Câu 1:
$\quad I = \displaystyle\iint\limits_D x^2(y-x)dxdy\qquad D:\begin{cases}y = x^2\\x = y^2\end{cases}$
Phương trình hoành độ giao điểm:
$+)\quad x^2 = -\sqrt x \Leftrightarrow x = 0$
$+)\quad x^2 = \sqrt x \Leftrightarrow \left[\begin{array}{l}x = 0\\x = 1\end{array}\right.$
Miền $D$ được biểu diễn:
$D =\left\{(x,y): 0 \leqslant x \leqslant 1;\ x^2 \leqslant y \leqslant \sqrt x\right\}$
Ta được:
$\quad I = \displaystyle\int\limits_0^1dx\displaystyle\int\limits_{x^2}^{\sqrt x}x^2(y - x)dy$
$\Leftrightarrow I = \displaystyle\int\limits_0^1\left[\left(-x^3y + \dfrac{x^2y}{2}\right)\Bigg|_{x^2}^{\sqrt x}\right]dx$
$\Leftrightarrow I = \displaystyle\int\limits_0^1\left(- x^3\sqrt x + \dfrac12x^3 + x^5 - \dfrac12x^6\right)dx$
$\Leftrightarrow I = \left(-\dfrac{3}{13}\sqrt[3]{x^{13}} - \dfrac{x^7}{14} + \dfrac{x^6}{6} + \dfrac{x^4}{8}\right)\Bigg|_0^1$
$\Leftrightarrow I = -\dfrac{23}{2184}$
Câu 2:
$\quad I = \displaystyle\oint\limits_L (xy + x + y)dx + (xy + x - y)dy,\quad L : x^2 + y^2 = 2x$
Đặt $\begin{cases}P = xy + x + y\\Q = xy + x - y\\x = r\cos\varphi\\y = r\sin\varphi\\-\dfrac{\pi}{2}\leqslant \varphi \leqslant \dfrac{\pi}{2}\end{cases}$
Miền $D$ được biểu diễn:
$D= \left\{(\varphi,r): -\dfrac{\pi}{2}\leqslant \varphi \leqslant \dfrac{\pi}{2};\ 0 \leqslant r \leqslant 2\cos\varphi\right\}$
Áp dụng công thức $Green$ ta có:
$\quad I = \displaystyle\iint\limits_D\left(\dfrac{\partial Q}{\partial x} - \dfrac{\partial P}{\partial y}\right)dxdy$
$\Leftrightarrow I = \displaystyle\iint\limits_D(y - x)dxdy$
$\Leftrightarrow I = \displaystyle\int\limits_{-\tfrac{\pi}{2}}^{\tfrac{\pi}{2}}d\varphi\displaystyle\int\limits_0^{2\cos\varphi}(r\sin\varphi - r\cos\varphi)rdr$
$\Leftrightarrow I = \displaystyle\int\limits_{-\tfrac{\pi}{2}}^{\tfrac{\pi}{2}}\left[\dfrac{r^3}{3}(\sin\varphi - \cos\varphi)\Bigg|_0^{2\cos\varphi}\right]d\varphi$
$\Leftrightarrow I = \displaystyle\int\limits_{-\tfrac{\pi}{2}}^{\tfrac{\pi}{2}}\dfrac83(\sin\varphi\cos^3\varphi - \cos^4\varphi)d\varphi$
$\Leftrightarrow I = -\dfrac{1}{12}\left(\sin4x + 8\sin2x + 8\cos^4x + 12x\right)\Bigg|_{-\tfrac{\pi}{2}}^{\tfrac{\pi}{2}}$
$\Leftrightarrow I = -\pi$
