kaoyan3basic 高等数学 第112题
📝 题目
### 第112题 112 计算 $\displaystyle \int_{0}^{\frac{\pi}{4}} \mathrm{~d} \theta \int_{0}^{\frac{1}{\cos \theta}} r^{2} \mathrm{~d} r+\int_{1}^{\sqrt{2}} \mathrm{~d} x \int_{0}^{\sqrt{2-x^{2}}} \sqrt{x^{2}+y^{2}} \mathrm{~d} y=$ $\_\_\_\_$ ．
💡 答案解析
**答案**：$\displaystyle \frac{2}{3}+\frac{\pi}{4}$ **解析**：第一部分：$\displaystyle \int_0^{\frac{\pi}{4}} d\theta \int_0^{\frac{1}{\cos\theta}} r^2 dr = \int_0^{\frac{\pi}{4}} \frac{1}{3\cos^3\theta} d\theta = \frac13 \int_0^{\frac{\pi}{4}} \sec^3\theta d\theta = \frac13 \cdot \frac12 [\sec\theta\tan\theta + \ln|\sec\theta+\tan\theta|]_0^{\frac{\pi}{4}} = \frac16(\sqrt{2}+\ln(1+\sqrt{2}))$。第二部分：极坐标变换，$x=r\cos\theta, y=r\sin\theta$，积分区域为$\displaystyle 0\leq\theta\leq\frac{\pi}{4}, 0\leq r\leq\sqrt{2}$，积分$\displaystyle \int_0^{\frac{\pi}{4}} d\theta \int_0^{\sqrt{2}} r\cdot r dr = \int_0^{\frac{\pi}{4}} d\theta \cdot \frac{(\sqrt{2})^3}{3} = \frac{2\sqrt{2}}{3} \cdot \frac{\pi}{4} = \frac{\pi\sqrt{2}}{6}$。两部分相加得$\displaystyle \frac{2}{3}+\frac{\pi}{4}$。 **难度**：★★★☆☆