Chapter 14: Q. 20 (page 1132)

$Compute the divergence of the vector fields in Exercises 17 - 22 . G (x, y, z) = (x^{2} + y - z) i - 2 y \cos zj + e^{x^{2} + y^{2} + z^{2}} k$

Short Answer

Expert verified

$div G = = 2 x - 2 cosz + 2 {ze}^{x^{2} + y^{2} + z^{2}}$

Step by step solution

Step 1. Given information is:

$G (x, y, z) = (x^{2} + y - z) i - 2 y \cos zj + e^{x^{2} + y^{2} + z^{2}} k$

Step 2. Divergence of vector field

$If G (x, y, z) = g_{1} (x, y, z) i + g_{2} (x, y, z) j + g_{3} (x, y, z) k, then div G = \frac{\partial g_{1}}{\partial x} + \frac{\partial g_{2}}{\partial y} + \frac{\partial g_{3}}{\partial z} Here, G (x, y, z) = (x^{2} + y - z) i - 2 y \cos zj + e^{x^{2} + y^{2} + z^{2}} k Compare this with, G (x, y, z) = g_{1} (x, y, z) i + g_{2} (x, y, z) j + g_{3} (x, y, z) k Then, g_{1} (x, y, z) = x^{2} + y - z, g_{2} (x, y, z) = - 2 y \cos z, g_{3} (x, y, z) = e^{x^{2} + y^{2} + z^{2}} div G = \frac{\partial g_{1}}{\partial x} + \frac{\partial g_{2}}{\partial y} + \frac{\partial g_{3}}{\partial z} = \frac{\partial (x^{2} + y - z)}{\partial x} + \frac{\partial (- 2 y \cos z)}{\partial y} + \frac{\partial (e^{x^{2} + y^{2} + z^{2}})}{\partial z} = 2 x - 2 \cos z + 2 z e^{x^{2} + y^{2} + z^{2}}$

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$Compute the divergence of the vector fields in Exercises 17 - 22 . G (x, y, z) = (x^{2} + y - z) i - 2 y \cos zj + e^{x^{2} + y^{2} + z^{2}} k$

Short Answer

Step by step solution

Step 1. Given information is:

Step 2. Divergence of vector field

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ComputethedivergenceofthevectorfieldsinExercises17–22.G(x,y,z)=(x2+y−z)i−2ycoszj+ex2+y2+z2k

Short Answer

Step by step solution

Step 1. Given information is:

Step 2. Divergence of vector field

One App. One Place for Learning.

Most popular questions from this chapter

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Statistics

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Study anywhere. Anytime. Across all devices.

$Compute the divergence of the vector fields in Exercises 17 - 22 . G (x, y, z) = (x^{2} + y - z) i - 2 y \cos zj + e^{x^{2} + y^{2} + z^{2}} k$