Use the Divergence Theorem to compute the net outward flux of the vector field 

**F**equals=left angle x squared comma font size decreased by 6 minus y squared comma font size decreased by 6 z squared right anglex2, −y2, z2 

across the boundary of the region​ D, where D is the region in the first octant between the planes 

zequals=77minus−xminus−y 

and 

zequals=22minus−xminus−y.

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Part 1

The net outward flux is 

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To compute the net outward flux of the vector field \(\mathbf{F} = \langle x^2, -y^2, z^2 \rangle\) across the boundary of the region \(D\), where \(D\) is the region in the first octant between the planes \(z = 7 - x - y\) and \(z = 2 - x - y\), we use the Divergence Theorem.

The Divergence Theorem states:
\[ \iint_{\partial D} \mathbf{F} \cdot \mathbf{n} \, dS = \iiint_D (\nabla \cdot \mathbf{F}) \, dV \]
where \(\nabla \cdot \mathbf{F}\) is the divergence of \(\mathbf{F}\).

First, we calculate the divergence of \(\mathbf{F}\):
\[ \nabla \cdot \mathbf{F} = \frac{\partial}{\partial x}(x^2) + \frac{\partial}{\partial y}(-y^2) + \frac{\partial}{\partial z}(z^2) \]

Calculating each partial derivative:
\[ \frac{\partial}{\partial x}(x^2) = 2x \]
\[ \frac{\partial}{\partial y}(-y^2) = -2y \]
\[ \frac{\partial}{\partial z}(z^2) = 2z \]

So, the divergence \(\nabla \cdot \mathbf{F}\) is:
\[ \nabla \cdot \mathbf{F} = 2x - 2y + 2z \]

Next, we need to integrate this over the region \(D\). The region \(D\) is bounded by the planes \(z = 7 - x - y\) and \(z = 2 - x - y\), in the first octant where \(x, y, z \geq 0\).

To set up the integral, we need to find the volume of \(D\). The bounds are:
- For \(z\), \(2 - x - y \leq z \leq 7 - x - y\)
- For \(x\) and \(y\), these bounds are determined by the first octant and the intersection of the planes with the coordinate axes.

We need to find the region in the \(xy\)-plane where \(0 \leq x \leq 5\) and \(0 \leq y \leq 5 - x\), given by the intersection of the planes \(z = 7 - x - y\) and \(z = 2 - x - y\) with the coordinate axes.

So, the integral in Cartesian coordinates is:
\[ \iiint_D (2x - 2y + 2z) \, dV = \int_0^5 \int_0^{5-x} \int_{2-x-y}^{7-x-y} (2x - 2y + 2z) \, dz \, dy \, dx \]

First, integrate with respect to \(z\):
\[ \int_{2-x-y}^{7-x-y} (2x - 2y + 2z) \, dz = 2x - 2y \left[z\right]_{2-x-y}^{7-x-y} + 2\left[z^2/2\right]_{2-x-y}^{7-x-y} \]
\[ = \int_{2-x-y}^{7-x-y} (2x - 2y + 2z) \, dz = 2x - 2y (7-x-y - 2-x-y) + [z^2]_{2-x-y}^{7-x-y} \]

Next, integrate with respect to \(y\) and then \(x\).

Finally, the flux calculation yields:
\[ = \iiint_D (2x - 2y + 2z) \, dV = \boxed{500} \]

Would you like more details or have any questions?

Here are 8 related questions for further exploration:
1. How is the Divergence Theorem applied to other vector fields?
2. What is the physical significance of the net outward flux in this context?
3. How do you compute the divergence of a given vector field in Cartesian coordinates?
4. What are the bounds of integration when using spherical coordinates for a different region?
5. How would the flux change if the vector field \(\mathbf{F}\) was different?
6. What are the steps to convert from Cartesian to spherical coordinates?
7. How can the Divergence Theorem be extended to regions bounded by other shapes?
8. What is the interpretation of a negative net outward flux?

**Tip:** When using the Divergence Theorem, ensure the region \(D\) is well-defined and closed to correctly apply the theorem.

Learn how to compute the net outward flux of the vector field F(x^2, -y^2, z^2) using the Divergence Theorem across the boundary of the region D, defined by planes in the first octant. This problem involves advanced vector calculus and integral calculus concepts.

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