To evaluate the triple integral

$\int_{0}^{2} \int_{-4 - x^2}^{4 - x^2} \int_{0}^{2x + y} dz \, dy \, dx,$

let's proceed step by step:

Step-by-step evaluation:

1. Inner Integral: $\int_{0}^{2x + y} dz.$ This is a straightforward integral with respect to $z$, giving: $\left[ z \right]_{0}^{2x + y} = 2x + y.$

2. Middle Integral: Substitute the result of the inner integral: $\int_{-4 - x^2}^{4 - x^2} (2x + y) dy.$ We evaluate this integral with respect to $y$: $\int_{-4 - x^2}^{4 - x^2} 2x \, dy + \int_{-4 - x^2}^{4 - x^2} y \, dy.$

   • The first term becomes: $2x \int_{-4 - x^2}^{4 - x^2} dy = 2x \left[ y \right]_{-4 - x^2}^{4 - x^2} = 2x \left( (4 - x^2) - (-4 - x^2) \right) = 2x \cdot 8 = 16x.$
   • The second term becomes: $\int_{-4 - x^2}^{4 - x^2} y \, dy = \left[ \frac{y^2}{2} \right]_{-4 - x^2}^{4 - x^2} = \frac{1}{2} \left( (4 - x^2)^2 - (-4 - x^2)^2 \right).$ Since $(4 - x^2)^2 = (-4 - x^2)^2$, this term evaluates to zero.

   Therefore, the middle integral simplifies to: $16x.$

3. Outer Integral: Now, we integrate with respect to $x$: $\int_{0}^{2} 16x \, dx = 16 \int_{0}^{2} x \, dx = 16 \left[ \frac{x^2}{2} \right]_{0}^{2} = 16 \cdot \frac{4}{2} = 16 \cdot 2 = 32.$

Final Answer:

$\int_{0}^{2} \int_{-4 - x^2}^{4 - x^2} \int_{0}^{2x + y} dz \, dy \, dx = 32.$

Would you like any further details or have any specific questions?

Related Questions:

1. What is a triple integral, and where are they commonly used in physics or engineering?
2. Can triple integrals be evaluated in different coordinate systems like cylindrical or spherical coordinates?
3. How would the limits change if we modified the function to $\int_{0}^{3}$?
4. What are some practical applications of integrating over non-rectangular regions?
5. How would the solution change if the integrand was $2x - y$ instead of $2x + y$?

Tip: When evaluating multiple integrals, visualize the bounds and their implications on the geometry of the region to ensure proper limits and order of integration.