The value of \[\int\limits_0^3 {\int\limits_0^{\rm{x}} {\left( {6 – {\rm{x}} – {\rm{y}}} \right)} {\rm{dx\,dy}}} \] is A. 13.5 B. 27.0 C. 40.5 D. 54.0

The value of the triple integral $\int_0^3 \int_0^x (6-x-y) \, dydx$ is $40.5$.

To evaluate this integral, we can first integrate with respect to $y$. This gives us

$$\int_0^3 \int_0^x (6-x-y) \, dydx = \int_0^3 \left[ 6y – xy – \frac{y^2}{2} \right]_0^x dx = \int_0^3 6x – x^2 – \frac{x^2}{2} dx$$

We can then integrate with respect to $x$. This gives us

$$\int_0^3 6x – x^2 – \frac{x^2}{2} dx = \left[ 3x^2 – \frac{x^3}{3} – \frac{x^3}{6} \right]_0^3 = 3 \cdot 9 – \frac{3 \cdot 9^2}{3} – \frac{3 \cdot 9^3}{6} = 40.5$$

Therefore, the value of the triple integral is $40.5$.

Here is a step-by-step solution to the integral:

1. Integrate with respect to $y$:

$$\int_0^3 \int_0^x (6-x-y) \, dydx = \int_0^3 \left[ 6y – xy – \frac{y^2}{2} \right]_0^x dx = \int_0^3 6x – x^2 – \frac{x^2}{2} dx$$

1. Integrate with respect to $x$:

$$\int_0^3 6x – x^2 – \frac{x^2}{2} dx = \left[ 3x^2 – \frac{x^3}{3} – \frac{x^3}{6} \right]_0^3 = 3 \cdot 9 – \frac{3 \cdot 9^2}{3} – \frac{3 \cdot 9^3}{6} = 40.5$$

1. The answer is $40.5$.