To evaluate the integral

$I = \int_0^4 \int_0^2 \int_0^{4-x-2y} y^3 \, dz \, dy \, dx,$

we can break it down step-by-step.

1. Inner Integral: First, we evaluate the integral with respect to $z$:

$\int_0^{4-x-2y} y^3 \, dz.$

Since $y^3$ is constant with respect to $z$, we get:

$= y^3 \cdot z \bigg|_0^{4-x-2y} = y^3 (4-x-2y).$

2. Substituting into the Middle Integral: Now we substitute this result into the next integral:

$I = \int_0^4 \int_0^2 y^3 (4-x-2y) \, dy \, dx.$

3. Distributing:

$I = \int_0^4 \left( \int_0^2 (4y^3 - 2xy^3 - 2y^4) \, dy \right) dx.$

4. Evaluating the Middle Integral:

$\int_0^2 (4y^3 - 2xy^3 - 2y^4) \, dy.$

Calculating each term separately:

• $\int_0^2 4y^3 \, dy = 4 \cdot \frac{y^4}{4} \bigg|_0^2 = 4 \cdot 4 = 16.$
• $\int_0^2 2xy^3 \, dy = 2x \cdot \frac{y^4}{4} \bigg|_0^2 = 2x \cdot 4 = 8x.$
• $\int_0^2 2y^4 \, dy = 2 \cdot \frac{y^5}{5} \bigg|_0^2 = 2 \cdot \frac{32}{5} = \frac{64}{5}.$

Combining these results, we have:

$\int_0^2 (4y^3 - 2xy^3 - 2y^4) \, dy = 16 - 8x - \frac{64}{5}.$

Now, express $16$ as $\frac{80}{5}$:

$= \frac{80}{5} - 8x - \frac{64}{5} = \frac{16}{5} - 8x.$

5. Final Integral: We now evaluate the outer integral:

$I = \int_0^4 \left( \frac{16}{5} - 8x \right) dx.$

Calculating this:

$\int_0^4 \frac{16}{5} \, dx = \frac{16}{5} \cdot x \bigg|_0^4 = \frac{64}{5},$ $\int_0^4 8x \, dx = 8 \cdot \frac{x^2}{2} \bigg|_0^4 = 8 \cdot 8 = 64.$

Thus,

$I = \frac{64}{5} - 64 = \frac{64}{5} - \frac{320}{5} = -\frac{256}{5}.$

So, the final value of the integral is

$I = -\frac{256}{5}.$

Summary:

The value of the integral is $-\frac{256}{5}$.

Do you want details on any specific part or have any questions? Here are some related questions you might consider:

1. What is the geometric interpretation of this integral?
2. How would the result change if the limits of integration were different?
3. Can you explain the process of evaluating a triple integral step-by-step?
4. What would happen if we switched the order of integration?
5. How does the integrand $y^3$ affect the overall integral compared to other functions?

Tip: When working with multiple integrals, always visualize the region of integration to better understand the limits.