Đáp án:
$I = \displaystyle\iiint\limits_V(x+y+z)dxdydz=\dfrac18$
Giải thích các bước giải:
$\ \ I = \displaystyle\iiint\limits_V(x+y+z)dxdydz\qquad V:\begin{cases}x = 0\\y = 0\\z = 0\\x+y+z= 1\end{cases}$
Hình chiếu của $V$ lên mặt $Oxy$ là miền $D:$
$D =\{(x,y): 0 \leqslant x \leqslant 1;\ 0 \leqslant y\leqslant 1 - x\}$
Giới hạn dưới: $z = 0$
Giới hạn trên: $z = 1 - x - y$
Ta được:
$\quad I = \displaystyle\int\limits_0^1dx\displaystyle\int\limits_0^{1-x}dy\displaystyle\int\limits_0^{1-x-y}(x+y+z)dz$
$\Leftrightarrow I = \displaystyle\int\limits_0^1dx\displaystyle\int\limits_0^{1-x}\left[\left(xz + yz +\dfrac{z^2}{2}\right)\Bigg|_0^{1-x-y}\right]dy$
$\Leftrightarrow I = \displaystyle\int\limits_0^1dx\displaystyle\int\limits_0^{1-x}\left(-\dfrac{x^2}{2} - \dfrac{y^2}{2} -xy + \dfrac12\right)dy$
$\Leftrightarrow I = \displaystyle\int\limits_0^1\left[\left(-\dfrac{x^2y}{2} - \dfrac{y^3}{6} - \dfrac{xy^2}{2} + \dfrac12y\right)\Bigg|_0^{1-x}\right]dx$
$\Leftrightarrow I = \displaystyle\int\limits_0^1\left(\dfrac{x^3}{6} -\dfrac x2 +\dfrac13\right)dx$
$\Leftrightarrow I = \left(\dfrac{x^4}{24} - \dfrac{x^2}{4} +\dfrac13x\right)\Bigg|_0^1$
$\Leftrightarrow I = \dfrac18$
