题目
1.计算下列曲线积分，并验证格林公式的正确性：(2xy-x²)dx+(x+y²)dy，其中L是由抛物线y=x²和y²=x所围成的区域的正向边界曲线
1.计算下列曲线积分，并验证格林公式的正确性：
(2xy-x²)dx+(x+y²)dy，其中L是由抛物线y=x²和y²=x所围成的区域的正向边界曲线
题目解答
答案
**解：**
1. **直接计算曲线积分：**
- **对于 $L_1: y = x^2$（从 $x=0$ 到 $x=1$）：**
\[
\int_{L_1} (2x^3 - x^2 + 2x^2 + 2x^5) \, dx = \frac{7}{6}
\]
- **对于 $L_2: x = y^2$（从 $y=1$ 到 $y=0$）：**
\[
\int_{L_2} (4y^4 - 2y^5 + 2y^2) \, dy = -\frac{17}{15}
\]
- **总和：**
\[
\frac{7}{6} - \frac{17}{15} = \frac{1}{30}
\]
2. **使用格林公式：**
- **计算偏导数：**
\[
\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 1 - 2x
\]
- **二重积分：**
\[
\iint_0^1 \int_{x^2}^{\sqrt{x}} (1 - 2x) \, dy \, dx = \frac{1}{30}
\]
**答案：**
\[
\boxed{\frac{1}{30}
\]
解析
步骤 1：直接计算曲线积分
- 对于 $L_1: y = x^2$（从 $x=0$ 到 $x=1$）：
\[ \int_{L_1} (2x^3 - x^2 + 2x^2 + 2x^5) \, dx = \int_{0}^{1} (2x^3 + x^2 + 2x^5) \, dx = \left[ \frac{1}{2}x^4 + \frac{1}{3}x^3 + \frac{1}{3}x^6 \right]_{0}^{1} = \frac{1}{2} + \frac{1}{3} + \frac{1}{3} = \frac{7}{6} \]
- 对于 $L_2: x = y^2$（从 $y=1$ 到 $y=0$）：
\[ \int_{L_2} (4y^4 - 2y^5 + 2y^2) \, dy = \int_{1}^{0} (4y^4 - 2y^5 + 2y^2) \, dy = -\int_{0}^{1} (4y^4 - 2y^5 + 2y^2) \, dy = -\left[ \frac{4}{5}y^5 - \frac{1}{3}y^6 + \frac{2}{3}y^3 \right]_{0}^{1} = -\left( \frac{4}{5} - \frac{1}{3} + \frac{2}{3} \right) = -\frac{17}{15} \]
- 总和：
\[ \frac{7}{6} - \frac{17}{15} = \frac{35}{30} - \frac{34}{30} = \frac{1}{30} \]
步骤 2：使用格林公式
- 计算偏导数：
\[ \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 1 - 2x \]
- 二重积分：
\[ \iint_0^1 \int_{x^2}^{\sqrt{x}} (1 - 2x) \, dy \, dx = \int_0^1 (1 - 2x)(\sqrt{x} - x^2) \, dx = \int_0^1 (\sqrt{x} - x^2 - 2x\sqrt{x} + 2x^3) \, dx = \int_0^1 (x^{1/2} - x^2 - 2x^{3/2} + 2x^3) \, dx = \left[ \frac{2}{3}x^{3/2} - \frac{1}{3}x^3 - \frac{4}{5}x^{5/2} + \frac{1}{2}x^4 \right]_0^1 = \frac{2}{3} - \frac{1}{3} - \frac{4}{5} + \frac{1}{2} = \frac{1}{30} \]
- 对于 $L_1: y = x^2$（从 $x=0$ 到 $x=1$）：
\[ \int_{L_1} (2x^3 - x^2 + 2x^2 + 2x^5) \, dx = \int_{0}^{1} (2x^3 + x^2 + 2x^5) \, dx = \left[ \frac{1}{2}x^4 + \frac{1}{3}x^3 + \frac{1}{3}x^6 \right]_{0}^{1} = \frac{1}{2} + \frac{1}{3} + \frac{1}{3} = \frac{7}{6} \]
- 对于 $L_2: x = y^2$（从 $y=1$ 到 $y=0$）：
\[ \int_{L_2} (4y^4 - 2y^5 + 2y^2) \, dy = \int_{1}^{0} (4y^4 - 2y^5 + 2y^2) \, dy = -\int_{0}^{1} (4y^4 - 2y^5 + 2y^2) \, dy = -\left[ \frac{4}{5}y^5 - \frac{1}{3}y^6 + \frac{2}{3}y^3 \right]_{0}^{1} = -\left( \frac{4}{5} - \frac{1}{3} + \frac{2}{3} \right) = -\frac{17}{15} \]
- 总和：
\[ \frac{7}{6} - \frac{17}{15} = \frac{35}{30} - \frac{34}{30} = \frac{1}{30} \]
步骤 2：使用格林公式
- 计算偏导数：
\[ \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 1 - 2x \]
- 二重积分：
\[ \iint_0^1 \int_{x^2}^{\sqrt{x}} (1 - 2x) \, dy \, dx = \int_0^1 (1 - 2x)(\sqrt{x} - x^2) \, dx = \int_0^1 (\sqrt{x} - x^2 - 2x\sqrt{x} + 2x^3) \, dx = \int_0^1 (x^{1/2} - x^2 - 2x^{3/2} + 2x^3) \, dx = \left[ \frac{2}{3}x^{3/2} - \frac{1}{3}x^3 - \frac{4}{5}x^{5/2} + \frac{1}{2}x^4 \right]_0^1 = \frac{2}{3} - \frac{1}{3} - \frac{4}{5} + \frac{1}{2} = \frac{1}{30} \]