Find the average value of $F(x,y,z) = 10x + 2y + 6z$ over the cube in the first octant bounded by the coordinate planes and the planes $x = 6$, $y = 6$, and $z = 6$.

Neetesh Kumar | December 11, 2024

Calculus definite integral Homework Help

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Step-by-step solution:

The average value of a function $F(x,y,z)$ over a region $R$ is given by the formula:

$\text{Average value} = \frac{1}{\text{Volume of } R} \int \int \int_R F(x,y,z) \, dV$

Here, the region $R$ is the cube in the first octant with bounds $0 \leq x \leq 6$, $0 \leq y \leq 6$, and $0 \leq z \leq 6$.

Thus, the volume of $R$ is:

$\text{Volume of } R = 6 \times 6 \times 6 = 216$

Now, we need to evaluate the triple integral of $F(x,y,z) = 10x + 2y + 6z$ over the cube:

$\int_0^6 \int_0^6 \int_0^6 (10x + 2y + 6z) \, dz \, dy \, dx$

Step 1: Evaluate the inner integral with respect to $z$

$\int_0^6 (10x + 2y + 6z) \, dz = \left[ (10x + 2y)z + 3z^2 \right]_0^6$

Evaluating at the bounds:

$= (10x + 2y)(6) + 3(6)^2 - 0$

$= 6(10x + 2y) + 108$

$= 60x + 12y + 108$

Step 2: Evaluate the middle integral with respect to $y$

Now integrate the result with respect to $y$:

$\int_0^6 (60x + 12y + 108) \, dy = \left[ 60xy + 6y^2 + 108y \right]_0^6$

Evaluating at the bounds:

$= 60x(6) + 6(6)^2 + 108(6) - 0$

$= 360x + 216 + 648$

$= 360x + 864$

Step 3: Evaluate the outer integral with respect to $x$

Finally, integrate the result with respect to $x$:

$\int_0^6 (360x + 864) \, dx = \left[ 180x^2 + 864x \right]_0^6$

Evaluating at the bounds:

$= 180(6)^2 + 864(6) - 0$

$= 180(36) + 5184$

$= 6480 + 5184 = 11664$

Step 4: Calculate the average value

Now, divide the result by the volume of the region:

$\text{Average value} = \frac{11664}{216} = 54$

Final Answer:

The average value of $F(x,y,z) = 10x + 2y + 6z$ over the cube is: $\boxed{54}$

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