考研高数 | 每日一题（1）

题目

已知 \(f(x, y)\) 为连续函数，则 $ \int_{0}^{\frac{\pi}{4}} \mathrm{d}\theta \int_{0}^{1} f(r\cos\theta, r\sin\theta) r \mathrm{d}r = ?$

(A) \(\int_{0}^{\frac{\sqrt{2}}{2}} \mathrm{d}x \int_{x}^{\sqrt{1-x^2}} f(x,y) \mathrm{d}y\) (B) \(\int_{0}^{\frac{\sqrt{2}}{2}} \mathrm{d}x \int_{0}^{\sqrt{1-x^2}} f(x,y) \mathrm{d}y\)
(C) \(\int_{0}^{\frac{\sqrt{2}}{2}} \mathrm{d}y \int_{y}^{\sqrt{1-y^2}} f(x,y) \mathrm{d}x\) (D) \(\int_{0}^{\frac{\sqrt{2}}{2}} \mathrm{d}y \int_{0}^{\sqrt{1-y^2}} f(x,y) \mathrm{d}x\)

知识点

• 二重积分积分区域的表示

答案

(C)

解析

由题可知，积分区域为：

\(D=\left\{(r,\theta)|0\leqslant\theta\leqslant\frac{\pi}{4},0\leqslant r\leqslant1\right\}.\)

转化成直角坐标系为：

\(D=\left\{(x,y)|0\leqslant y\leqslant\frac{\sqrt{2}}{2},y\leqslant x\leqslant\sqrt{1-y^2}\right\}.\)

则 \(\int_{0}^{\frac{\pi}{4}} \mathrm{d}\theta \int_{0}^{1} f(r\cos\theta,r\sin\theta) r \mathrm{d}r = \int_{0}^{\frac{\sqrt{2}}{2}} \mathrm{d}y \int_{y}^{\sqrt{1-y^2}} f(x,y) \mathrm{d}x.\)

综上可知，本题应选 (C).