To find the average value of the function $f(x, y, z) = y e^{-xy}$ over the rectangular prism defined by $0 \leq x \leq 2$, $0 \leq y \leq 2$, and $0 \leq z \leq 5$, we can use the formula for the average value of a function over a region $D$:

$\text{Average value} = \frac{1}{\text{Volume of } D} \iiint_D f(x, y, z) \, dV$

Step 1: Calculate the Volume of $D$

The volume of the rectangular prism $D$ is given by the product of the lengths of its sides: $\text{Volume of } D = (2 - 0)(2 - 0)(5 - 0) = 2 \times 2 \times 5 = 20$

Step 2: Set Up the Integral

The average value is then: $\text{Average value} = \frac{1}{20} \iiint_D y e^{-xy} \, dV$ Since $f(x, y, z) = y e^{-xy}$ does not depend on $z$, we can separate the integral with respect to $z$ as follows: $\iiint_D y e^{-xy} \, dV = \int_0^5 \int_0^2 \int_0^2 y e^{-xy} \, dx \, dy \, dz$

The integral with respect to $z$ will contribute a factor of $5$ (as it is simply an integral over a constant with bounds $0$ to $5$): $\iiint_D y e^{-xy} \, dV = 5 \int_0^2 \int_0^2 y e^{-xy} \, dx \, dy$

Step 3: Evaluate the Inner Integrals

Now, we focus on the double integral: $5 \int_0^2 \int_0^2 y e^{-xy} \, dx \, dy$

1. Inner integral (with respect to $x$): $\int_0^2 y e^{-xy} \, dx = y \int_0^2 e^{-xy} \, dx$ The integral of $e^{-xy}$ with respect to $x$ is: $\int e^{-xy} \, dx = -\frac{1}{y} e^{-xy}$ Evaluating from $x = 0$ to $x = 2$ gives: $y \left( -\frac{1}{y} e^{-2y} + \frac{1}{y} \right) = y \cdot \frac{1 - e^{-2y}}{y} = 1 - e^{-2y}$

2. Outer integral (with respect to $y$): $5 \int_0^2 (1 - e^{-2y}) \, dy = 5 \left( \int_0^2 1 \, dy - \int_0^2 e^{-2y} \, dy \right)$

   • The first part is straightforward: $\int_0^2 1 \, dy = y \Big|_0^2 = 2$
   • For the second part: $\int_0^2 e^{-2y} \, dy = \left( -\frac{1}{2} e^{-2y} \right) \Big|_0^2 = -\frac{1}{2} (e^{-4} - 1) = \frac{1}{2} (1 - e^{-4})$

   So the double integral becomes: $5 \left( 2 - \frac{1}{2} (1 - e^{-4}) \right) = 5 \left( \frac{4 + e^{-4}}{2} \right) = \frac{5}{2} (4 + e^{-4})$

Step 4: Compute the Average Value

Now we can plug this result back into our formula for the average value: $\text{Average value} = \frac{1}{20} \cdot \frac{5}{2} (4 + e^{-4}) = \frac{1}{8} (4 + e^{-4}) = \frac{1}{2} + \frac{e^{-4}}{8}$

Thus, the average value of $f(x, y, z) = y e^{-xy}$ over the specified region is: $\boxed{\frac{1}{2} + \frac{e^{-4}}{8}}$

Would you like further details on any of these steps or have additional questions?

1. What is the significance of finding an average value for a function in a region?
2. How would the result change if the limits of integration were different?
3. What if the function $f(x, y, z)$ depended on $z$ as well?
4. Can we generalize this approach to find the average value of any function in any region?
5. What if $f(x, y, z)$ involved more complex exponential terms?

Tip: For multiple integrals, carefully evaluate each integral step-by-step, starting with the innermost integral to ensure accuracy.