Đáp án:

\(\begin{array}{l}
\text{Câu 1:}\quad I = \displaystyle\oint\limits_L(ydx + xdy) = 0\\
\text{Câu 3:}\quad I = \displaystyle\oint\limits_C(y^2dx + (x+2xy)dy) = \pi
\end{array}\) 

Giải thích các bước giải:

\(\begin{array}{l}
\text{Câu 1:}\quad I = \displaystyle\oint\limits_L(ydx + xdy)\qquad \text{với:}\ (L): \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1\\
+)\quad \text{Cách 1: Viết phương trình tham số}\\
\text{Đặt}\ \begin{cases}x = a\cos t\\y = b\sin t\\0 \leqslant t \leqslant 2\pi\end{cases}\\
\text{Ta được:}\\
\quad I = \displaystyle\int\limits_0^{2\pi}(-b\sin t.a\sin t + a\cos t.b\cos t)dt\\
\Leftrightarrow I = \displaystyle\int\limits_0^{2\pi}ab\cos2tdt\\
\Leftrightarrow I = ab\sin t\cos t\Bigg|_0^{2\pi}\\
\Leftrightarrow I = 0\\
+)\quad \text{Cách 2: Dùng công thức Green}\\
\text{Đặt}\ \begin{cases}P = y\\Q = x\end{cases}\\
\ \Rightarrow \begin{cases}\dfrac{\partial Q}{\partial x} = 1\\\dfrac{\partial P}{\partial y} = 1\end{cases}\\
\text{Ta được:}\\
\quad I = \displaystyle\iint\limits_D\left(\dfrac{\partial Q}{\partial x} - \dfrac{\partial P}{\partial y}\right)dxdy\quad \text{với}\ D:\left\{ \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1\right\}\\
\Leftrightarrow I = 0\\
\text{Câu 3:}\quad I = \displaystyle\oint\limits_C(y^2dx + (x+2xy)dy)\qquad \text{với:}\ (C): x^2 + y^2 = 1\\
+)\quad \text{Cách 1: Viết phương trình tham số}\\
\text{Đặt}\ \begin{cases}x = \cos t\\y = \sin t\\0 \leqslant t \leqslant 2\pi\end{cases}\\
\text{Ta được:}\\
\quad I = \displaystyle\int\limits_0^{2\pi}(-\sin^2t.\sin t + (\cos t + 2\sin t\cos t).\cos t)dt\\
\Leftrightarrow I = \displaystyle\int\limits_0^{2\pi}(-\sin^3t + \cos^2t + 2\sin t\cos^2t)dt\\
\Leftrightarrow I = \dfrac14(2t + \sin2t + \cos t - \cos3t)\Bigg|_0^{2\pi}\\
\Leftrightarrow I = \pi\\
+)\quad \text{Cách 2: Dùng công thức Green}\\
\text{Đặt}\ \begin{cases}P = y^2\\Q = x + 2xy\end{cases}\\
\ \Rightarrow \begin{cases}\dfrac{\partial Q}{\partial x} = 1 + 2y\\\dfrac{\partial P}{\partial y} = 2y\end{cases}\\
\text{Ta được:}\\
\quad I = \displaystyle\iint\limits_D\left(\dfrac{\partial Q}{\partial x} - \dfrac{\partial P}{\partial y}\right)dxdy\quad \text{với}\ D: \{x^2 + y^2 = 1\}\\
\Leftrightarrow I = \displaystyle\int\limits_0^{2\pi}d\varphi\displaystyle\int\limits_0^1rdr\\
\Leftrightarrow I = \displaystyle\int\limits_0^{2\pi}\left(\dfrac{r^2}{2}\Bigg|_0^1\right)d\varphi\\
\Leftrightarrow I = \dfrac12\displaystyle\int\limits_0^{2\pi}d\varphi\\
\Leftrightarrow I = \dfrac12\varphi\Bigg|_0^{2\pi}\\
\Leftrightarrow I = \pi
\end{array}\)