Question

4. The curl of vector field
f(x,y,z)=x^2+2zj-yk is​

Answer 1

Given: Vector field f (x,y,z) = x^2 + 2zj - yk

To find:  The curl of vector field ?

Solution:

• Now we have given the Vector field f as : x^2 + 2zj - yk

• The curl is a vector operator that describes the infinitesimal rotation of a vector field in three-dimensional Euclidean space. At every point in the field, the curl of that point is represented by a vector.

• Formula for curl is:

                 Curl (f) = det        |     i        j        k      |

                                              | d/dx  d/dy  d/dz   |

                                              |   x^2      2z      -y   |

• Expanding the determinant, we get:

                 Curl (f) = i { d(-y)/dy - d(2z)/dz } - j { d(-y)/dx - d(x^2)/dz } + k { d(2z)/dx - d(x^2)/dy  }

                 Curl (f) = i ( -1-2 ) - j ( 0 ) + k ( 0 )

                 Curl (f) = -3i

Answer:

           So, the curl of given vector field is -3i.

Answer 2

please mark as Brainlliest if helpful bye bye