高数强化NO18|交换积分次序与坐标系的转换

$$\begin{array}{rl}& \text{解：}{\iint }_{D}xy\mathrm{d}\sigma \\ & ={\int }_0^{\pi }d\theta {\int }_0^{1+\cos \theta }(r\cos \theta )(r\sin \theta )\cdot rdr\\ & ={\int }_0^{\pi }\cos \theta \sin \theta \cdot \frac{(1+\cos \theta {)}^4}{4}d\theta \\ & =-{\int }_0^{\pi }\cos \theta \cdot \frac{(1+\cos \theta {)}^4}{4}d(\cos \theta )\\ & \\ & \text{令}u=\cos \theta ,\text{则当}\theta =0\text{时}u=1,\theta =\pi \text{时}u=-1,\\ & =-{\int }_1^{-1}u\cdot \frac{(1+u{)}^4}{4}du=\frac{1}{4}{\int }_{-1}^{1}u(1+u{)}^{4}du\\ & \\ & \text{展开}(1+u{)}^4=u^4+4u^3+6u^2+4u+1,\\ & =\frac{1}{4}{\int }_{-1}^{1}u(u^4+4u^3+6u^2+4u+1)du\\ & =\frac{1}{4}{\int }_{-1}^1(u^5+4u^4+6u^3+4u^2+u)du\\ & \\ & \text{利用奇偶函数积分性质（奇函数在对称区间积分得0），}\\ & \text{剩余偶次项：}4u^4+4u^2,\\ & =\frac{1}{4}\cdot 2{\int }_0^1(4u^4+4u^2)du=\frac{1}{2}\cdot 4{\int }_0^1(u^4+u^2)du\\ & =2\left({\frac{u^5}{5}+\frac{u^3}{3}|}_0^1\right)=2\left(\frac{1}{5}+\frac{1}{3}\right)=\frac{16}{15}\end{array}$$
$$\begin{array}{rl}& \text{法2：令}t=1+u,\text{则}u=t-1,\text{当}u=-1\text{时}t=0,u=1\text{时}t=2,\\ & \frac{1}{4}{\int }_{-1}^{1}u(1+u{)}^{4}du=\frac{1}{4}{\int }_0^2(t-1)\cdot t^{4}dt\\ & =\frac{1}{4}{\int }_0^2(t^5-t^4)dt\\ & =\frac{1}{4}\left({\frac{t^6}{6}-\frac{t^5}{5}|}_0^2\right)\\ & =\frac{1}{4}\left(\frac{64}{6}-\frac{32}{5}\right)\\ & =8\left(\frac{2}{6}-\frac{1}{5}\right)\\ & =\frac{16}{15}\end{array}$$
$$\begin{array}{rl}& \text{拓：}{\int }_0^{\pi }\sin \theta \cos \theta \cdot \frac{(1+\sin \theta {)}^4}{4}d\theta \\ & ={\int }_0^{\pi }\sin \theta \cdot \frac{(1+\sin \theta {)}^4}{4}d(\sin \theta )\\ & \\ & \text{正确：将积分拆分为}{\int }_0^{\frac{\pi }{2}}+{\int }_{\frac{\pi }{2}}^{\pi }\\ & \\ & \text{① 计算}{\int }_0^{\frac{\pi }{2}}\sin \theta \cdot \frac{(1+\sin \theta {)}^4}{4}d(\sin \theta ):\\ & \text{令}u=\sin \theta ,\text{则}{\int }_0^{1}u\cdot \frac{(1+u{)}^4}{4}du\\ & \text{再令}t=u+1,\text{则}u=t-1,\text{积分变为：}\\ & \frac{1}{4}{\int }_1^2(t-1)t^{4}dt=\frac{1}{4}{\int }_1^2(t^5-t^4)dt\\ & =\frac{1}{4}\left(\frac{t^6}{6}-\frac{t^5}{5}\right){|}_1^2=\frac{1}{4}\left(\frac{64}{6}-\frac{32}{5}-\frac{1}{6}+\frac{1}{5}\right)\\ & =\frac{1}{4}\left(\frac{63}{6}-\frac{31}{5}\right)=\frac{129}{120}\\ & \\ & \text{② 计算}{\int }_{\frac{\pi }{2}}^{\pi }\sin \theta \cdot \frac{(1+\sin \theta {)}^4}{4}d(\sin \theta ):\\ & \text{令}u=\sin \theta ,\text{则}{\int }_1^{0}u\cdot \frac{(1+u{)}^4}{4}du=-{\int }_0^{1}u\cdot \frac{(1+u{)}^4}{4}du\\ & =-\frac{129}{120}\\ & \\ & \text{故原式}=\frac{129}{120}-\frac{129}{120}=0\end{array}$$
$$\begin{array}{rl}& {\int }_0^{\pi }\sin \theta \cos \theta \cdot \frac{(1+\sin \theta {)}^4}{4}d\theta ={\int }_0^{\pi }\sqrt{1-{\cos }^2\theta }\cdot \cos \theta \cdot \frac{(1+\sqrt{1-{\cos }^2\theta }{)}^4}{4}d\theta \\ & \\ & \text{令}u=\cos \theta ,\text{则}d\theta =\frac{-1}{\sqrt{1-u^2}}du,\text{当}\theta =0\text{时}u=1,\theta =\pi \text{时}u=-1,\\ & ={\int }_1^{-1}u\sqrt{1-u^2}\cdot \frac{(1+\sqrt{1-u^2}{)}^4}{4}\cdot \frac{-1}{\sqrt{1-u^2}}du\\ & ={\int }_1^{-1}u\cdot \frac{(1+\sqrt{1-u^2}{)}^4}{4}\cdot (-1)du\\ & \\ & \text{被积函数是关于}u\text{的奇函数（对称区间}[-1,1]\text{），故积分结果为：}\\ & =0\end{array}$$
$$\begin{array}{rl}& 【小结】若题目中以极坐标直接给出积分区域，且被积函数以直角坐标给出，\\ & 一般计算比较简单，直接在极坐标下计算积分即可。\end{array}$$

解：记D1={(x,y)∣−1≤y≤0, −y≤x≤y},D2={(x,y)∣0≤y≤1, −y≤x≤y}.在D1上，yxe12(x2+y2)关于x为奇函数，故∬D1yxe12(x2+y2)dxdy=0.在D2上，yxe12(x2+y2)关于y为奇函数，故∬D2yxe12(x2+y2)dxdy=0∬D2ydxdy=0.∬Dydxdy=∬D1ydxdy=2∬D1ydxdy=2∫−10dy∫0−yydx=2∫−10−y2dy=2⋅(−y33)∣−10=−23 \begin{aligned} &\text{解：记} \\ &D_1 = \left\{ (x,y) \mid -1 \leq y \leq 0,\ -y \leq x \leq y \right\}, \\ &D_2 = \left\{ (x,y) \mid 0 \leq y \leq 1,\ -y \leq x \leq y \right\}. \\ \\ &\text{在}D_1\text{上，}yxe^{\frac{1}{2}(x^2+y^2)}\text{关于}x\text{为奇函数，} \\ &\text{故}\iint_{D_1} yxe^{\frac{1}{2}(x^2+y^2)} dxdy = 0. \\ \\ &\text{在}D_2\text{上，}yxe^{\frac{1}{2}(x^2+y^2)}\text{关于}y\text{为奇函数，} \\ &\text{故}\iint_{D_2} yxe^{\frac{1}{2}(x^2+y^2)} dxdy = 0 \\& \iint_{D_{2}} y dxdy = 0. \\ \\ &\iint_{D} y dxdy = \iint_{D_1} y dxdy = 2\iint_{D_1} y dxdy = 2\int_{-1}^{0} dy \int_{0}^{-y} ydx \\ &= 2\int_{-1}^{0} -y^2 dy = 2 \cdot \left. \left( -\frac{y^3}{3} \right) \right|_{-1}^{0} = -\frac{2}{3} \end{aligned} ​解：记D1​={(x,y)∣−1≤y≤0, −y≤x≤y},D2​={(x,y)∣0≤y≤1, −y≤x≤y}.在D1​上，yxe21​(x2+y2)关于x为奇函数，故∬D1​​yxe21​(x2+y2)dxdy=0.在D2​上，yxe21​(x2+y2)关于y为奇函数，故∬D2​​yxe21​(x2+y2)dxdy=0∬D2​​ydxdy=0.∬D​ydxdy=∬D1​​ydxdy=2∬D1​​ydxdy=2∫−10​dy∫0−y​ydx=2∫−10​−y2dy=2⋅(−3y3​)​−10​=−32​​
$$\begin{array}{rl}& \text{对称点的函数值互为相反数，积分必为}0.\end{array}$$

$$\begin{array}{rl}& \iint x{y}^{3}d\sigma =0\end{array}$$

$$\begin{array}{rl}& \iint x^3{y}^{2}d\sigma =0\end{array}$$

$$\begin{array}{rl}& \iint (x+y)d\sigma =0\end{array}$$

$$\begin{array}{rl}& \iint (x^3+y)d\sigma =0\end{array}$$

$$\begin{array}{rl}& \text{【小结】}(1)\text{对于本身关于两个坐标轴都不对称的区域，也可以考虑先将积分区域分割}\\ & \text{成为具有对称性的若干个子区域，然后再逐一运用对称性；}\\ & \\ & (2)\text{当积分区域关于某一个坐标轴对称时，可以考虑通过函数的奇偶性化简积分,如果}\\ & \text{我们遇到的函数整体不具备奇偶性，则可以逐项讨论每一项的奇偶性,多数情况下，我们}\\ & \text{更加关注的是奇函数的部分，因为这部分的积分为零；}\\ & \\ & (3)\text{积分区域关于}y\text{轴对称的时候，讨论被积函数关于}x\text{的奇偶性；积分区域}D\text{关于}\\ & \text{轴}x\text{对称的时候，讨论被积函数关于}y\text{的奇偶性。}\end{array}$$

$$\begin{array}{rl}& \text{解：}I={\int }_0^{1}dx{\int }_0^{1}xy{f}_{xy}^{\prime \prime }(x,y)dy\\ & \\ & ={\int }_0^{1}xdx{\int }_0^{1}y\cdot \frac{\partial f_x^{\prime }(x,y)}{\partial y}dy\\ & ={\int }_0^{1}x\left(y\cdot f_x^{\prime }(x,y){|}_{y=0}^{y=1}-{\int }_0^1{f}_x^{\prime }(x,y)\cdot \frac{\partial y}{\partial y}dy\right)dx\\ & \\ & ={\int }_0^1\left(x{f}_x^{\prime }(x,1)-x{\int }_0^1{f}_x^{\prime }(x,y)dy\right)dx\\ & \\ & ={\int }_0^{1}x{f}_x^{\prime }(x,1)dx-{\int }_0^1\left({\int }_0^{1}x{f}_x^{\prime }(x,y)dy\right)dx\\ & \stackrel{\text{交换积分次序}}{\Rightarrow }\\ & ={xf(x,1)|}_0^1-{\int }_0^{1}f(x,1)dx-{\int }_0^1\left({\int }_0^{1}x{f}_x^{\prime }(x,y)dx\right)dy\\ & \\ & =f(1,1)-{\int }_0^{1}f(x,1)dx-{\int }_0^1\left({xf(x,y)|}_{x=0}^{x=1}-{\int }_0^{1}f(x,y)dx\right)dy\\ & \\ & =f(1,1)-{\int }_0^{1}f(x,1)dx-{\int }_0^{1}f(1,y)dy+{\int }_0^1{\int }_0^{1}f(x,y)dxdy\\ & \\ & \text{（结合边界条件，最终化简得）}\\ & ={\int }_0^1{\int }_0^{1}f(x,y)dxdy=\alpha \end{array}$$
$$\begin{array}{rl}& \text{【小结】本题的关键是由被积函数为}xy\text{与二阶混合偏导数}{f}_{xy}^{\prime \prime }(x,y)\text{的乘积，想到使用}\\ & \text{分部积分来将}{f}_{xy}^{\prime \prime }(x,y)\text{化成}f(x,y)\text{。}\end{array}$$

$$\begin{array}{rl}& \text{解：}{\int }_0^{2\pi }dx{\int }_0^{y(x)}(x+2y)dy={\int }_0^{2\pi }{\left(xy+y^2\right)|}_{y=0}^{y=y(x)}dx\\ & ={\int }_0^{2\pi }\left(xy+y^2\right)dx={\int }_0^{2\pi }\left(x\cdot y(x)+y^2(x)\right)dx\\ & ={\int }_0^{2\pi }\left(x(t)\cdot y(t)+y^2(t)\right)d\left(x(t)\right)\end{array}$$
$$\begin{array}{rl}& ={\int }_0^{2\pi }\left[(t-\sin t)(1-\cos t)+(1-\cos t{)}^2\right](1-\cos t)dt\\ & ={\int }_0^{2\pi }\left[(t-\sin t)(1-\cos t{)}^2+(1-\cos t{)}^3\right]dt\\ & ={\int }_0^{2\pi }t(1-\cos t{)}^{2}dt-{\int }_0^{2\pi }\sin t(1-\cos t{)}^{2}dt+{\int }_0^{2\pi }(1-3\cos t+3{\cos }^{2}t-{\cos }^{3}t)dt\\ & ={\int }_0^{2\pi }t(1-\cos t{)}^{2}dt-\frac{1}{3}(1-\cos t{)}^3{|}_0^{2\pi }+{\int }_0^{2\pi }(1-3\cos t+3{\cos }^{2}t-{\cos }^{3}t)dt\\ & =\pi {\int }_0^{2\pi }(1-\cos t{)}^{2}dt+{\int }_0^{2\pi }(1+3{\cos }^{2}t)dt\\ & =\pi {\int }_0^{2\pi }(1-2\cos t+{\cos }^{2}t)dt+{\int }_0^{2\pi }(1+3{\cos }^{2}t)dt\\ & =\pi \cdot \left(2\pi +4\cdot \frac{\pi }{2}\right)+\left(2\pi +3\cdot \frac{\pi }{2}\right)\\ & =\pi \cdot 3\pi +5\pi =3{\pi }^2+5\pi \end{array}$$
$$\begin{array}{rl}& \text{【小结】若题目中以参数方程给出积分区域，可以将}y\text{看作}x\text{的函数，进而将二重积分}\\ & \text{化为定积分，然后用所给参数方程换元计算出积分。}\end{array}$$

交换积分次序与坐标系的转换

$${\int }_0^{\frac{1}{2}}dx{\int }_{x^2}^{x}f(x,y)dy$$

$$\begin{array}{rl}& 0\le r\le \cos \theta ,r=\cos \theta ,r^2=r\cos \theta ,x^2+y^2=x\\ & \\ & {\left(x-\frac{1}{2}\right)}^2+y^2=\frac{1}{4}\end{array}$$

$$\begin{array}{rl}& 【小结】将极坐标转化为直角坐标，这类问题的求解思路和交换积分次序的问题类似，\\ & 也是先通过积分上下限画出积分区域，同时将被积表达式转化为直角坐标形式，\\ & 再选择合适的积分次序确定出积分的上下限。\end{array}$$
$$\{\begin{array}{l}r\cos \theta =x\\ r\sin \theta =y\end{array}\{\begin{array}{l}{r}^2=x^2+y^2\\ \theta =\arctan \frac{y}{x}\end{array}$$

$$\begin{array}{rl}& \text{注：}r=\sec \theta \to r=\frac{1}{\cos \theta }⟹r\cos \theta =1⟹x=1\\ & \\ & \text{解：}I={\iint }_{D}r\sin \theta \sqrt{1-r^2{\cos }^2\theta -{\sin }^2\theta }\cdot rdrd\theta \\ & ={\iint }_{D}r\sin \theta \sqrt{1-r^2{\cos }^2\theta +r^2{\sin }^2\theta }\cdot rdrd\theta \\ & ={\iint }_{D}y\sqrt{1-x^2+y^2}dxdy={\int }_0^{1}dx{\int }_0^{x}y\sqrt{1-x^2+y^2}dy\\ & \\ & ={\int }_0^1{\frac{1}{3}\cdot (1-x^2+y^2{)}^{\frac{3}{2}}|}_{y=0}^{y=x}dx\\ & ={\int }_0^1\frac{1}{3}\left[1-(1-x^2{)}^{\frac{3}{2}}\right]dx=\frac{1}{3}-\frac{1}{3}{\int }_0^{\frac{\pi }{2}}{\cos }^{3}td\sin t\\ & \\ & =\frac{1}{3}-\frac{1}{3}\cdot \frac{3!!}{4!!}\cdot \frac{\pi }{2}=\frac{1}{3}-\frac{\pi }{16}\end{array}$$
$$\begin{array}{rl}& 【小结】若题目中以极坐标直接给出积分区域，且被积函数也以极坐标给出，\\ & 一般计算比较复杂，需要转化为直角坐标系计算。\end{array}$$