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Curl of a vector field at M(1;1;1)

This problem asks to find the curl of a vector field at a given point.

Problem

Find the curl of the vector field:

a → = x 2 y i → − 2 y 3 z j → + 3 z 2 x k → , M ( 1 ; 1 ; 1 )

Solution

Step 1

Write the formula for the curl of the vector field ā = P ī + Q j̄ + R k̄:

rot a → = ( ∂ R ∂ y − ∂ Q ∂ z ) i → − ( ∂ R ∂ x − ∂ P ∂ z ) j → + ( ∂ Q ∂ x − ∂ P ∂ y ) k →

Step 2

Write out the components of the vector field:

P = x 2 y , Q = − 2 y 3 z , R = 3 z 2 x

Step 3

Compute the partial derivatives appearing in the curl formula:

Why are some derivatives zero?

A derivative is zero when the corresponding variable does not appear in the component of the field. For example, ∂R/∂y = 0 because R = 3z²x does not depend on y. This is typical for polynomial fields and significantly simplifies the computation.

∂ R ∂ y = 0 , ∂ Q ∂ z = − 2 y 3 ∂ R ∂ x = 3 z 2 , ∂ P ∂ z = 0 ∂ Q ∂ x = 0 , ∂ P ∂ y = x 2

Step 4

Substitute the found derivatives into the curl formula:

rot a → = ( 0 − ( − 2 y 3 ) ) i → − ( 3 z 2 − 0 ) j → + ( 0 − x 2 ) k → = 2 y 3 i → − 3 z 2 j → − x 2 k →

Step 5

Substitute the coordinates of the point M(1; 1; 1) into the expression for the curl:

rot a → ( M ) = 2 i → − 3 j → − k → = ( 2 ; − 3 ; − 1 )

Answer

rot ā(M) = (2; −3; −1).

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