I=∫
0
4
​
∫
0
2
4−x
​

​
∫
0
4−x−2y
​
y
3
dzdydx

To evaluate the triple integral 

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

we will proceed step by step.

### Step 1: Evaluate the innermost integral

The innermost integral is:

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

Since \(y^3\) is independent of \(z\), this simplifies to:

\[
y^3 \int_0^{4-x-2y} dz = y^3 [z]_0^{4-x-2y} = y^3 (4-x-2y).
\]

### Step 2: Substitute and evaluate the middle integral

Now substitute this result into the middle integral:

\[
\int_0^2 y^3 (4 - x - 2y) \, dy.
\]

Distributing \(y^3\):

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

Now, evaluate each term:

1. \(\int_0^2 4y^3 \, dy = 4 \left[ \frac{y^4}{4} \right]_0^2 = [y^4]_0^2 = 16\).
2. \(\int_0^2 xy^3 \, dy = x \left[ \frac{y^4}{4} \right]_0^2 = x \cdot 4 = 4x\).
3. \(\int_0^2 2y^4 \, dy = 2 \left[ \frac{y^5}{5} \right]_0^2 = 2 \cdot \frac{32}{5} = \frac{64}{5}\).

Putting this together, we get:

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

Combine the constant terms:

\[
16 - \frac{64}{5} = \frac{80}{5} - \frac{64}{5} = \frac{16}{5}.
\]

Thus, 

\[
\int_0^2 y^3 (4 - x - 2y) \, dy = \frac{16}{5} - 4x.
\]

### Step 3: Evaluate the outer integral

Now, we need to evaluate:

\[
\int_0^4 \left( \frac{16}{5} - 4x \right) \, dx.
\]

This splits into two integrals:

1. \(\int_0^4 \frac{16}{5} \, dx = \frac{16}{5} [x]_0^4 = \frac{16}{5} \cdot 4 = \frac{64}{5}\).
2. \(\int_0^4 4x \, dx = 4 \left[ \frac{x^2}{2} \right]_0^4 = 4 \cdot 8 = 32\).

Putting these together gives:

\[
I = \frac{64}{5} - 32 = \frac{64}{5} - \frac{160}{5} = -\frac{96}{5}.
\]

### Final Result

Thus, the value of the triple integral is 

\[
I = -\frac{96}{5}.
\]

If you have any questions or need further clarification, feel free to ask! Here are some related questions you might consider:

1. What is the geometric interpretation of this triple integral?
2. How would changing the limits of integration affect the result?
3. Can you explain the steps of evaluating a double integral?
4. What are the applications of triple integrals in physics?
5. How would this integral change if the function to integrate was different? 

**Tip:** Always visualize the region of integration to better understand the limits and the function being integrated!

Discover how to evaluate the triple integral ∫0^4∫0^2∫0^(4-x-2y)y^3dzdydx with a comprehensive step-by-step solution. This integral involves key concepts in calculus and is suitable for students in Grades 11-12.

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