Đáp án:
$6)\quad I = \displaystyle\iint\limits_D(x^2 + y)dxdy=\dfrac43$
$7)\quad I = \displaystyle\iint\limits_Ddxdy=\dfrac{5\sqrt5}{6}$
Giải thích các bước giải:
$6)\quad I = \displaystyle\iint\limits_D(x^2 + y)dxdy\quad \text{với}\quad D:\begin{cases}y = -\dfrac{x}{2} -1\\y = \dfrac x2 +1\\x = 0\end{cases}$
Phương trình hoành độ giao điểm:
$\quad -\dfrac x2 - 1=\dfrac x2 + 1\Leftrightarrow x = -2$
Miền $D$ được biểu diễn:
$D:\left\{(x,y): - 2 \leqslant x \leqslant 0;\ -\dfrac x2 - 1 \leqslant y \leqslant \dfrac x2 +1\right\}$
Ta được:
$\quad I = \displaystyle\int\limits_{-2}^0dx\displaystyle\int\limits_{-\tfrac x2 -1}^{\tfrac x2 +1}(x^2 + y)dy$
$\Leftrightarrow I = \displaystyle\int\limits_{-2}^0\left[\left(x^2y + \dfrac{y^2}{2}\right)\Bigg|_{-\tfrac x2 -1}^{\tfrac x2 +1}\right]dx$
$\Leftrightarrow I = \displaystyle\int\limits_{-2}^0(x^3 + 2x^2)dx$
$\Leftrightarrow I = \left(\dfrac{x^4}{4} + \dfrac{2x^3}{3}\right)\Bigg|_{-2}^0$
$\Leftrightarrow I = \dfrac43$
$7)\quad I = \displaystyle\iint\limits_Ddxdy\quad \text{với}\quad D:\begin{cases}y = x -1\\x = y^2\end{cases}$
Phương trình tung độ giao điểm:
$\quad y+1 = y^2\Leftrightarrow \left[\begin{array}{l}y = \dfrac{1-\sqrt5}{2}\\y = \dfrac{1 +\sqrt5}{2}\end{array}\right.$
Miền $D$ được biểu diễn:
$D:\left\{(x,y): \dfrac{1 -\sqrt5}{2}\leqslant y \leqslant \dfrac{1 +\sqrt5}{2}; y^2 \leqslant x \leqslant y +1\right\}$
Ta được:
$\quad I = \displaystyle\int\limits_{\tfrac{1-\sqrt5}{2}}^{\tfrac{1+\sqrt5}{2}}dy\displaystyle\int\limits_{y^2}^{y+1}dx$
$\Leftrightarrow I = \displaystyle\int\limits_{\tfrac{1-\sqrt5}{2}}^{\tfrac{1+\sqrt5}{2}}\left(x\Bigg|_{y^2}^{y+1}\right)dy$
$\Leftrightarrow I = \displaystyle\int\limits_{\tfrac{1-\sqrt5}{2}}^{\tfrac{1+\sqrt5}{2}}(y+1- y^2)dy$
$\Leftrightarrow I = \left(\dfrac{y^2}{2} + y - \dfrac{y^3}{3}\right)\Bigg|_{\tfrac{1-\sqrt5}{2}}^{\tfrac{1+\sqrt5}{2}}$
$\Leftrightarrow I = \dfrac{5\sqrt5}{6}$
