Math Problem Statement

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

Fequals=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.

Question content area bottom

Part 1

The net outward flux is

enter your response here.

Solution

To compute the net outward flux of the vector field F=x2,y2,z2\mathbf{F} = \langle x^2, -y^2, z^2 \rangle across the boundary of the region DD, where DD is the region in the first octant between the planes z=7xyz = 7 - x - y and z=2xyz = 2 - x - y, we use the Divergence Theorem.

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

First, we calculate the divergence of F\mathbf{F}: F=x(x2)+y(y2)+z(z2)\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: x(x2)=2x\frac{\partial}{\partial x}(x^2) = 2x y(y2)=2y\frac{\partial}{\partial y}(-y^2) = -2y z(z2)=2z\frac{\partial}{\partial z}(z^2) = 2z

So, the divergence F\nabla \cdot \mathbf{F} is: F=2x2y+2z\nabla \cdot \mathbf{F} = 2x - 2y + 2z

Next, we need to integrate this over the region DD. The region DD is bounded by the planes z=7xyz = 7 - x - y and z=2xyz = 2 - x - y, in the first octant where x,y,z0x, y, z \geq 0.

To set up the integral, we need to find the volume of DD. The bounds are:

  • For zz, 2xyz7xy2 - x - y \leq z \leq 7 - x - y
  • For xx and yy, 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 xyxy-plane where 0x50 \leq x \leq 5 and 0y5x0 \leq y \leq 5 - x, given by the intersection of the planes z=7xyz = 7 - x - y and z=2xyz = 2 - x - y with the coordinate axes.

So, the integral in Cartesian coordinates is: D(2x2y+2z)dV=0505x2xy7xy(2x2y+2z)dzdydx\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 zz: [ \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} ] =2xy7xy(2x2y+2z)dz=2x2y(7xy2xy)+[z2]2xy7xy= \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 yy and then xx.

Finally, the flux calculation yields: =D(2x2y+2z)dV=500= \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 F\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 DD is well-defined and closed to correctly apply the theorem.

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Math Problem Analysis

Mathematical Concepts

Vector Calculus
Divergence Theorem
Integral Calculus

Formulas

Divergence Theorem

Theorems

Divergence Theorem

Suitable Grade Level

Advanced Undergraduate