Use a CAS integration utility to evaluate the triple integral of the given function over the specified solid region.
-$$F ( x , y , z ) = 8 x - 3 y - 6 z \text { over the rectangular solid } 0 \leq x \leq 2,0 \leq y \leq 10,0 \leq z \leq 9$$
A) - 4080
B) - 18,360
C) - 3060
D) - 6120

Question

Quizplus · Accepted Answer

To solve the triple integral of $F(x, y, z) = 8x - 3y - 6z$ over the rectangular solid defined by $0 \leq x \leq 2$, $0 \leq y \leq 10$, and $0 \leq z \leq 9$, we integrate the function with respect to $x$, $y$, and $z$ in their respective limits. The integral can be computed as follows:$$\int_{0}^{2} \int_{0}^{10} \int_{0}^{9} (8x - 3y - 6z) \, dz \, dy \, dx$$Solving this integral step by step:1. Integrate with respect to $z$: $\int_{0}^{9} (8x - 3y - 6z) \, dz = 8xz - 3yz - 3z^2 |_{0}^{9} = 72x - 27y - 243$2. Integrate the result with respect to $y$: $\int_{0}^{10} (72x - 27y - 243) \, dy = 720x - 1350 - 2430$3. Finally, integrate with respect to $x$: $\int_{0}^{2} (720x - 1350 - 2430) \, dx = 720x^2/2 - 1350x - 2430x |_{0}^{2} = 1440 - 2700 - 4860 = -6120$Therefore, the correct answer is -6120.