Q. 25 Compute聽the聽curl聽of聽the聽vec... [FREE SOLUTION]

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Chapter 14: Q. 25 (page 1132)

$Compute the curl of the vector fields in Exercises 23 鈥� 28 . F (x, y, z) = {xe}^{yz} i + {ye}^{xz} j + {ze}^{xy} k$

Short Answer

Expert verified

$curl F = = ({xze}^{xy} - {xye}^{xz}) i + ({xye}^{yz} - {yze}^{xy}) j + ({zye}^{xz} - {zxe}^{yz}) k$

Step by step solution

Step 1. Given information is:

$F (x, y, z) = {xe}^{yz} i + {ye}^{xz} j + {ze}^{xy} k$

Step 2. Curl of vector field

$Here, F (x, y, z) = {xe}^{yz} i + {ye}^{xz} j + {ze}^{xy} k Compare this with, F (x, y, z) = f_{1} (x, y, z) i + f_{2} (x, y, z) j + f_{3} (x, y, z) k Then, f_{1} (x, y, z) = {xe}^{yz}, f_{2} (x, y, z) = {ye}^{xz}, f_{3} (x, y, z) = {ze}^{xy} Now, curl F = |\begin{matrix} i & j & k \\ \frac{鈭�}{鈭� x} & \frac{鈭�}{鈭� y} & \frac{鈭�}{鈭� z} \\ {xe}^{yz} & {ye}^{xz} & {ze}^{xy} \end{matrix}| = i (\frac{鈭� ({ze}^{xy})}{鈭� y} - \frac{鈭� ({ye}^{xz})}{鈭� z}) + j (\frac{鈭� ({xe}^{yz})}{鈭� z} - \frac{鈭� ({ze}^{xy})}{鈭� x}) + k (\frac{鈭� ({ye}^{xz})}{鈭� x} - \frac{鈭� ({xe}^{yz})}{鈭� y}) = ({xze}^{xy} - {xye}^{xz}) i + ({xye}^{yz} - {yze}^{xy}) j + ({zye}^{xz} - {zxe}^{yz}) k$

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$Compute the curl of the vector fields in Exercises 23 鈥� 28 . F (x, y, z) = {xe}^{yz} i + {ye}^{xz} j + {ze}^{xy} k$

Short Answer

Step by step solution

Step 1. Given information is:

Step 2. Curl of vector field

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ComputethecurlofthevectorfieldsinExercises23鈥�28.F(x,y,z)=xeyzi+yexzj+zexyk

Short Answer

Step by step solution

Step 1. Given information is:

Step 2. Curl of vector field

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Most popular questions from this chapter

Recommended explanations on Math Textbooks

Applied Mathematics

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Study anywhere. Anytime. Across all devices.

$Compute the curl of the vector fields in Exercises 23 鈥� 28 . F (x, y, z) = {xe}^{yz} i + {ye}^{xz} j + {ze}^{xy} k$