D {x \over {x^2 + y^2}} dxdy = \int_{-1}^1 {dy} \int_{\sqrt {1 - y^2}}^1 {x \over {x^2 + y^2}} dx$
$= \int_{-1}^1 {dy} \left( {1 \over y} - {1 \over {\sqrt {1 - y^2}}} \right)$
$= 2\int_0^1 {dy \over y} - 2\int_0^1 {dy \over {\sqrt {1 - y^2}}}$
$= 2\left( \ln \left| {y} \right| - \ln \left| {\sqrt {1 - y^2}} \right| \right) \left| \begin{matrix} {1 \\ 0} \end{matrix} \right.$
$= 2\ln 2$