Question

10．$\underset{∬}{D}$$\frac{y}{{x}^{2}+{y}^{2}}$dσ，D是由y²=x，y=x及y=$\sqrt{3}$圍成的區(qū)域．

Answer 1

分析 根據(jù)二重積分的性質(zhì)和法則計(jì)算即可．

解答 解：由$\left\{\begin{array}{l}{{y}^{2}=x}\\{y=x}\end{array}\right.$解得$\left\{\begin{array}{l}{x=1}\\{y=1}\end{array}\right.$，由$\left\{\begin{array}{l}{{y}^{2}=x}\\{y=\sqrt{3}}\end{array}\right.$，解得$\left\{\begin{array}{l}{x=3}\\{y=\sqrt{3}}\end{array}\right.$，
∴$\underset{∬}{D}$$\frac{y}{{x}^{2}+{y}^{2}}$dσ=${∫}_{1}^{\sqrt{3}}$dy${∫}_{y}^{{y}^{2}}$$\frac{y}{{x}^{2}+{y}^{2}}$dx=$\frac{\sqrt{3}π}{12}$-$\frac{1}{2}$ln2

點(diǎn)評(píng) 本題考查了二重積分的性質(zhì)和計(jì)算，屬于基礎(chǔ)題．