大连理工大学 2024年数学分析第0题

曲线 $C$ 是球面 $z = \sqrt{a^2 - x^2 - y^2}$（上半球面）与平面 $x = y$ 的交线。代入 $x = y$ 得 $z = \sqrt{a^2 - 2x^2}$，$x$ 的取值范围为 $\left[-\frac{a}{\sqrt{2}}, \frac{a}{\sqrt{2}}\right]$。方向从 $\left(\frac{a}{\sqrt{2}}, \frac{a}{\sqrt{2}}, 0\right)$ 到 $\left(-\frac{a}{\sqrt{2}}, -\frac{a}{\sqrt{2}}, 0\right)$，即 $x$ 从 $\frac{a}{\sqrt{2}}$ 减小到 $-\frac{a}{\sqrt{2}}$。令 $x = t$，则 $y = t$，$z = \sqrt{a^2 - 2t^2}$，$t$ 从 $\frac{a}{\sqrt{2}}$ 到 $-\frac{a}{\sqrt{2}}$。

由参数化 $x = t$，$y = t$，$z = \sqrt{a^2 - 2t^2}$，得 $dx = dt$，$dy = dt$，$dz = \frac{-2t}{\sqrt{a^2 - 2t^2}} dt = -\frac{2t}{z} dt$。

将 $x = t$，$y = t$，$z = \sqrt{a^2 - 2t^2}$ 及微分代入曲线积分： \[ \begin{aligned} & (z^3 + 3x^2 y) dx + (x^3 + 3y^2 z) dy + (y^3 + 3z^2 x) dz \\ = & (z^3 + 3t^3) dt + (t^3 + 3t^2 z) dt + (t^3 + 3z^2 t) \left(-\frac{2t}{z}\right) dt \\ = & \left[ z^3 + 3t^3 + t^3 + 3t^2 z - \frac{2t}{z}(t^3 + 3z^2 t) \right] dt \\ = & \left[ z^3 + 4t^3 + 3t^2 z - \frac{2t^4}{z} - 6t^2 z \right] dt \\ = & \left[ z^3 + 4t^3 - 3t^2 z - \frac{2t^4}{z} \right] dt \end{aligned} \]

考虑向量场 $\mathbf{F} = (z^3 + 3x^2 y, x^3 + 3y^2 z, y^3 + 3z^2 x)$。计算旋度： \[ \nabla \times \mathbf{F} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \partial_x & \partial_y & \partial_z \\ z^3 + 3x^2 y & x^3 + 3y^2 z & y^3 + 3z^2 x \end{vmatrix} \] 各分量： - $\frac{\partial}{\partial y}(y^3 + 3z^2 x) - \frac{\partial}{\partial z}(x^3 + 3y^2 z) = 3y^2 - 3y^2 = 0$ - $\frac{\partial}{\partial z}(z^3 + 3x^2 y) - \frac{\partial}{\partial x}(y^3 + 3z^2 x) = 3z^2 - 3z^2 = 0$ - $\frac{\partial}{\partial x}(x^3 + 3y^2 z) - \frac{\partial}{\partial y}(z^3 + 3x^2 y) = 3x^2 - 3x^2 = 0$ 旋度为零，故向量场是保守场。

设势函数 $f$ 满足 $\nabla f = \mathbf{F}$。由 $\frac{\partial f}{\partial x} = z^3 + 3x^2 y$，积分得 $f = x z^3 + x^3 y + g(y, z)$。由 $\frac{\partial f}{\partial y} = x^3 + 3y^2 z$，得 $x^3 + \frac{\partial g}{\partial y} = x^3 + 3y^2 z$，故 $\frac{\partial g}{\partial y} = 3y^2 z$，积分得 $g = y^3 z + h(z)$，所以 $f = x z^3 + x^3 y + y^3 z + h(z)$。由 $\frac{\partial f}{\partial z} = y^3 + 3z^2 x$，得 $3x z^2 + y^3 + h'(z) = y^3 + 3z^2 x$，故 $h'(z) = 0$，取 $h(z) = 0$，则势函数 $f(x,y,z) = x z^3 + x^3 y + y^3 z$。

由于向量场保守，曲线积分等于势函数在终点与起点的差： \[ \int_C \mathbf{F} \cdot d\mathbf{r} = f\left(-\frac{a}{\sqrt{2}}, -\frac{a}{\sqrt{2}}, 0\right) - f\left(\frac{a}{\sqrt{2}}, \frac{a}{\sqrt{2}}, 0\right) \] 在 $z=0$ 时，$f(x,y,0) = x^3 y$。所以： \[ f\left(\frac{a}{\sqrt{2}}, \frac{a}{\sqrt{2}}, 0\right) = \left(\frac{a}{\sqrt{2}}\right)^3 \cdot \frac{a}{\sqrt{2}} = \frac{a^4}{4} \] \[ f\left(-\frac{a}{\sqrt{2}}, -\frac{a}{\sqrt{2}}, 0\right) = \left(-\frac{a}{\sqrt{2}}\right)^3 \cdot \left(-\frac{a}{\sqrt{2}}\right) = \frac{a^4}{4} \] 差值为 $\frac{a^4}{4} - \frac{a^4}{4} = 0$。

曲线积分为 0。