8.1 第6次作业评讲

8.1.1 解答和证明部分

解 $f(z)$在有界闭区域$D=\{(x,y,z)\mid 0\le z\le y\le x\le 1\}$上Riemann可积，积分域可转写为$$\begin{array}{}\text{(3)}& \{\begin{array}{l}0\le x\le 1\\ 0\le y\le x\\ 0\le z\le y\end{array}⟹0\le z\le y\le x\le 1⟹\{\begin{array}{l}0\le z\le 1\\ z\le y\le 1\\ y\le x\le 1\end{array}\end{array}$$ 故有$$\begin{array}{}\text{(4)}& {\int }_0^1\mathrm{d}x{\int }_0^x\mathrm{d}y{\int }_0^{y}f(z)\mathrm{d}z={\int }_0^{1}f(z)\mathrm{d}z{\int }_z^1\mathrm{d}y{\int }_y^1\mathrm{d}x={\int }_0^1(\frac{1}{2}-z+\frac{z^2}{2})f(z)\mathrm{d}z=0\end{array}$$ $◻$

解 积分域可转写为$$\begin{array}{}\text{(6)}& \{\begin{array}{l}\frac{1}{4}\le y\le \frac{1}{2}\\ \frac{1}{2}\le x\le \sqrt{y}\end{array}\cup \{\begin{array}{l}\frac{1}{2}\le y\le 1\\ y\le x\le \sqrt{y}\end{array}⟹\frac{1}{4}\le max\{y,\frac{1}{2}\}\le x\le \sqrt{y}\le 1⟹\{\begin{array}{l}\frac{1}{2}\le x\le 1\\ x^2\le y\le x\end{array}\end{array}$$ 故有$$\begin{array}{}\text{(7)}& I={\int }_{\frac{1}{2}}^1\mathrm{d}x{\int }_{x^2}^x{\mathrm{e}}^{\frac{y}{x}}\mathrm{d}y\stackrel{t=\frac{y}{x}}{=}{\int }_{\frac{1}{2}}^1\mathrm{d}x{\int }_x^1{\mathrm{e}}^{t}x\mathrm{d}t={\int }_{\frac{1}{2}}^{1}x(\mathrm{e}-{\mathrm{e}}^x)\mathrm{d}x=\frac{3\mathrm{e}}{8}-\frac{\sqrt{e}}{2}\end{array}$$ $◻$

解 由题可得$$\begin{array}{}\text{(9)}& \frac{y^2-5}{2}\le x\le y-1\text{或}y-1\le x\le \frac{y^2-5}{2}\end{array}$$ 只有前一个不等式组解得的$y$是有界的，解得$-1\le y\le 3$。故有$$\begin{array}{}\text{(10)}& {\int }_{-1}^3\mathrm{d}y{\int }_{y-1}^{\frac{y^2-5}{2}}xy\mathrm{d}x=0\end{array}$$ $◻$

解 设$u=\frac{x^2}{y}$、$v=\frac{y^2}{x}$，则$1\le u\le 4$、$1\le v\le 3$，上述变换的Jacobi行列式为$$\begin{array}{}\text{(12)}& det\left(\begin{array}{cc}\frac{2x}{y}& -\frac{x^2}{y^2}\\ -\frac{y^2}{x^2}& \frac{2y}{x}\end{array}\right)=3\end{array}$$ 故有$$\begin{array}{}\text{(13)}& {\int }_{[1,4]\times [1,3]}uv\cdot \frac{1}{\left|\det \frac{\partial (u,v)}{\partial (x,y)}\right|}\mathrm{d}u\mathrm{d}v=\frac{1}{3}\cdot \frac{4^2-1}{2}\cdot \frac{3^2-1}{2}=10\end{array}$$ $◻$

解 先对被积函数进行展开$$\begin{array}{}\text{(15)}& {\int }_D(x+y{)}^2\mathrm{d}x\mathrm{d}y={\int }_D(x^2+y^2)\mathrm{d}x\mathrm{d}y+2{\int }_{D}xy\mathrm{d}x\mathrm{d}y\end{array}$$ 再用对称性$(x,y)\in D⟺(-x,y)\in D$，得到$$\begin{array}{}\text{(16)}& {\int }_{D}xy\mathrm{d}x\mathrm{d}y=0\end{array}$$ 然后用极坐标$x=r\cos \theta ,y=r\sin \theta$，区域$D$可表示为$$\begin{array}{}\text{(17)}& \frac{x^2+y^2}{4}\le y\le \frac{x^2+y^2}{2}⟹2\sin \theta \le r\le 4\sin \theta ,\sin \theta \ge 0(0\le \theta \le \pi )\end{array}$$ 于是极坐标下，原积分为$$\begin{array}{}\text{(18)}& {\int }_{0\le \theta \le \pi ,2\sin \theta \le r\le 4\sin \theta }{r}^{2}r\mathrm{d}r\mathrm{d}\theta =\frac{45\pi }{2}\end{array}$$ $◻$

解 正确，参考例7.3.14。 $◻$

解 积分域可转写为$$\begin{array}{}\text{(21)}& D_1:\{\begin{array}{l}0\le x\le a\\ 0\le y\le x\\ y\le z\le a\end{array}⟹0\le y\le min\{x,z\}\le a\end{array}$$ 本题需要利用积分区域的对称性，即$(x,y,z)$的任意排列都可以得到相同的积分值，故有$$\begin{array}{}\text{(22)}& \bigcup \{\begin{array}{l}{D}_1:0\le y\le min\{x,z\}\le a\\ D_2:0\le z\le min\{x,y\}\le a\\ D_3:0\le x\le min\{y,z\}\le a\end{array}⟹[0,a{]}^3\end{array}$$ 被积函数在$D_1,D_2,D_3$上的积分值相同，故$A=3$。 $◻$

解 首先需要利用对称性$(x,y)\in \mathrm{\Omega }⟺(-x,-y)\in \mathrm{\Omega }$，得到$$\begin{array}{}\text{(24)}& {\int }_{\mathrm{\Omega }}(x+y)\mathrm{d}x\mathrm{d}y\mathrm{d}z=0\end{array}$$ 剩下的积分使用柱坐标系进行计算，区域$\mathrm{\Omega }$可表示为$$\begin{array}{}\text{(25)}& \sqrt{r^2}\le z\le \sqrt{1-r^2}⟹0\le \theta <2\pi ,0\le r\le \frac{\sqrt{2}}{2},0\le z\le \sqrt{1-r^2}\end{array}$$ 因此$$\begin{array}{}\text{(26)}& {\int }_{\mathrm{\Omega }}8z\mathrm{d}x\mathrm{d}y\mathrm{d}z={\int }_0^{2\pi }\mathrm{d}\theta {\int }_0^{\frac{\sqrt{2}}{2}}\mathrm{d}r{\int }_r^{\sqrt{1-r^2}}8zr\mathrm{d}z=8\pi {\int }_0^{\frac{\sqrt{2}}{2}}r(1-2r^2)\mathrm{d}r=4\pi {(r^2-r^4)}_0^{\frac{\sqrt{2}}{2}}=\pi \end{array}$$ $◻$

解 利用球坐标系进行计算，可得$$\begin{array}{}\text{(28)}& F(t)={\int }_0^{2\pi }\mathrm{d}\varphi {\int }_0^{\pi }\sin \theta \mathrm{d}\theta {\int }_0^{t}f(r^2)r^2\mathrm{d}r=4\pi {\int }_0^{t}f(r^2)r^2\mathrm{d}r\end{array}$$ 再利用变上限积分求导法则，可得$$\begin{array}{}\text{(29)}& F^{\prime }(t)=4\pi f(1)=4\pi \end{array}$$ $◻$

解 本题如果直接对$z$积分，由于被积函数不存在初等原函数，无法继续操作；但注意到积分域是直角棱长为$\frac{\pi }{2}$的直角三棱锥，可以预见先对$x,y$积分会得到因子${(\frac{\pi }{2}-z)}^2$，从而化简被积函数。积分域可转写为$$\begin{array}{}\text{(31)}& \{\begin{array}{l}0\le x\le \frac{\pi }{2}\\ 0\le y\le x\\ 0\le z\le y\end{array}⟹0\le z\le y\le x\le \frac{\pi }{2}⟹\{\begin{array}{l}0\le z\le \frac{\pi }{2}\\ z\le y\le \frac{\pi }{2}\\ y\le x\le \frac{\pi }{2}\end{array}\end{array}$$ 故有$$\begin{array}{}\text{(32)}& I={\int }_0^{\pi /2}\frac{\cos z}{(\frac{\pi }{2}-z)}\mathrm{d}z{\int }_z^{\frac{\pi }{2}}{\int }_y^{\frac{\pi }{2}}\mathrm{d}x\mathrm{d}y={\int }_0^{\pi /2}\frac{\cos z}{(\frac{\pi }{2}-z)}\mathrm{d}z\cdot \frac{1}{2}{(\frac{\pi }{2}-z)}^2=\frac{1}{2}\end{array}$$ $◻$