The problem asks to evaluate the volume integral of the vector field F over a specified region V. Let's break down the problem:

Given:

• F = $2xzi - xj - y^2k$
• The region V is bounded by:
  • $x = 0$
  • $y = 0$
  • $y = 6$
  • $z = x^2$
  • $z = 4$

You are asked to evaluate the volume integral $\iiint_V F \, dV$.

Steps to solve:

1. Identify the bounds of integration:

   • From the boundaries:
     • $x$ varies from 0 to 2 (since $z = 4$ and $z = x^2$ implies $x = 2$).
     • $y$ varies from 0 to 6.
     • $z$ varies from $x^2$ to 4.

2. Decompose the vector field:

   • The integral to evaluate is $\iiint_V F \cdot dV$, where the vector field components are integrated accordingly over the volume.
   • You will integrate the components of F in each direction.

Now, for the integral itself:

$\iiint_V F \, dV = \iiint_V (2xz \, dx + (-x) \, dy + (-y^2) \, dz)$

Let's evaluate this integral step-by-step:

First Component (2xz):

$\int_0^2 \int_0^6 \int_{x^2}^4 2xz \, dz \, dy \, dx$ The $z$-integral yields: $\int_{x^2}^4 2xz \, dz = x \left[ z^2 \right]_{x^2}^4 = x(16 - x^4)$ Thus, the integral becomes: $\int_0^2 \int_0^6 x(16 - x^4) \, dy \, dx$ The $y$-integral simply introduces a factor of 6: $6 \int_0^2 x(16 - x^4) \, dx$ Evaluating this: $6 \left[ 8x^2 - \frac{x^6}{6} \right]_0^2 = 6 \left( 32 - \frac{64}{6} \right) = 6 \times \frac{128}{6} = 128$

Second Component (-x):

$\int_0^2 \int_0^6 \int_{x^2}^4 (-x) \, dz \, dy \, dx$ The $z$-integral gives: $-x \left[ z \right]_{x^2}^4 = -x(4 - x^2)$ Thus, the integral becomes: $\int_0^2 \int_0^6 -x(4 - x^2) \, dy \, dx$ Again, the $y$-integral introduces a factor of 6: $6 \int_0^2 -x(4 - x^2) \, dx$ Evaluating this: $6 \left[ -2x^2 + \frac{x^4}{4} \right]_0^2 = 6 \left( -8 + 4 \right) = 6 \times -4 = -24$

Third Component (-y²):

$\int_0^2 \int_0^6 \int_{x^2}^4 (-y^2) \, dz \, dy \, dx$ The $z$-integral introduces a factor of $(4 - x^2)$: $(4 - x^2) \int_0^6 -y^2 \, dy = (4 - x^2) \left[ -\frac{y^3}{3} \right]_0^6 = (4 - x^2) \times -72$ Thus, the integral becomes: $-72 \int_0^2 (4 - x^2) \, dx$ Evaluating this: $-72 \left[ 4x - \frac{x^3}{3} \right]_0^2 = -72 \left( 8 - \frac{8}{3} \right) = -72 \times \frac{16}{3} = -384$

Final Answer:

Summing the three components: $128 + (-24) + (-384) = -280$

Thus, the value of the volume integral is $\boxed{-280}$.

Let me know if you'd like further clarification or details on any part of this solution! Here are some related questions you could explore:

1. How do the limits of integration change if the surfaces are altered?
2. What is the physical significance of this volume integral in vector calculus?
3. Can this result be verified using the Divergence Theorem?
4. How would the problem change if a different vector field was used?
5. How is the method of cylindrical or spherical coordinates applicable to integrals like this?

Tip: When setting up triple integrals, always sketch the region or carefully analyze the boundaries to avoid errors in setting up the limits of integration.