Đáp án:
$1)\quad I =\displaystyle\int\limits_0^2dx\displaystyle\int\limits_0^{x^2}60ydy=192$
$ 2)\quad I = \displaystyle\iint\limits_D(x^2 + y^2)dxdy = \dfrac{4}{3}$
Giải thích các bước giải:
\(\begin{array}{l}
1)\quad I =\displaystyle\int\limits_0^2dx\displaystyle\int\limits_0^{x^2}60ydy\\
\Leftrightarrow I = \displaystyle\int\limits_0^230y^2\Bigg|_0^{x^2}dx\\
\Leftrightarrow I = \displaystyle\int\limits_0^230x^4dx\\
\Leftrightarrow I = 6x^5\Bigg|_0^2\\
\Leftrightarrow I = 192\\
2)\quad I = \displaystyle\iint\limits_D(x^2 + y^2)dxdy\quad \text{với}\quad D: \begin{cases}x = 0\\x - y = 0\\x + y - 2 =0\end{cases}\\
\text{Phương trình hoành độ giao điểm:}\\
\quad x = 2- x \Leftrightarrow x = 1\\
\text{Miền D được biểu diễn:}\\
D = \{(x,y):0\leqslant x \leqslant 1;\ x \leqslant y \leqslant 2- x\}\\
\text{Ta được:}\\
\quad I = \displaystyle\int\limits_0^1dx\displaystyle\int\limits_x^{2-x}(x^2 + y^2)dy\\
\Leftrightarrow I = \displaystyle\int\limits_0^1\left[\left(x^2y + \dfrac{y^3}{3}\right)\Bigg|_x^{2-x}\right]dx\\
\Leftrightarrow I = \displaystyle\int\limits_0^1\left(-\dfrac{8}{3}x^3+4x^2 -4x + \dfrac{8}{3}\right)dx\\
\Leftrightarrow I = \left(-\dfrac{2x^4}{3} + \dfrac{4x^3}{3} - 2x^2 + \dfrac{8}{3}x\right)\Bigg|_0^1\\
\Leftrightarrow I = \dfrac{4}{3}
\end{array}\)
