Chapter 14: Q. 19 (page 1132)

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

Short Answer

Expert verified

$div G = 0$

Step by step solution

Step 1. Given information is:

$G (x, y, z) = {yz}^{2} i - x sinz j + x^{2} e^{xy} k$

Step 2. Divergence of vector field

$If G (x, y, z) = g_{1} (x, y, z) i + g_{2} (x, y, z) j + g_{3} (x, y, z) k, then div G = \frac{\partial g_{1}}{\partial x} + \frac{\partial g_{2}}{\partial y} + \frac{\partial g_{3}}{\partial z} Here, G (x, y, z) = {yz}^{2} i - x sinz j + x^{2} e^{xy} k Compare this with, G (x, y, z) = g_{1} (x, y, z) i + g_{2} (x, y, z) j + g_{3} (x, y, z) k Then, g_{1} (x, y, z) = {yz}^{2}, g_{2} (x, y, z) = - x sinz, g_{3} (x, y, z) k = x^{2} e^{xy} div G = \frac{\partial g_{1}}{\partial x} + \frac{\partial g_{2}}{\partial y} + \frac{\partial g_{3}}{\partial z} = \frac{\partial ({yz}^{2})}{\partial x} + \frac{\partial (- x sinz)}{\partial y} + \frac{\partial (x^{2} e^{xy})}{\partial z} = 0$

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

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)=yz2i−xsinzj+x2exyk

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