Math Problem Statement

To verify the Divergence Theorem for the vector field 𝐹=π‘₯2 𝑖+𝑦2 𝑗+𝑧2β€‰π‘˜ over the closed surface 𝑆 in the first octant bounded by the coordinate planes and the plane π‘₯+𝑦+𝑧=π‘Ž

Solution

To verify the Divergence Theorem for the vector field F=x2i+y2j+z2k\mathbf{F} = x^2 \mathbf{i} + y^2 \mathbf{j} + z^2 \mathbf{k} over the closed surface SS in the first octant bounded by the coordinate planes and the plane x+y+z=ax + y + z = a, we will follow these steps:

  1. Compute the Divergence of F\mathbf{F}: βˆ‡β‹…F=βˆ‚βˆ‚x(x2)+βˆ‚βˆ‚y(y2)+βˆ‚βˆ‚z(z2)=2x+2y+2z\nabla \cdot \mathbf{F} = \frac{\partial}{\partial x}(x^2) + \frac{\partial}{\partial y}(y^2) + \frac{\partial}{\partial z}(z^2) = 2x + 2y + 2z

  2. Set Up the Volume Integral: We need to evaluate the volume integral of the divergence over the region VV defined by the plane x+y+z≀ax + y + z \leq a in the first octant.

    The volume integral is: ∭V(2x+2y+2z) dV\iiint_V (2x + 2y + 2z) \, dV

  3. Define the Limits of Integration: In the first octant, the bounds for zz will depend on xx and yy: z ranges from 0 to aβˆ’xβˆ’yz \text{ ranges from } 0 \text{ to } a - x - y For the yy integral: y ranges from 0 to aβˆ’xy \text{ ranges from } 0 \text{ to } a - x For the xx integral: x ranges from 0 to ax \text{ ranges from } 0 \text{ to } a

  4. Compute the Volume Integral: The integral becomes: ∫0a∫0aβˆ’x∫0aβˆ’xβˆ’y(2x+2y+2z) dz dy dx\int_0^a \int_0^{a-x} \int_0^{a-x-y} (2x + 2y + 2z) \, dz \, dy \, dx

    Evaluate the innermost integral: ∫0aβˆ’xβˆ’y(2x+2y+2z) dz=(2x+2y)z+z2∣0aβˆ’xβˆ’y=(2x+2y)(aβˆ’xβˆ’y)+(aβˆ’xβˆ’y)2\int_0^{a-x-y} (2x + 2y + 2z) \, dz = (2x + 2y)z + z^2 \Big|_0^{a-x-y} = (2x + 2y)(a-x-y) + (a-x-y)^2

    Expanding this gives: (2x+2y)(aβˆ’xβˆ’y)+(aβˆ’xβˆ’y)2(2x + 2y)(a - x - y) + (a - x - y)^2

    Now, this needs to be integrated with respect to yy and xx.

  5. Calculate the Surface Integral: By the Divergence Theorem, the surface integral over the closed surface SS is: ∬SFβ‹…dS\iint_S \mathbf{F} \cdot d\mathbf{S}

    This includes contributions from the three coordinate planes (the xyxy-plane, xzxz-plane, and yzyz-plane) and the plane x+y+z=ax + y + z = a.

  6. Evaluate Each Surface Integral:

    • On the xyxy-plane (z=0z = 0): F=x2i+y2j\mathbf{F} = x^2 \mathbf{i} + y^2 \mathbf{j}
    • On the xzxz-plane (y=0y = 0): F=x2i+z2k\mathbf{F} = x^2 \mathbf{i} + z^2 \mathbf{k}
    • On the yzyz-plane (x=0x = 0): F=y2j+z2k\mathbf{F} = y^2 \mathbf{j} + z^2 \mathbf{k}
    • On the plane x+y+z=ax + y + z = a: Evaluate Fβ‹…dS\mathbf{F} \cdot d\mathbf{S} appropriately.
  7. Combine Results: After computing the volume and surface integrals, you should verify that they are equal as per the Divergence Theorem: ∬SFβ‹…dS=∭Vβˆ‡β‹…F dV\iint_S \mathbf{F} \cdot d\mathbf{S} = \iiint_V \nabla \cdot \mathbf{F} \, dV

If you'd like detailed calculations for any specific integral or step, let me know!

Questions for Further Exploration:

  1. How do you derive the limits of integration for other shapes?
  2. What is the significance of the Divergence Theorem in physics?
  3. Can you provide an example where the Divergence Theorem does not hold?
  4. What other theorems are similar to the Divergence Theorem?
  5. How does this theorem apply in three-dimensional fluid dynamics?

Tip:

Always sketch the region of integration to visualize bounds and surfaces for triple integrals, especially in complex geometries.

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Math Problem Analysis

Mathematical Concepts

Vector Calculus
Divergence
Triple Integrals

Formulas

βˆ‡ Β· F = 2x + 2y + 2z
∫∫_S F Β· dS = ∫∫∫_V (βˆ‡ Β· F) dV

Theorems

Divergence Theorem

Suitable Grade Level

Grades 11-12