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Find the value of divergence theorem for the field D = 2xy i + x2 j for the rectangular parallelepiped given by x = 0 and 1, y = 0 and 2, z = 0 and 3.
- a)10
- b)12
- c)14
- d)16
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
Find the value of divergence theorem for the field D = 2xy i + x2j for...
Answer: b
Explanation: While evaluating surface integral, there has to be two variables and one constant compulsorily. ∫∫D.ds = ∫∫Dx=0 dy dz + ∫∫Dx=1 dy dz + ∫∫Dy=0 dx dz + ∫∫Dy=2 dx dz + ∫∫Dz=0 dy dx + ∫∫Dz=3 dy dx. Put D in equation, the integral value we get is 12.
Explanation: While evaluating surface integral, there has to be two variables and one constant compulsorily. ∫∫D.ds = ∫∫Dx=0 dy dz + ∫∫Dx=1 dy dz + ∫∫Dy=0 dx dz + ∫∫Dy=2 dx dz + ∫∫Dz=0 dy dx + ∫∫Dz=3 dy dx. Put D in equation, the integral value we get is 12.
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Most Upvoted Answer
Find the value of divergence theorem for the field D = 2xy i + x2j for...
Understanding the Divergence Theorem
The Divergence Theorem relates a vector field's divergence over a volume to the field's flux across the surface bounding that volume.
Given Vector Field
For the vector field \( \mathbf{D} = 2xy \, \mathbf{i} + x^2 \, \mathbf{j} \):
- \( D_x = 2xy \)
- \( D_y = x^2 \)
- \( D_z = 0 \)
Calculate the Divergence
The divergence of a vector field \( \mathbf{D} \) is given by:
\[
\nabla \cdot \mathbf{D} = \frac{\partial D_x}{\partial x} + \frac{\partial D_y}{\partial y} + \frac{\partial D_z}{\partial z}
\]
Calculating each term:
- \( \frac{\partial D_x}{\partial x} = \frac{\partial (2xy)}{\partial x} = 2y \)
- \( \frac{\partial D_y}{\partial y} = \frac{\partial (x^2)}{\partial y} = 0 \)
- \( \frac{\partial D_z}{\partial z} = 0 \)
Thus,
\[
\nabla \cdot \mathbf{D} = 2y + 0 + 0 = 2y
\]
Volume Integral
Next, integrate the divergence over the volume of the rectangular parallelepiped defined by \( 0 \leq x \leq 1 \), \( 0 \leq y \leq 2 \), and \( 0 \leq z \leq 3 \):
\[
\int_0^1 \int_0^2 \int_0^3 (2y) \, dz \, dy \, dx
\]
Calculating the integrals step-by-step:
1. Integrate with respect to \( z \):
\[
\int_0^3 1 \, dz = 3
\]
2. Next, integrate with respect to \( y \):
\[
\int_0^2 (2y) \cdot 3 \, dy = 3 \cdot \left[ y^2 \right]_0^2 = 3 \cdot 4 = 12
\]
3. Finally, integrate with respect to \( x \):
\[
\int_0^1 12 \, dx = 12 \cdot 1 = 12
\]
Conclusion
Thus, the value of the divergence theorem for the given field and volume is:
Option (b) 12.
The Divergence Theorem relates a vector field's divergence over a volume to the field's flux across the surface bounding that volume.
Given Vector Field
For the vector field \( \mathbf{D} = 2xy \, \mathbf{i} + x^2 \, \mathbf{j} \):
- \( D_x = 2xy \)
- \( D_y = x^2 \)
- \( D_z = 0 \)
Calculate the Divergence
The divergence of a vector field \( \mathbf{D} \) is given by:
\[
\nabla \cdot \mathbf{D} = \frac{\partial D_x}{\partial x} + \frac{\partial D_y}{\partial y} + \frac{\partial D_z}{\partial z}
\]
Calculating each term:
- \( \frac{\partial D_x}{\partial x} = \frac{\partial (2xy)}{\partial x} = 2y \)
- \( \frac{\partial D_y}{\partial y} = \frac{\partial (x^2)}{\partial y} = 0 \)
- \( \frac{\partial D_z}{\partial z} = 0 \)
Thus,
\[
\nabla \cdot \mathbf{D} = 2y + 0 + 0 = 2y
\]
Volume Integral
Next, integrate the divergence over the volume of the rectangular parallelepiped defined by \( 0 \leq x \leq 1 \), \( 0 \leq y \leq 2 \), and \( 0 \leq z \leq 3 \):
\[
\int_0^1 \int_0^2 \int_0^3 (2y) \, dz \, dy \, dx
\]
Calculating the integrals step-by-step:
1. Integrate with respect to \( z \):
\[
\int_0^3 1 \, dz = 3
\]
2. Next, integrate with respect to \( y \):
\[
\int_0^2 (2y) \cdot 3 \, dy = 3 \cdot \left[ y^2 \right]_0^2 = 3 \cdot 4 = 12
\]
3. Finally, integrate with respect to \( x \):
\[
\int_0^1 12 \, dx = 12 \cdot 1 = 12
\]
Conclusion
Thus, the value of the divergence theorem for the given field and volume is:
Option (b) 12.
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