多重积分

极坐标转换直角坐标系

$$计算 I = \iint r^2 \sin\theta \sqrt{1-r^2\cos2\theta} \mathrm{d}r\mathrm{d}\theta，其中 D = \{(r, \theta) | 0 \leqslant r \leqslant \sec\theta, \, 0 \leqslant \theta \leqslant \frac{\pi}{4} \}.$$

$$\begin{aligned} \text{解：}I &= \iint\limits_{D} r^2 \sin\theta \sqrt{1-r^2(\cos^2\theta-\sin^2\theta)} \mathrm{d}r\mathrm{d}\theta, \\ &= \int_{0}^{1}\mathrm{d}x \int_{0}^{x} y \sqrt{1-x^2+y^2} \mathrm{d}y, \\ &= \frac{1}{3} - \frac{\pi}{16}. \end{aligned}$$

含参未定积分

$$设 f(x) 满足 f(x) = x^2 + x \int_{0}^{x^2} f(x^2-t)\mathrm{d}t + \iint\limits_{D} f(xy)\mathrm{d}x \mathrm{d}y，其中 D 是以 (-1, -1), \, (1, -1), \, (1, 1) 为顶点的三角形，且 f(1) = 0，求 \int_{0}^{1}f(x)\mathrm{d}x$$

$$解：设 A = \iint\limits_{D} f(xy)\mathrm{d}x \mathrm{d}y, \, 则 f(xy) = x^2y^2 + xy \int_{0}^{x^2y^2}f(u)\mathrm{d}u + A，两边同时在 D 上积分得$$

$$\begin{aligned} A &= \iint\limits_{D}x^2y^2\mathrm{d}x\mathrm{d}y + 0 + 2A, \\ A &= -\frac{2}{9}. \end{aligned}$$

$$即 f(x) = x^2 + x\int_{0}^{x^2}f(u)\mathrm{d}u - \frac{2}{9}，令 x = 1 得$$

$$\int_{0}^{1} f(x) \mathrm{d}x = -\frac{7}{9}$$