高等微积分（2） 第8次作业
Chasse_neige
1.给定 \(0 < a < b, 0 < c < d\)，设四条曲线 \(xy = a, xy = b, y = cx\) 以及 \(y = dx\) 在第一象限内围成的平面区域为 \(D\)。计算 \(D\) 的面积。
作换元 $$ \begin{aligned} u &= xy \\ v &= \frac{y}{x} \end{aligned} $$ 此时积分区域 \(\Phi^{-1} (D)\) 变为 \(u \in [a,b], v \in [c,d]\)
利用换元公式 $$ \begin{aligned} S &= \iint_{D} dx dy = \iint_{\Phi^{-1} (D)} \begin{vmatrix} y & - \frac{y}{x^{2}} \\ x & \frac{1}{x} \end{vmatrix}^{-1} du dv \\ &= \iint_{\Phi^{-1} (D)} \begin{vmatrix} \sqrt{uv} & - \frac{v}{\sqrt{\frac{u}{v}}} \\ \sqrt{\frac{u}{v}} & \frac{1}{\sqrt{\frac{u}{v}}} \end{vmatrix}^{-1} du dv = \iint_{\Phi^{-1} (D)} \frac{1}{2v} du dv \\ &= \frac{b - a}{2} \ln \frac{d}{c} \end{aligned} $$ 2.设四条曲线 \(x^2 - y^2 = 1, x^2 - y^2 = 4, \frac{x^2}{4} + y^2 = 1\) 以及 \(\frac{x^2}{4} + y^2 = 4\) 在第一象限内围成的平面区域为 \(D\)。计算积分 $$ \iint_D \frac{xy}{x^2 - y^2} dxdy $$
作换元 $$ \begin{aligned} u &= x^{2} - y^{2} \\ v &= \frac{x^{2}}{4} + y{2} \end{aligned} $$ 此时积分区间变为 \(\Phi^{-1} (D)\) ，即 \(u \in [1,4], v \in [1,4]\)
利用换元公式 $$ \begin{aligned} \iint_D \frac{xy}{x^2 - y^2} dxdy &= \iint_{\Phi^{-1} (D)} (f \circ \Phi) \det (J_{\Phi}) du dv \\ &= \iint_{\Phi^{-1} (D)} \frac{\frac{4}{5} \sqrt{(u + v) (v - \frac{u}{4})}}{u} \begin{vmatrix} \frac{1}{\sqrt{5 (u + v)}} & - \frac{1}{4} \frac{1}{\sqrt{5 (v - \frac{u}{4})}} \\ \frac{1}{\sqrt{5 (u + v)}} & \frac{1}{\sqrt{5 (v - \frac{u}{4})}} \end{vmatrix} du dv \\ &= \iint_{\Phi^{-1} (D)} \frac{1}{5u} du dv = \frac{6}{5} \ln 2 \end{aligned} $$ 3.给定 \(a, b, c > 0\)，令 $$ V = {(x, y, z) | \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} \leq 1, z \geq 0} $$ 计算三重积分 $$ \iiint_V zdxdydz $$
作换元 $$ \begin{aligned} x &= a r \sin \theta \cos \phi \\ y &= b r \sin \theta \sin \phi \\ z &= c r \cos \theta \end{aligned} $$ 其中 \(r \in [0,1], \theta \in [0, \frac{\pi}{2}] , \phi \in [0, 2 \pi)\)
利用换元公式 $$ \begin{aligned} \iiint_V zdxdydz &= \iiint_{\Phi^{-1} (V)} (f \circ \Phi) \det (J_{\Phi}) dr d\theta d\phi \\ &= \iiint_{\Phi^{-1} (V)} c r \cos \theta abc r^{2} \sin \theta dr d \theta d \phi \\ &= \iiint_{\Phi^{-1} (V)}abc^{2} r^{3} dr \sin \theta \cos \theta d \theta d \phi \\ &= abc^{2} \cdot \frac{1}{4} \cdot \frac{1}{2} \cdot 2 \pi = \frac{\pi}{4} abc^{2} \end{aligned} $$ 4.考虑三维区域 $$ V = {(x, y, z) | x^2 + y^2 + z^2 + xy + yz + zx \leq 1} $$ 计算 \(V\) 的体积 (提示: 把 \(V\) 的定义式配方, 然后适当换元) $$ \begin{aligned} &x^2 + y^2 + z^2 + xy + yz + zx \leq 1 \\ &(\frac{\sqrt{2}}{2} x + \frac{\sqrt{2}}{2} y)^{2} + (\frac{\sqrt{2}}{2} y + \frac{\sqrt{2}}{2} z)^{2} + (\frac{\sqrt{2}}{2} x + \frac{\sqrt{2}}{2} z)^{2} \leq 1 \end{aligned} $$ 所以作换元 $$ \begin{aligned} u &= \frac{\sqrt{2}}{2} y + \frac{\sqrt{2}}{2} z \\ v &= \frac{\sqrt{2}}{2} x + \frac{\sqrt{2}}{2} z \\ w &= \frac{\sqrt{2}}{2} x + \frac{\sqrt{2}}{2} y \end{aligned} $$ 利用换元公式 $$ \begin{aligned} \iiint_V dxdydz &= \iiint_{\Phi^{-1} (V)} (f \circ \Phi) \det (J_{\Phi}) du dv dw \\ &= \iiint_{\Phi^{-1} (V)} \begin{vmatrix} 0 & \frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} \\ \frac{\sqrt{2}}{2} & 0 & \frac{\sqrt{2}}{2} \\ \frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} & 0 \end{vmatrix}^{-1} du dv dw \\ &= \frac{1}{\frac{\sqrt{2}}{2}} \cdot \frac{4 \pi}{3} = \frac{4 \sqrt{2} \pi}{3} \end{aligned} $$ 5.设 \(f\) 在矩形区域 \([a, b] \times [c, d]\) 上连续，且有连续的偏导函数 \(\frac{\partial f}{\partial x}\)。
(1) 证明: 对任何 \(x_0 \in [a, b]\)，有 $$ \int_c^d f(x_0, y) dy = \int_c^d f(a, y) dy + \int_a^{x_0} dx \int_c^d \frac{\partial f(x, y)}{\partial x} dy $$
证明
定义函数 \(g (x) = f (x, y_{0})\)
容易得知 $$ \left. \frac{d}{dx} g (x) \right|_{x = x_{0}} = \left. \frac{\partial f}{\partial x} \right|_{ (x,y) = (x_{0}, y_{0})} $$ 由于偏导数 \(\frac{\partial f}{\partial x}\) 连续，所以 \(g'\) 连续，对 \(g (x)\) 使用一元的 N-L 公式 $$ g (x_{0}) = g (a) + \int_{a}^{x_{0}} g' (x) dx $$ 所以 $$ f (x_{0}, y) = f (a, y) + \int_{a}^{x_{0}} \frac{\partial f (x, y)}{\partial x} dx $$
由于偏导数 \(\frac{\partial f}{\partial x}\) 连续，所以它可积。利用 Fubini 定理 $$ \int_{c}^{d} dy \int_{a}^{x_{0}} \frac{\partial f (x, y)}{\partial x} dx = \int_a^{x_0} dx \int_c^d \frac{\partial f(x, y)}{\partial x} dy $$ 所以 $$ \int_c^d f(x_0, y) dy = \int_c^d f(a, y) dy + \int_a^{x_0} dx \int_c^d \frac{\partial f(x, y)}{\partial x} dy $$ (2) 对每个 \(x \in [a, b]\)，定义函数 \(g(x) = \int_c^d f(x, y) dy\)。证明: \(g'(x) = \int_c^d \frac{\partial f(x, y)}{\partial x} dy\)。
证明 $$ \begin{aligned} g' (x) &= \lim_{h \to 0} \frac{g (x + h) - g (x)}{h} = \lim_{h \to 0} \frac{1}{h} \left( \int_{c}^{d} f (x + h, y) dy - \int_{c}^{d} f (x, y) dy \right) \\ &= \lim_{h \to 0} \int_{c}^{d} \frac{f (x + h, y) - f(x, y)}{h} dy \end{aligned} $$ 现在证明积分和极限符号的可交换性
由微分中值定理，存在 \(\theta \in [0, 1]\)，使得 $$ \frac{f (x + h, y) - f(x, y)}{h} = \frac{\partial}{\partial x} f (x + \theta h, y) $$ 由于偏导 \(\frac{\partial f}{\partial x}\) 在闭区间上连续，所以原函数一定可积 $$ g' (x) = \lim_{h \to 0} \int_{c}^{d} \frac{\partial}{\partial x} f (x + \theta h, y) dy $$ 利用 \(\frac{\partial f}{\partial x}\) 的连续性与（1）中的结论 $$ \lim_{h \to 0} \int_{c}^{d} \frac{\partial}{\partial x} f (x + \theta h, y) dy = \int_{c}^{d} \frac{\partial}{\partial x} f (x, y) dy + \lim_{h \to 0} \int_{x}^{x + \theta h} dx \int_{c}^{d} \frac{\partial f}{\partial x} (x, y) dy $$ 二最后一项显然为 \(0\) 。所以 $$ g'(x) = \int_c^d \frac{\partial f(x, y)}{\partial x} dy $$ 6.设 \(D\) 是 \(Oxy\) 平面中的区域，其面积为 \(S\)。定义以 \(D\) 为底面，\((0,0,1)\) 为顶点的锥体为 $$ V = {(1-t)(u,v,0) + t(0,0,1)|(u,v,0) \in D, t \in [0,1]} $$ 求 \(V\) 的体积，要求答案用 \(S\) 表示。 $$ V = \iiint_{V} dx dy dz $$ 现在作如下换元 $$ \begin{aligned} x &= (1 - t) u \\ y &= (1 - t) v \\ z &= t \end{aligned} $$ 且满足 \((u,v,0) \in D, t \in [0,1]\)
利用换元公式 $$ \begin{aligned} V &= \iiint_{\Phi^{-1} (V)} \det (J_{\Phi}) du dv dt \\ &= \iiint_{\Phi^{-1} (V)} \begin{vmatrix} 1 - t & 0 & 0 \\ 0 & 1 - t & 0 \\ -u & -v & 1 \end{vmatrix} du dv dt \\ &= \iiint_{\Phi^{-1} (V)} (1 - t)^{2} du dv dt = \iint_{D} du dv \int_{0}^{1} (1 - t)^{2} dt = \frac{1}{3} S \end{aligned} $$
7.给定非负整数 \(a,b,c\)。令 $$ Q = {(x,y,z) \in R^3 | x+y+z \leq 1, x \geq 0, y \geq 0, z \geq 0} $$ 计算三重积分 $$ \iiint_Q x^ay^bz^c dxdydz $$ 利用 Fubini 定理的推论 $$ \begin{aligned} \iiint_{Q} x^{a} y^{b} z^{c} dx dy dz &= \int_{0}^{1} z^{c} dz \iint_{D_{z}} x^{a} y^{b} dx dy \\ &= \int_{0}^{1} z^{c} dz \int_{0}^{1 - z} y^{b} dy \int_{0}^{1 - z - y} x^{a} dx \\ &= \int_{0}^{1} z^{c} dz \int_{0}^{1 - z} \frac{1}{1 + a} y^{b} (1 - z -y)^{a + 1} dy \\ &= \frac{1}{a + 1} \int_{0}^{1} z^{c} dz \int_{0}^{1 - z} y^{b} \sum_{n = 0}^{a + 1} C_{a + 1}^{n} (1 - z)^{n} (- y)^{a + 1 - n} dy \\ &= \frac{1}{a + 1} \int_{0}^{1} z^{c} dz \sum_{n = 0}^{a + 1} C_{a + 1}^{n} (1 - z)^{n} (-1)^{a + 1 - n} \frac{1}{a + b + 2 - n} (1 - z)^{a + b + 2 - n} \\ &= \frac{1}{a + 1} \sum_{n = 0}^{a + 1} C_{a + 1}^{n} \frac{(-1)^{a + 1 - n}}{a + b + 2 - n} \int_{0}^{1} z^{c} (1 - z)^{a + b + 2} dz \end{aligned} $$ 因为 $$ \begin{aligned} \int_{0}^{1} z^{c} (1 - z)^{a + b + 2} &= \left. \frac{1}{c + 1} z^{c + 1} (1 - z)^{a + b + 2} \right|_{z = 0}^{z = 1} + \frac{a + b + 2}{c + 1} \int_{0}^{1} z^{c + 1} (1 - z)^{a + b + 1} dz \\ &= \frac{a + b + 2}{c + 1} \int_{0}^{1} z^{c + 1} (1 - z)^{a + b + 1} dz \end{aligned} $$ 一直迭代下去 $$ \int_{0}^{1} z^{c} (1 - z)^{a + b + 2} = \frac{a + b + 2}{c + 1} \cdot \frac{a + b + 1}{c + 2} \cdots \frac{1}{a + b + c + 2} \int_{0}^{1} z^{a + b + c + 2} dz = \frac{(a + b + 2)! c!}{(a + b + c + 3)!} $$ 所以原积分等于 $$ \begin{aligned} &\iiint_{Q} x^{a} y^{b} z^{c} dx dy dz \\ &= \sum_{n = 0}^{a + 1} \frac{1}{a + 1} C_{a + 1}^{n} \frac{(-1)^{a + 1 - n}}{a + b + 2 - n} \frac{(a + b + 2)! c!}{(a + b + c + 3)!} \\ &= \sum_{n = 0}^{a + 1} \frac{(a + 1)!}{n! (a + 1 - n)!} \frac{(a + b + 2)! c!}{(a + b + c + 3)!} \frac{(-1)^{a + 1 - n}}{(a + 1) (a + b + 2 - n)} \\ &= \frac{a! c! (a + b + 2)!}{(a + b + c + 3)!} \sum_{n = 0}^{a + 1} \frac{(-1)^{n}}{n! (a + 1 - n)! (b + 1 + n)} \\ &= \frac{a! c!}{(a + b + c + 3)!} \sum_{n = 0}^{a + 1} \frac{(-1)^{n} (b + n)!}{n!} C_{a + b + 2}^{a + 1 - n} \\ &= \frac{a! b! c!}{(a + b + c + 3)!} \sum_{n = 0}^{a + 1} (-1)^{n} C_{b + n}^{b} C_{a + b + 2}^{b + 1 + n} \end{aligned} $$ 在计算最后的求和，注意到 $$ \sum_{k = 0}^{\infty} (-1)^{k} C_{b + k}^{b} x^{k} = \frac{1}{(1 + x)^{b + 1}} $$
所以 $$ \sum_{k = 0}^{\infty} \sum_{l = 0}^{a + b + 2} (-1)^{k} C_{b + k}^{b} C_{a + b + 2}^{l} x^{k + l} = \sum_{n = 0}^{\infty} \sum_{k = \max {0, n - (a + b + 2) }}^{n} (-1)^{k} C_{b + k}^{b} C_{a + b + 2}^{n - k} x^{n} = (1 + x)^{a + 1} = \sum_{n = 0}^{a + 1} C_{a + 1}^{n} x^{n} $$ 所以令 \(n= a + 1\)，对比系数 $$ C_{a + 1}^{a + 1} = \sum_{k = 0}^{a + 1} (-1)^{k} C_{b + k}^{b} C_{a + b + 2}^{a + 1 - k} = 1 $$ 所以最终结果为 $$ \iiint_{Q} x^{a} y^{b} z^{c} dx dy dz = \frac{a! b! c!}{(a + b + c + 3)!} $$