格林公式

解题思路： 利用面积公式 $S = \frac{1}{2}\oint_L x \, dy - y \, dx$。

详细步骤：

1. 椭圆参数方程：$x = a\cos t, \, y = b\sin t, \, t: 0 \to 2\pi$（逆时针）
2. $dx = -a\sin t \, dt, \, dy = b\cos t \, dt$
3. $S = \frac{1}{2}\oint_L x \, dy - y \, dx = \frac{1}{2}\int_0^{2\pi} [a\cos t \cdot b\cos t - b\sin t \cdot (-a\sin t)] dt = \frac{1}{2}\int_0^{2\pi} ab(\cos^2 t+\sin^2 t) dt = \frac{ab}{2} \cdot 2\pi = \pi ab$

答案：$\pi ab$

解题思路： 用格林公式转化为二重积分，$\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}$ 可能很简单。

详细步骤：

1. $P = x^2 - 2y, \, Q = 3x + ye^y$
2. $\frac{\partial Q}{\partial x} = 3, \quad \frac{\partial P}{\partial y} = -2$
3. $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 3 - (-2) = 5$
4. 由格林公式：$\oint_L Pdx+Qdy = \iint_D 5 \, dx\,dy = 5 \times \text{圆面积} = 5 \times 4\pi = 20\pi$

答案：$20\pi$

解题思路： 验证 $\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}$，然后用折线路径求原函数。

详细步骤：

1. $P = x+y, \, Q = x-y$
2. $\frac{\partial P}{\partial y} = 1, \quad \frac{\partial Q}{\partial x} = 1$
3. $\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}$，在全平面（单连通）成立，故积分与路径无关。
4. 求原函数，取 $(x_0,y_0) = (0,0)$，折线路径： $u(x,y) = \int_0^x P(x,0) dx + \int_0^y Q(x,y) dy = \int_0^x x \, dx + \int_0^y (x-y) dy = \frac{x^2}{2} + [xy - \frac{y^2}{2}]_0^y = \frac{x^2}{2} + xy - \frac{y^2}{2}$
5. 验证：$\frac{\partial u}{\partial x} = x+y = P, \quad \frac{\partial u}{\partial y} = x-y = Q$ ✓

答案：原函数为 $u(x,y) = \frac{x^2}{2} + xy - \frac{y^2}{2} + C$

解题思路： 用格林公式转化为二重积分。

详细步骤：

1. $P = 2xy - x^2, \, Q = x + y^2$
2. $\frac{\partial Q}{\partial x} = 1, \quad \frac{\partial P}{\partial y} = 2x$
3. $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 1 - 2x$
4. 区域 $D: 0 \leq x \leq 1, \, x^2 \leq y \leq \sqrt{x}$
5. $\iint_D (1-2x) dx\,dy = \int_0^1 dx \int_{x^2}^{\sqrt{x}} (1-2x) dy = \int_0^1 (1-2x)(\sqrt{x}-x^2) dx$ $= \int_0^1 (x^{1/2} - x^2 - 2x^{3/2} + 2x^3) dx = [\frac{2}{3}x^{3/2} - \frac{x^3}{3} - \frac{4}{5}x^{5/2} + \frac{x^4}{2}]_0^1 = \frac{2}{3} - \frac{1}{3} - \frac{4}{5} + \frac{1}{2} = \frac{1}{30}$

答案：$\frac{1}{30}$

解题思路： 验证 $\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}$，求原函数后代入端点。

详细步骤：

1. $P = e^y + x, \, Q = xe^y - 2y$
2. $\frac{\partial P}{\partial y} = e^y, \quad \frac{\partial Q}{\partial x} = e^y$，相等，故与路径无关。
3. 原函数：$u(x,y) = \int_0^x (1+x)dx + \int_0^y (xe^y-2y)dy = x + \frac{x^2}{2} + [xe^y - y^2]_0^y = x + \frac{x^2}{2} + xe^y - y^2 - x = \frac{x^2}{2} + xe^y - y^2$
4. 从 $(0,0)$ 到 $(1,2)$：$u(1,2) - u(0,0) = (\frac{1}{2} + e^2 - 4) - 0 = e^2 - \frac{7}{2}$

答案：$e^2 - \frac{7}{2}$