Answer
$\dfrac{3}{2}$
Work Step by Step
We have the triple integral:
$\int^1_0 \int^{3-3x}_0 \int^{3-3x-y}_0 dz \space dy \space dx=\int^1_0 \int^{3-3x}_0 (3-3x-y) dy \space dx $
or, $=\int^1_0[(3-3x)^2-\dfrac{1}{2}(3-3x)^2]dx $
or, $=\dfrac{9}{2} \times \int^1_0 (1-x)^2 \space dx $
Now, $\int^1_0 \int^{3-3x}_0 \int^{3-3x-y}_0 dz \space dy \space dx=\dfrac{-3}{2}[(1-x)^3]^1_0=\dfrac{3}{2}$
