MA127-2025 春期末考参考答案

一、选择题

(1) D; (2) C; (3) A; (4) D; (5) A

二、填空题

(1) $2x-y-z=1$

(2) $\frac{\pi }{2}$

(3) $\sqrt{5}$

(4) $\frac{1}{2}$

(5) $2$

三

当 $t=\frac{\pi }{3}$ 时，对应点为 $(\frac{\pi }{3}-\frac{\sqrt{3}}{2},\frac{1}{2})$。

3-1

$$\frac{dy}{dx}=\frac{\sin t}{1-\cos t}$$$$\sqrt{3}x-y=\frac{\pi }{\sqrt{3}}-2$$

3-2

$$\frac{d^{2}y}{d{x}^2}=-\frac{1}{(1-\cos t{)}^2}$$

四

4-1

Applying the root test, we obtain that the series converges absolutely.

4-2

The alternating series test yields that the series converges, while the integral test leads to the conclusion that the series does not converge absolutely. Therefore, the series converges conditionally.

五

We first identify critical points by solving the gradient equations:

$$\frac{\delta f}{\delta x}=(1-x^2)e^{-\frac{x^2+y^2}{2}}=0$$$$\frac{\delta f}{\delta y}=-xy{e}^{-\frac{x^2+y^2}{2}}=0$$

Solving these equations yields two critical points: $(1,0)$ and $(-1,0)$.

Next, we employ the second derivative test. Computing the Hessian components:

$$f_{xx}(x,y)=(x^3-3x)e^{-\frac{x^2+y^2}{2}}$$$$f_{xy}(x,y)=(x^2-1)y{e}^{-\frac{x^2+y^2}{2}}$$$$f_{yy}(x,y)=x(y^2-1)e^{-\frac{x^2+y^2}{2}}$$

For the critical point $(1,0)$:

$$f_{xx}(1,0)=-2e^{\frac{-1}{2}}<0$$$$f_{xy}(1,0)=0$$$$f_{yy}(1,0)=-e^{\frac{-1}{2}}$$

Then, $\mathrm{\Delta }=f_{xx}{f}_{yy}-f_{xy}^2=2e^{-1}>0$

Therefore, $(1,0)$ is a local minimum point with $f(1,0)=e^{-\frac{1}{2}}$.

Similarly, at $(-1,0)$:

$$f_{xx}(-1,0)=-2e^{\frac{-1}{2}}<0$$$$f_{xy}(-1,0)=0$$$$f_{yy}(-1,0)=-e^{\frac{-1}{2}}$$

Then, $\mathrm{\Delta }=f_{xx}{f}_{yy}-f_{xy}^2=2e^{-1}>0$

$(-1,0)$ is a local minimum point with $f(-1,0)=-e^{-\frac{1}{2}}$.

In summary, the function attains its local maximum value $e^{-\frac{1}{2}}$ at $(1,0)$, and the local minimum value $-e^{-\frac{1}{2}}$ at $(-1,0)$.

六

From the symmetry, we know that $\overline{x}=\overline{y}=0$

$$M_{xy}={\iiint }_{D}zdxdydz={\int }_0^{2\pi }{\int }_0^1{\int }_{r^2}^{1}zrd\theta drdt$$$$={\int }_0^{2\pi }d\theta {\int }_0^{1}r\frac{1-r^4}{2}dr=\frac{\pi }{3}$$$$M={\iiint }_{D}dxdydz={\int }_0^{2\pi }{\int }_0^1{\int }_{r^2}^{1}rd\theta drdt$$$$={\int }_0^{2\pi }d\theta {\int }_0^{1}r(1-r^2)dr==\frac{\pi }{2}$$

$\overline{z}=\frac{M_{xy}}{M}=\frac{2}{3}$

Therefore, the centroid is $(0,0,\frac{2}{3})$

七

$$\mathrm{\nabla }\cdot \mathbf{F}=6x^2+3y^2$$

Thus, the flux is given by:

$${\iiint }_D\mathrm{\nabla }\cdot \mathbf{F}dV={\iiint }_{0\le 2x^2+y^2\le 1}(1-(2x^2+y^2))(6x^3+3y^2)dxdy$$

Apply the transformation $x=\frac{1}{\sqrt{2}}u$, $y=v$, $z=r\cos \varphi$, we obtain:

$$\frac{3}{\sqrt{2}}{\iint }_{0\le u^2+v^2\le 1}(1-u^2-v^2)(u^2+v^2)dudv$$$$=\frac{3}{\sqrt{2}}{\int }_0^{2\pi }{\int }_0^1(1-r^2)r^{3}drd\theta$$$$=\frac{\sqrt{2}}{4}\pi$$

$\mathrm{\nabla }\cdot \mathbf{F}=<0,-1,2>$

八

Let $S$ be the region $\{(x,y,z)\mid x^2+y^2\le 2y,x-y+2=2\}$. Obviously, $C$ is the boundary of $S$ with the corresponding outward normal vector $\mathbf{n}=\frac{1}{\sqrt{3}}(1,-1,1)$. For the surface $S$, we have $dS=\sqrt{3}dxdy$.

Applying Stokes' Theorem, the equivalent surface integral is obtained as:

$${\iint }_S(\mathrm{\nabla }\times \mathbf{F})\cdot \mathbf{n}dS={\iint }_{x^2+y^2\le 1}3dxdy=3\pi$$