Analysis quiz

1. 设$v(n)$ 是$n$ 维球体$\lbrace x^2_1 +\cdots +x_n^2\leqslant 1\rbrace$ 的体积，求$v(n)$ 的最大值

$$B_n(a)=\lbrace (x_1,\cdots ,x_n):x^2_1 +\cdots +x_n^2\leqslant 1\rbrace$$
易证 $$\mu(B_n(a))=\int_{B_n(a)}\mathop{}\mathrm{d}\mu=a^n\int_{B_n(1))}\mathop{}\mathrm{d}\mu=a^n\mu(B_n(1))$$
$$\begin{aligned} \mu(B_n(1))&=\int_{B_n(1))}\mathop{}\mathrm{d}\mu=\iint_{u^2+v^2}\mathop{}\mathrm{d}u\mathop{}\mathrm{d}v\idotsint_{B_{n-2}}(\sqrt{1-u^2-v^2})\mathop{}\mathrm{d}x_1\cdots\mathop{}\mathrm{d}x_{n-2}\\ &=\mu(B_{n-2}(1))\iint_{u^2+v^2\leqslant 1}(1-u^2-v^2)^{(n-2)/2}\\ &=\frac{2\pi}{n}\mu(B_{n-2}(1)) \end{aligned}$$
当$V(n)$最大时，$n$只可能取$5,6,7(\frac{2\pi}{n}\approx 1)$

$$V(5)=\frac{8\pi^2}{15},V(6)=\frac{\pi^2}{6},V(7)=\frac{16\pi^3}{105}\\ \therefore V_{max}=V(5)=\frac{8\pi^2}{15}$$

2. 设$(r,\theta,\varphi)$ 是$\mathbb{R}^3$ 的球坐标 ，$S: r=r(\theta ,\varphi)$$[(\theta ,\varphi \in D)]$是$\mathbb{R}^3$ 曲面方程，求该曲面面积

3. 设$a>b>0$ ，求椭圆盘$\frac{x^2}{a^2}+\frac{y^2}{b^2}\leqslant 1$ 与 $\frac{x^2}{b^2}+\frac{y^2}{a^2}\leqslant 1$公共部分的面积(要求用第二型曲线积分计算)

$$\int_D\mathop{}\mathrm{d}S=\oint_{\partial D}x\mathop{}\mathrm{d}y$$
$$\therefore S=4\left(\int_{0}^{\arctan a/b}ab\cos^2\theta\mathop{}\mathrm{d}\theta+\int_{\arctan b/a}^{\frac{\pi}{2}}ab\cos^2\theta\mathop{}\mathrm{d}\theta\right)=\pi+2ab(\arctan\frac{a}{b}-\arctan\frac{b}{a})$$

4. 设$f,g\ \mathbb{R}^3 \mapsto \mathbb{R}$ 一阶连续可微，且 $\nabla f\parallel \nabla g$,$\nabla f\neq 0$,$\nabla g\neq 0$,$P_0\in\mathbb{R}^3$, $f(P_0)=g(P_0)=0$，证明在$P_0$ 附近，等值面$\lbrace f=0\rbrace$ 与等值面 $\lbrace g=0\rbrace$相等

设$P_0(x_0,y_0,z_0)$,不妨设$\frac{\partial f}{\partial z}\neq 0\Longrightarrow$在$P_0$的邻域$f(P)=f(P_0)=0$有解$z=z_1(x,y)\Longrightarrow$等值面${r}_1(x,y)=(x,y,z_1(x,y))$,由$\nabla f\parallel \nabla g\Longrightarrow\frac{\partial g}{\partial z}\neq 0\Longrightarrow\lbrace g=0\rbrace$等值面${r}_2(x,y)=(x,y,z_2(x,y))$要证$z_1(x,y)=z_2(x,y)$
$$\begin{aligned} z_0&=z_1(x_0,y_0)\mapsto {r}_1(x_0,y_0)=P_0\\ z_0&=z_2(x_0,y_0)\mapsto {r}_2(x_0,y_0)=P_0\\ \frac{\partial z_1}{\partial x}&=-\frac{\partial f}{\partial x}/\frac{\partial f}{\partial z}\\ \frac{\partial z_2}{\partial x}&=-\frac{\partial g}{\partial x}/\frac{\partial g}{\partial z}\\ \frac{\partial z_1}{\partial x}&=\frac{\partial z_2}{\partial x}\\ \text{同理:}\frac{\partial z_1}{\partial y}&=\frac{\partial z_2}{\partial y}\\ \therefore \text{等值面相等} \end{aligned}$$