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$\color{blue}{\S 第七章 第七节 二重积分的概念与性质}$

一、二重积分的定义域性质
二、二重积分的几何意义
三、计算曲顶柱体的体积

$\color{blue}{一、引例}$

曲顶柱体的体积
$给定曲顶柱体:\\ 底:xoy面上的闭区域D\\ 顶:连续曲面z = f(x, y) \geq 0\\ 侧面:以D的边界为准线,母线平行于z轴的柱面,求其体积.$
$解法:类似定积分解决问题的思想:$
$大化小，常代变，近似和，求极限$
$1)分割\\ 用任意曲线网分D为n个区域\\ \Delta{\sigma_1}, \Delta{\sigma_2}, \cdots, \Delta{\sigma_n} \\ 以他们为低把曲顶柱体分为n个小曲顶柱体.\\ 2)近似\\ 在每个\Delta{\sigma_k}中任取一点(\xi_k, \eta_k), 以f(\xi_k, \eta_k)为高，则\\ \Delta{V_k} \approx f(\xi_k, \eta_k)\Delta{\rho_k} \quad (k = 1, 2, \cdots, n) \\ 3)作和\\ V = \sum_{k = 1}^{n}{\Delta{V_k}} \approx \sum_{k = 1}{n}{f(\xi_k, \eta_k)\Delta{\sigma_k}} \\ 4)取极限\\ 定义\Delta{\sigma_k}的直径为\\ \lambda(\Delta{\sigma_k}) = max \lbrace |P_1P_2| |P_1, P_2 \in \Delta{\sigma_k} \rbrace \\ 令 \lambda = max_{1 \leq k \leq n}{\lbrace \lambda (\Delta{\sigma_k}) \rbrace } \\ V = \lim_{\lambda \rightarrow 0}{ \sum_{k = 1}{n}{f(\xi_k, \eta_k) \Delta{\sigma_k} } }$

$\color{blue}{二、二重积分的定义及可积性}$

$定义:设f(x, y)是定义在有界区域D上的有界函数,将区域D\\ 任意分割成n个小区域\Delta{\sigma_k}(k = 1, 2, \cdots, n), 任取一点\\ (\xi_k, \eta_k) \in \Delta{\sigma_k},若存在一个常数I,使\\ I = \lim_{\lambda \rightarrow 0}{\sum_{k = 1}{n}{f(\xi_k, \eta_k)\Delta{\sigma_k}}} \stackrel{记作}{\rightarrow} \iint_D{f(x,y)d{\sigma}} \\ 则称f(x, y)可积,称I为f(x, y)在D上的二重积分(黎曼积分).$
$f(x, y):被积分函数\\ x,y:积分变量\\ d{\sigma}:面积元素\\ f(x, y)d{\sigma}:积分表达式\\ D:积分域 \\ \sum_{k = 1}{n}{f(\xi_k, \eta_k)\Delta{\sigma_k}}:积分和$

$如果f(x, y)在D上可积,可用平行坐标轴的直线来划分区域D,\\ 这时\Delta{\sigma_k} = \Delta{x_k}\Delta{y_k},因此面积元素d\sigma也常记作dxdy,二重积分记作\\ \iint_D{f(x, y)dxdy}$

$引例中曲顶柱体体积:V = \iint_D{f(x, y)d\sigma}$

二重积分存在定理:
$定理1.若函数f(x, y)在有界闭区域D上连续,则f(x, y)在D上可积.$
$定理2.若有界函数f(x, y)在有界闭区域D上除去有限个点或有限个光滑曲线外都连续,则f(x, y)在D上可积.$

$\color{blue}{三、二重积分的性质}$

$1.\iint_D{kf(x, y)d\sigma} = k\iint_D{f(x, y)d\sigma} \quad (k为常数)$
$2.\iint_D{[f(x, y) \pm g(x, y)]d\sigma} = \iint_D{f(x, y)d\sigma} \pm \iint_D{g(x, y)d\sigma}$
$3.\iint_D{f(x, y)d\sigma} = \iint_{D_1}{f(x, y)d\sigma} + \iint_{D_2}{f(x, y)d\sigma}\\ (D = D_1 + D_2,D_1, D_2无公共内点)\\ 即:二重积分对积分区域具有可加性$
$4.若在D上f(x, y) \equiv 1,\sigma为D的面积,则\\ \sigma = \iint_D{1\cdot d\sigma} = \iint_D{\sigma}$
$5.若在D上f(x, y) \leq g(x, y),则\iint_D{f(x, y)d\sigma} \leq \iint_D{g(x, y) d\sigma}$
$特别,由于-|f(x, y)| \leq f(x, y) \leq |f(x, y)| \\ \therefore |\iint_D{f(x, y)d\sigma}| \leq \iint_D{|f(x, y)|d\sigma}$
$6.[估值定理]设M= max_D{f(x, y)}, m = min_D{f(x, y)},D的面积为\sigma,\\ 则有 \quad m\sigma \leq \iint_D{f(x, y)d\sigma} \leq M\sigma$
$7.[积分中值定理]设函数f(x, y)在闭区域D上连续,\\ \sigma为D的面积,则至少存在一点(\xi, \eta) \in D,使\\ \iint_D{f(x, y)d\sigma} = f(\xi, \eta)\sigma$
$证:由性质6可知,\\ m \leq \dfrac{1}{\sigma}\iint_D{f(x, y)d\sigma} \leq M \\ 由连续函数介值定理,至少有一点(\xi, \eta) \in D使\\ f(\xi, \eta) = \dfrac{1}{\sigma}\iint_D{f(x, y)d\sigma} \\ 即：\iint_D{f(x, y)d\sigma} = f(\xi, \eta)\sigma$
$8.[二重积分对称性]设函数f(x, y)在闭区域上连续,域D关于x轴对称,\\ D位于x轴上方的部分为D_1,在D上\\ (1)f(x, -y) = f(x, y),则 \iint_D{f(x, y)d\sigma} = 2\iint_{D_1}{f(x, y)d\sigma}\\ (2)f(x, -y) = -f(x, y),则\iint_D{f(x, y)d\sigma} = 0$
$当区域关于y轴对称,函数关于变量x有奇偶函性时,仍有类似结果.$
$如,D_1为圆域D:x^2 + y^2 \leq 1在第一象限部分,则有\\ \iint_D{(x^2 + y^2)dxdy} = 4\iint_{D_1}{(x^2 + y^2)dxdy}$

练习
$1.比较下列积分值得大小关系: \\ I_1 = \iint_{x^2 + y^2 \leq 1}{|xy|dxdy} \\ I_2 = \iint_{|x| + |y| \leq 1}{|xy|dxdy} \\ I_3 = \int_{-1}^{1}\int_{-1}^{1}{|xy|dxdy}$
$解:I_1的积分区域为半径为1的圆，积分区域面积A_1 = \pi\\ I_2的积分区域为半径为 \sqrt{1 - 2|xy|}的圆，积分区域面积A_2 < \pi\\ I_3的积分区域为边长为2的正方形, 积分区域面积A_3 = 4\\ 积分函数f(x, y) = |xy| \geq 0 \\ \therefore I_3 > I_1 > I_2$

$2.设D是第二象限的一个有界闭域,且0 < y < 1, 则\\ I_1 = \iint_D{yx^3d\sigma}, I_2 = \iint_D{y^2x^3d\sigma}, I_3 = \iint_D{y^{\frac{1}{2}}x^3d\sigma}\\ 的大小顺序为(\ \ {\color{blue}{D} }\ \ )$
$A.I_1 \leq I_2 \leq I_3; \quad B.I_2 \leq I_1 \leq I_3; \\ C.I_3 \leq I_2 \leq I_1; \quad D.I_3 \leq I_1 \leq I_2.$
$解:y > 0, x^3 < 0\\ y^{\frac{1}{2}} > y > y^2 \\ y^{\frac{1}{2}}x^3 < yx^3 < y^2x^3 \\ \therefore I_3 < I_1 < I_2$

$3.计算I= \int_0^{\frac{\pi}{2}}\int_0^{\frac{\pi}{2}}{\sin{(x + y)}dxdy}$
$解:\\ I= \int_0^{\frac{\pi}{2}}\int_0^{\frac{\pi}{2}}{\sin{(x + y)}dxdy} \\ = \int_0^{\frac{\pi}{2}}dy\int_0^{\frac{\pi}{2}}{d\sin{(x + y)}dx} \\ = \int_0^{\frac{\pi}{2}}{[-\cos{(x + y)}]_0^{\frac{\pi}{2}} dy} \\ = \int_0^{\frac{\pi}{2}}{[\sin{y} + \cos{y}] dy} \\ = [\sin{y} - \cos{y}]_0^{\frac{\pi}{2}} \\ = 2$